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Complex Analysis

A complex number is z=a+biz = a + bi where a,b∈Ra, b \in \mathbb{R} and i2=−1i^2 = -1. We call a=Re(z)a = \mathrm{Re}(z) the real part and b=Im(z)b = \mathrm{Im}(z) the imaginary part.

Arithmetic: (a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i and (a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i.

Proposition 1.1 (Properties of Complex Arithmetic). For all z,w∈Cz, w \in \mathbb{C}:

  1. z+w=w+zz + w = w + z and zw=wzzw = wz (commutativity)
  2. (z+w)+u=z+(w+u)(z + w) + u = z + (w + u) and (zw)u=z(wu)(zw)u = z(wu) (associativity)
  3. z(w+u)=zw+zuz(w + u) = zw + zu (distributivity)
  4. There exist additive identity 00 and multiplicative identity 11.
  5. Every z≠0z \neq 0 has a multiplicative inverse 1z=zˉ∣z∣2\frac{1}{z} = \frac{\bar{z}}{|z|^2}.

Remark. The complex field C\mathbb{C} cannot be ordered: there is no total ordering on C\mathbb{C} Compatible with the field operations. In particular, i2=−1i^2 = -1 precludes any such ordering.

Definition. The complex conjugate of z=a+biz = a + bi is zˉ=a−bi\bar{z} = a - bi.

Proposition 1.2. For all z,w∈Cz, w \in \mathbb{C}:

  1. z+w‾=zˉ+wˉ\overline{z + w} = \bar{z} + \bar{w} and zw‾=zˉwˉ\overline{zw} = \bar{z}\bar{w}
  2. zzˉ=∣z∣2z\bar{z} = |z|^2
  3. z+zˉ=2 Re(z)z + \bar{z} = 2\,\mathrm{Re}(z) and z−zˉ=2i Im(z)z - \bar{z} = 2i\,\mathrm{Im}(z)
  4. zˉˉ=z\bar{\bar{z}} = z

Definition. The modulus (or absolute value) of z=a+biz = a + bi is ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}.

Proposition 1.3 (Modulus Properties). For all z,w∈Cz, w \in \mathbb{C}:

  1. ∣z∣≥0|z| \geq 0 with equality iff z=0z = 0
  2. ∣zw∣=∣z∣∣w∣|zw| = |z||w|
  3. ∣z+w∣≤∣z∣+∣w∣|z + w| \leq |z| + |w| (triangle inequality)
  4. ∣∣z∣−∣w∣∣≤∣z−w∣\bigl||z| - |w|\bigr| \leq |z - w| (reverse triangle inequality)

Proof of (3). ∣z+w∣2=(z+w)(zˉ+wˉ)=∣z∣2+zwˉ+zˉw+∣w∣2=∣z∣2+2 Re(zwˉ)+∣w∣2≤∣z∣2+2∣z∣∣w∣+∣w∣2=(∣z∣+∣w∣)2|z + w|^2 = (z + w)(\bar{z} + \bar{w}) = |z|^2 + z\bar{w} + \bar{z}w + |w|^2 = |z|^2 + 2\,\mathrm{Re}(z\bar{w}) + |w|^2 \leq |z|^2 + 2|z||w| + |w|^2 = (|z| + |w|)^2. The inequality follows from Re(zwˉ)≤∣zwˉ∣=∣z∣∣w∣\mathrm{Re}(z\bar{w}) \leq |z\bar{w}| = |z||w|. ■\blacksquare

Every non-zero complex number can be written in polar form:

z=r(cos⁡θ+isin⁡θ)=reiθz = r(\cos\theta + i\sin\theta) = re^{i\theta}

Where r=∣z∣=a2+b2r = |z| = \sqrt{a^2 + b^2} is the modulus and θ=arg⁡(z)\theta = \arg(z) is the argument.

Definition. The principal argument Arg(z)\mathrm{Arg}(z) is the unique θ∈(−π,π]\theta \in (-\pi, \pi] Such that z=∣z∣eiθz = |z|e^{i\theta}. The argument arg⁡(z)\arg(z) is multi-valued: arg⁡(z)=Arg(z)+2πk\arg(z) = \mathrm{Arg}(z) + 2\pi k for k∈Zk \in \mathbb{Z}.

Proposition 1.4. If z1=r1eiθ1z_1 = r_1 e^{i\theta_1} and z2=r2eiθ2z_2 = r_2 e^{i\theta_2}Then z1z2=r1r2ei(θ1+θ2)z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)} and z1/z2=(r1/r2) ei(θ1−θ2)z_1/z_2 = (r_1/r_2)\, e^{i(\theta_1 - \theta_2)}.

Solution

Problem. Convert z=−1+3 iz = -1 + \sqrt{3}\,i to polar form and find all arguments.

∣z∣=(−1)2+(3)2=1+3=2|z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2.

Re(z)=−1<0\mathrm{Re}(z) = -1 \lt 0 and Im(z)=3>0\mathrm{Im}(z) = \sqrt{3} \gt 0So zz is in the second quadrant.

θ=arctan⁡ ⁣(3−1)=2π3\theta = \arctan\!\left(\frac{\sqrt{3}}{-1}\right) = \frac{2\pi}{3} (adjusting to second quadrant).

Polar form: z=2 e2πi/3z = 2\,e^{2\pi i/3}.

All arguments: arg⁡(z)=2π3+2πk\arg(z) = \frac{2\pi}{3} + 2\pi k for k∈Zk \in \mathbb{Z}.

Problem. Convert z=3e−iπ/4z = 3e^{-i\pi/4} to rectangular form.

z=3(cos⁡ ⁣(−π4)+isin⁡ ⁣(−π4))=3(22−i 22)=322−322 iz = 3\left(\cos\!\left(-\frac{\pi}{4}\right) + i\sin\!\left(-\frac{\pi}{4}\right)\right) = 3\left(\frac{\sqrt{2}}{2} - i\,\frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2}\,i.

Problem. Express z=−3−4iz = -3 - 4i in polar form.

∣z∣=9+16=5|z| = \sqrt{9 + 16} = 5.

Both real and imaginary parts are negative, so zz is in the third quadrant.

θ=arctan⁡(4/3)+π=π+arctan⁡(4/3)\theta = \arctan(4/3) + \pi = \pi + \arctan(4/3).

z=5 ei(π+arctan⁡(4/3))z = 5\,e^{i(\pi + \arctan(4/3))}.

1.4 Euler”s Formula and De Moivre’s Theorem

Section titled “1.4 Euler”s Formula and De Moivre’s Theorem”

Euler’s formula: eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta.

De Moivre’s theorem: (eiθ)n=einθ(e^{i\theta})^n = e^{in\theta}So

(cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)

Proposition 1.5. De Moivre’s theorem holds for all integers nnIncluding negative values.

Proof. For n≥0n \geq 0It follows by induction from the multiplication law eiαeiβ=ei(α+β)e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)}. For n<0n \lt 0Write n=−mn = -m with m>0m \gt 0: (cos⁡θ+isin⁡θ)n=1(cos⁡θ+isin⁡θ)m=1cos⁡(mθ)+isin⁡(mθ)=cos⁡(−mθ)+isin⁡(−mθ)=cos⁡(nθ)+isin⁡(nθ)(\cos\theta + i\sin\theta)^n = \frac{1}{(\cos\theta + i\sin\theta)^m} = \frac{1}{\cos(m\theta) + i\sin(m\theta)} = \cos(-m\theta) + i\sin(-m\theta) = \cos(n\theta) + i\sin(n\theta). ■\blacksquare

Example. Compute (1+i)20(1 + i)^{20}.

1+i=2 eiπ/41 + i = \sqrt{2}\,e^{i\pi/4}So (1+i)20=(2)20 e20πi/4=210 e5πi=1024 eπi=−1024(1 + i)^{20} = (\sqrt{2})^{20}\, e^{20\pi i/4} = 2^{10}\, e^{5\pi i} = 1024\,e^{\pi i} = -1024.

Solution

Problem. Express cos⁡(5θ)\cos(5\theta) in terms of cos⁡θ\cos\theta using de Moivre.

By de Moivre: cos⁡(5θ)+isin⁡(5θ)=(cos⁡θ+isin⁡θ)5\cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5.

Expanding the right side by the binomial theorem and equating real parts:

cos⁡(5θ)=cos⁡5θ−10cos⁡3θsin⁡2θ+5cos⁡θsin⁡4θ\cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta.

Using sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta:

cos⁡(5θ)=cos⁡5θ−10cos⁡3θ(1−cos⁡2θ)+5cos⁡θ(1−cos⁡2θ)2\cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2 =cos⁡5θ−10cos⁡3θ+10cos⁡5θ+5cos⁡θ−10cos⁡3θ+5cos⁡5θ= \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta =16cos⁡5θ−20cos⁡3θ+5cos⁡θ= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta.

Problem. Show that ∑k=0n−1cos⁡(kθ)=sin⁡(nθ/2)sin⁡(θ/2)cos⁡ ⁣((n−1)θ2)\sum_{k=0}^{n-1} \cos(k\theta) = \frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos\!\left(\frac{(n-1)\theta}{2}\right) For θ∉2πZ\theta \notin 2\pi\mathbb{Z}.

Consider S=∑k=0n−1eikθ=1−einθ1−eiθS = \sum_{k=0}^{n-1} e^{ik\theta} = \frac{1 - e^{in\theta}}{1 - e^{i\theta}} (geometric series with r=eiθ≠1r = e^{i\theta} \neq 1).

S=einθ/2(e−inθ/2−einθ/2)eiθ/2(e−iθ/2−eiθ/2)=ei(n−1)θ/2⋅sin⁡(nθ/2)sin⁡(θ/2)S = \frac{e^{in\theta/2}(e^{-in\theta/2} - e^{in\theta/2})}{e^{i\theta/2}(e^{-i\theta/2} - e^{i\theta/2})} = e^{i(n-1)\theta/2} \cdot \frac{\sin(n\theta/2)}{\sin(\theta/2)}.

Taking real parts gives the result.

Definition. An nn-th root of w∈Cw \in \mathbb{C} is a complex number zz such that zn=wz^n = w.

Proposition 1.6. Every non-zero w∈Cw \in \mathbb{C} has exactly nn distinct nn-th roots. If w=ρ eiϕw = \rho\, e^{i\phi}Then

zk=ρ1/n ei(ϕ+2πk)/n,k=0,1,…,n−1z_k = \rho^{1/n}\, e^{i(\phi + 2\pi k)/n}, \quad k = 0, 1, \ldots, n - 1

Where ρ1/n>0\rho^{1/n} \gt 0 is the positive real nn-th root of ρ\rho.

Proof. If zn=wz^n = wWrite z=r eiθz = r\,e^{i\theta}. Then rneinθ=ρ eiϕr^n e^{in\theta} = \rho\, e^{i\phi} So r=ρ1/nr = \rho^{1/n} and nθ=ϕ+2πkn\theta = \phi + 2\pi k. For k=0,1,…,n−1k = 0, 1, \ldots, n-1 these give distinct Values of θ\theta; for k≥nk \geq n they repeat. ■\blacksquare

Remark. The nn-th roots of ww lie equally spaced on a circle of radius ρ1/n\rho^{1/n}Forming a Regular nn-gon.

The nn-th roots of unity are the solutions of zn=1z^n = 1:

zk=e2πik/n,k=0,1,…,n−1z_k = e^{2\pi i k / n}, \quad k = 0, 1, \ldots, n - 1

They form a regular nn-gon on the unit circle in the complex plane.

Proposition 1.7. If ω=e2πi/n\omega = e^{2\pi i/n} is a primitive nn-th root of unity, then ∑k=0n−1ωk=0\sum_{k=0}^{n-1} \omega^k = 0 and ∑k=0n−1ωjk=0\sum_{k=0}^{n-1} \omega^{jk} = 0 for any jj not divisible by nn.

Proof. The sum ∑k=0n−1ωk=1−ωn1−ω=1−11−ω=0\sum_{k=0}^{n-1} \omega^k = \frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0 Provided ω≠1\omega \neq 1. For jj not divisible by nn, ωj\omega^j is a non-trivial root of unity, So the same argument applies. ■\blacksquare

Solution

Problem. Find all cube roots of −8-8.

−8=8 eiπ-8 = 8\,e^{i\pi}. The cube roots are: zk=81/3 ei(π+2πk)/3=2 ei(π+2πk)/3z_k = 8^{1/3}\, e^{i(\pi + 2\pi k)/3} = 2\, e^{i(\pi + 2\pi k)/3} for k=0,1,2k = 0, 1, 2.

z0=2 eiπ/3=2(12+i 32)=1+i3z_0 = 2\,e^{i\pi/3} = 2\left(\frac{1}{2} + i\,\frac{\sqrt{3}}{2}\right) = 1 + i\sqrt{3}. z1=2 eiπ=−2z_1 = 2\,e^{i\pi} = -2. z2=2 ei5π/3=2(12−i 32)=1−i3z_2 = 2\,e^{i5\pi/3} = 2\left(\frac{1}{2} - i\,\frac{\sqrt{3}}{2}\right) = 1 - i\sqrt{3}.

Problem. Find all fourth roots of z=16iz = 16i.

16i=16 eiπ/216i = 16\,e^{i\pi/2}. The fourth roots are: zk=161/4 ei(π/2+2πk)/4=2 ei(π/8+πk/2)z_k = 16^{1/4}\, e^{i(\pi/2 + 2\pi k)/4} = 2\, e^{i(\pi/8 + \pi k/2)} for k=0,1,2,3k = 0, 1, 2, 3.

z0=2 eiπ/8z_0 = 2\,e^{i\pi/8}, z1=2 ei5π/8z_1 = 2\,e^{i5\pi/8}, z2=2 ei9π/8z_2 = 2\,e^{i9\pi/8}, z3=2 ei13π/8z_3 = 2\,e^{i13\pi/8}.

Problem. Show that the nn-th roots of any non-zero ww are in geometric progression.

The roots are zk=ρ1/n ei(ϕ+2πk)/n=z0⋅(e2πi/n)k=z0⋅ωkz_k = \rho^{1/n}\, e^{i(\phi + 2\pi k)/n} = z_0 \cdot \left(e^{2\pi i/n}\right)^k = z_0 \cdot \omega^k Where ω=e2πi/n\omega = e^{2\pi i/n} is a primitive nn-th root of unity. This is a geometric sequence With ratio ω\omega.

A complex function is a function f:D⊆C→Cf : D \subseteq \mathbb{C} \to \mathbb{C}. We can write f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + iv(x, y) where z=x+iyz = x + iy and u,vu, v are real-valued functions.

Example. f(z)=z2=(x+iy)2=(x2−y2)+i(2xy)f(z) = z^2 = (x + iy)^2 = (x^2 - y^2) + i(2xy). Here u=x2−y2u = x^2 - y^2 and v=2xyv = 2xy.

Example. f(z)=zˉ=x−iyf(z) = \bar{z} = x - iy. Here u=xu = x and v=−yv = -y.

Example. f(z)=∣z∣2=x2+y2f(z) = |z|^2 = x^2 + y^2. Here u=x2+y2u = x^2 + y^2 and v=0v = 0.

The limit lim⁡z→z0f(z)=L\lim_{z \to z_0} f(z) = L means: for every ε>0\varepsilon \gt 0There exists δ>0\delta \gt 0 Such that 0<∣z−z0∣<δ0 \lt |z - z_0| \lt \delta implies ∣f(z)−L∣<ε|f(z) - L| \lt \varepsilon.

Unlike the real case, zz can approach z0z_0 from any direction in C\mathbb{C}. This makes limits More restrictive.

Proposition 2.1. lim⁡z→z0f(z)=L\lim_{z \to z_0} f(z) = L if and only if lim⁡(x,y)→(x0,y0)u(x,y)=a\lim_{(x,y) \to (x_0, y_0)} u(x, y) = a And lim⁡(x,y)→(x0,y0)v(x,y)=b\lim_{(x,y) \to (x_0, y_0)} v(x, y) = b where L=a+biL = a + bi.

Definition. ff is continuous at z0z_0 if lim⁡z→z0f(z)=f(z0)\lim_{z \to z_0} f(z) = f(z_0).

Solution

Problem. Show that lim⁡z→0zˉz\lim_{z \to 0} \frac{\bar{z}}{z} does not exist.

Let z=reiθz = re^{i\theta}. Then zˉz=e−2iθ\frac{\bar{z}}{z} = e^{-2i\theta}. As z→0z \to 0 along different Rays (θ=0,π/2,π/4\theta = 0, \pi/2, \pi/4Etc.), the ratio takes different values (1,−1,−i1, -1, -iEtc.). Since the limit depends on the direction of approach, it does not exist.

Problem. Determine whether f(z)=z2−1z−1f(z) = \frac{z^2 - 1}{z - 1} is continuous at z=1z = 1.

For z≠1z \neq 1: f(z)=z+1f(z) = z + 1. The limit as z→1z \to 1 is 22But f(1)f(1) is undefined (division by zero). If we define f(1)=2f(1) = 2Then ff becomes continuous at z=1z = 1.

Definition. ff is differentiable at z0z_0 if

f′(z0)=lim⁡h→0f(z0+h)−f(z0)hf'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}

Exists (and is independent of how h→0h \to 0 in C\mathbb{C}).

Remark. The requirement that the limit be the same for all directions of approach of hh is what Makes complex differentiability far more restrictive than real differentiability.

Definition. A function ff is analytic (or holomorphic) on an open set U⊆CU \subseteq \mathbb{C} if ff is differentiable at every point of UU. A function that is analytic On all of C\mathbb{C} is called entire.

Examples of entire functions: znz^n, eze^z, sin⁡z\sin z, cos⁡z\cos zPolynomials.

Example of a non-analytic function: f(z)=zˉf(z) = \bar{z} is nowhere differentiable (except at z=0z = 0 if we define it, but still not analytic there).

Solution

Problem. Show that f(z)=∣z∣2f(z) = |z|^2 is differentiable only at z=0z = 0.

f(z)=x2+y2f(z) = x^2 + y^2So u=x2+y2u = x^2 + y^2 and v=0v = 0. ux=2xu_x = 2x, uy=2yu_y = 2y, vx=0v_x = 0, vy=0v_y = 0. The Cauchy-Riemann equations require 2x=02x = 0 and 2y=02y = 0So x=y=0x = y = 0. Thus ff satisfies CR only at z=0z = 0.

At z=0z = 0: f′(0)=lim⁡h→0∣h∣2h=lim⁡h→0hˉ=0f'(0) = \lim_{h \to 0} \frac{|h|^2}{h} = \lim_{h \to 0} \bar{h} = 0So ff is Differentiable at 00 but not analytic anywhere (no neighbourhood of 00 is analytic).

Problem. Show that f(z)=zzˉ+zf(z) = z\bar{z} + z is differentiable only at z=0z = 0.

f(z)=∣z∣2+z=(x2+y2+x)+iyf(z) = |z|^2 + z = (x^2 + y^2 + x) + iy. ux=2x+1u_x = 2x + 1, uy=2yu_y = 2y, vx=0v_x = 0, vy=1v_y = 1. CR equations: 2x+1=1⇒x=02x + 1 = 1 \Rightarrow x = 0And 2y=0⇒y=02y = 0 \Rightarrow y = 0. At (0,0)(0, 0): f′(0)=lim⁡h→0hhˉ+hh=lim⁡h→0(hˉ+1)=1f'(0) = \lim_{h \to 0} \frac{h\bar{h} + h}{h} = \lim_{h \to 0} (\bar{h} + 1) = 1. So ff is differentiable at z=0z = 0 only, hence nowhere analytic.

2.5 Branch Cuts and Multi-Valued Functions

Section titled “2.5 Branch Cuts and Multi-Valued Functions”

Many important functions in complex analysis are inherently multi-valued. To work with them as Single-valued functions, we must restrict the domain.

Definition. A branch of a multi-valued function ff is a single-valued analytic function gg Defined on a domain DD such that g(z)∈f(z)g(z) \in f(z) for all z∈Dz \in D.

The Complex Logarithm. We define log⁡z=ln⁡∣z∣+iarg⁡(z)\log z = \ln|z| + i\arg(z)Which is multi-valued because arg⁡(z)=Arg(z)+2πk\arg(z) = \mathrm{Arg}(z) + 2\pi k for k∈Zk \in \mathbb{Z}. The principal branch is

Log z=ln⁡∣z∣+i Arg(z)\mathrm{Log}\, z = \ln|z| + i\,\mathrm{Arg}(z)

Defined on C∖(−∞,0]\mathbb{C} \setminus (-\infty, 0]. The negative real axis is called the branch cut.

Proposition 2.2. The principal branch Log z\mathrm{Log}\, z is analytic on C∖(−∞,0]\mathbb{C} \setminus (-\infty, 0] and ddz Log z=1z\frac{d}{dz}\,\mathrm{Log}\, z = \frac{1}{z}.

Complex Powers. For z,α∈Cz, \alpha \in \mathbb{C} with z≠0z \neq 0:

zα=eαlog⁡zz^\alpha = e^{\alpha \log z}

This is multi-valued . When α\alpha is rational with reduced form p/qp/qThere are exactly qq distinct values.

Solution

Problem. Find all values of (−1)i(-1)^i.

(−1)i=eilog⁡(−1)=ei(iπ+2πik)=e−π−2πk(-1)^i = e^{i \log(-1)} = e^{i(i\pi + 2\pi i k)} = e^{-\pi - 2\pi k} for k∈Zk \in \mathbb{Z}.

These are all positive real numbers: …,e3π,eπ,e−π,e−3π,…\ldots, e^{3\pi}, e^{\pi}, e^{-\pi}, e^{-3\pi}, \ldots. The principal value (using the principal branch) is e−πe^{-\pi}.

Problem. Find all values of i1/2i^{1/2}.

i1/2=e(1/2)log⁡i=e(1/2)(iπ/2+2πik)=eiπ/4+iπki^{1/2} = e^{(1/2)\log i} = e^{(1/2)(i\pi/2 + 2\pi i k)} = e^{i\pi/4 + i\pi k}.

For k=0k = 0: eiπ/4=22(1+i)e^{i\pi/4} = \frac{\sqrt{2}}{2}(1 + i). For k=1k = 1: ei5π/4=−22(1+i)e^{i5\pi/4} = -\frac{\sqrt{2}}{2}(1 + i). These are the two square roots of ii.

Problem. Find the domain of analyticity of f(z)=Log(z2+1)f(z) = \mathrm{Log}(z^2 + 1).

Log w\mathrm{Log}\, w is analytic on C∖(−∞,0]\mathbb{C} \setminus (-\infty, 0]So we need z2+1∉(−∞,0]z^2 + 1 \notin (-\infty, 0].

z2+1≤0z^2 + 1 \leq 0 when z2≤−1z^2 \leq -1I.e., z∈[−i,0]∪[0,i]z \in [-i, 0] \cup [0, i] (the imaginary axis Segment from −i-i to ii). Also z2+1=0z^2 + 1 = 0 at z=±iz = \pm i.

Domain: C∖{z:z=iy, y∈[−1,1]}\mathbb{C} \setminus \{z : z = iy,\, y \in [-1, 1]\}.

Theorem 3.1 (Cauchy-Riemann Equations). If f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + iv(x, y) is differentiable at z=x+iyz = x + iyThen

∂u∂x=∂v∂y,∂u∂y=−∂v∂x\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}

Proof. Compute the limit along the real axis (h∈Rh \in \mathbb{R}, h→0h \to 0):

f′(z)=lim⁡h→0u(x+h,y)−u(x,y)h+ilim⁡h→0v(x+h,y)−v(x,y)h=∂u∂x+i∂v∂xf'(z) = \lim_{h \to 0} \frac{u(x+h, y) - u(x, y)}{h} + i\lim_{h \to 0} \frac{v(x+h, y) - v(x, y)}{h} = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}

Compute along the imaginary axis (h=ikh = ik, k∈Rk \in \mathbb{R}, k→0k \to 0):

f′(z)=lim⁡k→0u(x,y+k)−u(x,y)ik+ilim⁡k→0v(x,y+k)−v(x,y)ik=−i∂u∂y+∂v∂yf'(z) = \lim_{k \to 0} \frac{u(x, y+k) - u(x, y)}{ik} + i\lim_{k \to 0} \frac{v(x, y+k) - v(x, y)}{ik} = -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}

Equating real and imaginary parts: ∂u∂x=∂v∂y\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} And ∂v∂x=−∂u∂y\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}. ■\blacksquare

Theorem 3.2. If uu and vv have continuous first partial derivatives on an open set UU and Satisfy the Cauchy-Riemann equations on UUThen f=u+ivf = u + iv is analytic on UU.

Proof. Since ux,uy,vx,vyu_x, u_y, v_x, v_y are continuous on UU, uu and vv are (real) differentiable. Let Δz=Δx+iΔy\Delta z = \Delta x + i\Delta y. By real differentiability:

u(x+Δx,y+Δy)−u(x,y)=ux Δx+uy Δy+ε1u(x + \Delta x, y + \Delta y) - u(x, y) = u_x\,\Delta x + u_y\,\Delta y + \varepsilon_1 v(x+Δx,y+Δy)−v(x,y)=vx Δx+vy Δy+ε2v(x + \Delta x, y + \Delta y) - v(x, y) = v_x\,\Delta x + v_y\,\Delta y + \varepsilon_2

Where ε1,ε2=o(∣Δz∣)\varepsilon_1, \varepsilon_2 = o(|\Delta z|). Therefore

f(z+Δz)−f(z)Δz=(ux+ivx)Δx+(uy+ivy)Δy+ε1+iε2Δx+iΔy\frac{f(z + \Delta z) - f(z)}{\Delta z} = \frac{(u_x + iv_x)\Delta x + (u_y + iv_y)\Delta y + \varepsilon_1 + i\varepsilon_2}{\Delta x + i\Delta y}

By CR: uy+ivy=−vx+iux=i(ux+ivx)u_y + iv_y = -v_x + iu_x = i(u_x + iv_x). Substituting:

=(ux+ivx)Δx+iΔyΔx+iΔy+o(∣Δz∣)Δz→ux+ivx= (u_x + iv_x)\frac{\Delta x + i\Delta y}{\Delta x + i\Delta y} + \frac{o(|\Delta z|)}{\Delta z} \to u_x + iv_x

As Δz→0\Delta z \to 0. ■\blacksquare

3.3 The Derivative in Terms of Partial Derivatives

Section titled “3.3 The Derivative in Terms of Partial Derivatives”

When the Cauchy-Riemann equations hold:

f′(z)=∂u∂x+i∂v∂x=∂v∂y−i∂u∂yf'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i\frac{\partial u}{\partial y}

Definition. A real-valued function ϕ(x,y)\phi(x, y) is harmonic if ϕxx+ϕyy=0\phi_{xx} + \phi_{yy} = 0 (Laplace’s equation).

Proposition 3.3. If f=u+ivf = u + iv is analytic, then uu and vv are harmonic.

Proof. From the Cauchy-Riemann equations: ux=vyu_x = v_y and uy=−vxu_y = -v_x. Differentiating: uxx=vyxu_{xx} = v_{yx} and uyy=−vxyu_{yy} = -v_{xy}. By equality of mixed partials, uxx+uyy=vyx−vxy=0u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0. Similarly for vv. ■\blacksquare

Definition. If uu and vv are harmonic on UU and satisfy the Cauchy-Riemann equations, then vv is the harmonic conjugate of uu.

Proposition 3.4. If UU is a connected domain and uu is harmonic on UUThen uu has A harmonic conjugate on UUUnique up to an additive constant.

Proof. Define v(x,y)=∫(x0,y0)(x,y)(−uy dx+ux dy)v(x, y) = \int_{(x_0, y_0)}^{(x, y)} (-u_y\, dx + u_x\, dy). The integrand is closed (since (−uy)y=−uyy=uxx=(ux)x(-u_y)_y = -u_{yy} = u_{xx} = (u_x)_x) and since UU is Connected, vv is well-defined (path-independent) by Green’s theorem. Then vx=−uyv_x = -u_y and vy=uxv_y = u_xWhich are the CR equations. ■\blacksquare

Solution

Problem. Find the harmonic conjugate of u(x,y)=x3−3xy2u(x, y) = x^3 - 3xy^2.

Verify uu is harmonic: uxx=6xu_{xx} = 6x, uyy=−6xu_{yy} = -6xSo uxx+uyy=0u_{xx} + u_{yy} = 0. ✓\checkmark

By CR: vy=ux=3x2−3y2v_y = u_x = 3x^2 - 3y^2So v=3x2y−y3+g(x)v = 3x^2 y - y^3 + g(x). Also vx=−uy=6xyv_x = -u_y = 6xySo 6xy=6xy+g′(x)6xy = 6xy + g'(x)Giving g′(x)=0g'(x) = 0So g(x)=Cg(x) = C.

Harmonic conjugate: v(x,y)=3x2y−y3+Cv(x, y) = 3x^2 y - y^3 + C.

Note: f(z)=u+iv=x3−3xy2+i(3x2y−y3)=(x+iy)3=z3f(z) = u + iv = x^3 - 3xy^2 + i(3x^2 y - y^3) = (x + iy)^3 = z^3.

Problem. Show that u(x,y)=ln⁡(x2+y2)u(x, y) = \ln(x^2 + y^2) is harmonic on R2∖{0}\mathbb{R}^2 \setminus \{0\} but Has no harmonic conjugate on R2∖{0}\mathbb{R}^2 \setminus \{0\}.

ux=2xx2+y2u_x = \frac{2x}{x^2 + y^2}, uxx=2(y2−x2)(x2+y2)2u_{xx} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2}. uy=2yx2+y2u_y = \frac{2y}{x^2 + y^2}, uyy=2(x2−y2)(x2+y2)2u_{yy} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2}. Δu=0\Delta u = 0. ✓\checkmark

However, ∮∣z∣=1(−uy dx+ux dy)=∮∣z∣=1−y dx+x dyx2+y2=∫02π1 dθ=2π≠0\oint_{|z|=1} (-u_y\, dx + u_x\, dy) = \oint_{|z|=1} \frac{-y\, dx + x\, dy}{x^2 + y^2} = \int_0^{2\pi} 1\, d\theta = 2\pi \neq 0.

Since R2∖{0}\mathbb{R}^2 \setminus \{0\} is not connected and this integral is non-zero, no Harmonic conjugate exists on this domain.

3.5 Worked Examples: Verifying CR Equations

Section titled “3.5 Worked Examples: Verifying CR Equations”
Solution

Problem. Verify that f(z)=ezf(z) = e^z satisfies the Cauchy-Riemann equations and find f′(z)f'(z).

Solution. ez=ex+iy=ex(cos⁡y+isin⁡y)e^z = e^{x+iy} = e^x(\cos y + i\sin y). So u=excos⁡yu = e^x \cos y and v=exsin⁡yv = e^x \sin y.

ux=excos⁡yu_x = e^x \cos y, uy=−exsin⁡yu_y = -e^x \sin y, vx=exsin⁡yv_x = e^x \sin y, vy=excos⁡yv_y = e^x \cos y.

Cauchy-Riemann: ux=excos⁡y=vyu_x = e^x \cos y = v_y and uy=−exsin⁡y=−vxu_y = -e^x \sin y = -v_x. Both satisfied.

f′(z)=ux+ivx=excos⁡y+iexsin⁡y=ezf'(z) = u_x + iv_x = e^x \cos y + ie^x \sin y = e^z. ■\blacksquare

Problem. Verify CR for f(z)=sin⁡zf(z) = \sin z and find f′(z)f'(z).

sin⁡z=sin⁡(x+iy)=sin⁡xcosh⁡y+icos⁡xsinh⁡y\sin z = \sin(x + iy) = \sin x \cosh y + i\cos x \sinh y.

u=sin⁡xcosh⁡yu = \sin x \cosh y, v=cos⁡xsinh⁡yv = \cos x \sinh y.

ux=cos⁡xcosh⁡yu_x = \cos x \cosh y, uy=sin⁡xsinh⁡yu_y = \sin x \sinh y. vx=−sin⁡xsinh⁡yv_x = -\sin x \sinh y, vy=cos⁡xcosh⁡yv_y = \cos x \cosh y.

CR: ux=cos⁡xcosh⁡y=vyu_x = \cos x \cosh y = v_y ✓\checkmark and uy=sin⁡xsinh⁡y=−vxu_y = \sin x \sinh y = -v_x ✓\checkmark.

f′(z)=ux+ivx=cos⁡xcosh⁡y−isin⁡xsinh⁡y=cos⁡zf'(z) = u_x + iv_x = \cos x \cosh y - i\sin x \sinh y = \cos z. ■\blacksquare

Problem. Show f(z)=1zf(z) = \frac{1}{z} satisfies CR on C∖{0}\mathbb{C} \setminus \{0\}.

1z=zˉ∣z∣2=x−iyx2+y2\frac{1}{z} = \frac{\bar{z}}{|z|^2} = \frac{x - iy}{x^2 + y^2}.

u=xx2+y2u = \frac{x}{x^2 + y^2}, v=−yx2+y2v = \frac{-y}{x^2 + y^2}.

ux=y2−x2(x2+y2)2u_x = \frac{y^2 - x^2}{(x^2 + y^2)^2}, vy=y2−x2(x2+y2)2v_y = \frac{y^2 - x^2}{(x^2 + y^2)^2}. So ux=vyu_x = v_y. ✓\checkmark

uy=−2xy(x2+y2)2u_y = \frac{-2xy}{(x^2 + y^2)^2}, vx=2xy(x2+y2)2v_x = \frac{2xy}{(x^2 + y^2)^2}. So uy=−vxu_y = -v_x. ✓\checkmark

f′(z)=ux+ivx=−(x2−y2+2ixy)(x2+y2)2=−1z2f'(z) = u_x + iv_x = \frac{-(x^2 - y^2 + 2ixy)}{(x^2 + y^2)^2} = \frac{-1}{z^2}. ■\blacksquare

A contour (or piecewise smooth path) in C\mathbb{C} is a continuous function γ:[a,b]→C\gamma : [a, b] \to \mathbb{C} that is differentiable except at finitely many points, with a Continuous derivative everywhere it exists.

A simple closed contour is a contour with γ(a)=γ(b)\gamma(a) = \gamma(b) and no other Self-intersections.

Definition. For a contour γ\gamma and a continuous function ff on γ\gamma:

∫γf(z) dz=∫abf(γ(t))γ′(t) dt\int_{\gamma} f(z)\, dz = \int_a^b f(\gamma(t))\gamma'(t)\, dt

Proposition 4.1. The complex integral is linear:

∫γ(af+bg) dz=a∫γf dz+b∫γg dz\int_\gamma (af + bg)\, dz = a\int_\gamma f\, dz + b\int_\gamma g\, dz

Proposition 4.2. Reversing orientation changes the sign:

∫−γf dz=−∫γf dz\int_{-\gamma} f\, dz = -\int_\gamma f\, dz

Proposition 4.3. Additivity over contours:

∫γ1+γ2f dz=∫γ1f dz+∫γ2f dz\int_{\gamma_1 + \gamma_2} f\, dz = \int_{\gamma_1} f\, dz + \int_{\gamma_2} f\, dz

Proposition 4.4 (ML Inequality). If ∣f(z)∣≤M|f(z)| \leq M for all zz on a contour γ\gamma of length LLThen

∣∫γf(z) dz∣≤ML\left|\int_\gamma f(z)\, dz\right| \leq ML

Proof. ∣∫abf(γ(t))γ′(t) dt∣≤∫ab∣f(γ(t))∣∣γ′(t)∣ dt≤M∫ab∣γ′(t)∣ dt=ML\left|\int_a^b f(\gamma(t))\gamma'(t)\, dt\right| \leq \int_a^b |f(\gamma(t))||\gamma'(t)|\, dt \leq M \int_a^b |\gamma'(t)|\, dt = ML. ■\blacksquare

Solution

Problem. Evaluate ∫γz2 dz\int_\gamma z^2\, dz where γ\gamma is the line segment from 00 to 1+i1 + i.

Solution. Parameterize γ(t)=t(1+i)\gamma(t) = t(1 + i) for 0≤t≤10 \leq t \leq 1. Then γ′(t)=1+i\gamma'(t) = 1 + i.

∫γz2 dz=∫01(t(1+i))2(1+i) dt=(1+i)3∫01t2 dt=(1+i)3⋅13\int_\gamma z^2\, dz = \int_0^1 (t(1+i))^2 (1+i)\, dt = (1+i)^3 \int_0^1 t^2\, dt = (1+i)^3 \cdot \frac{1}{3}

(1+i)3=(1+i)(1+i)2=(1+i)(2i)=2i+2i2=2i−2=−2+2i(1+i)^3 = (1+i)(1+i)^2 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i.

∫γz2 dz=−2+2i3\int_\gamma z^2\, dz = \frac{-2 + 2i}{3}. ■\blacksquare

Problem. Evaluate ∫γzˉ dz\int_\gamma \bar{z}\, dz where γ\gamma is the unit circle traversed once Counterclockwise.

γ(t)=eit\gamma(t) = e^{it}, 0≤t≤2π0 \leq t \leq 2\pi, γ′(t)=ieit\gamma'(t) = ie^{it}. zˉ=e−it\bar{z} = e^{-it} on γ\gamma.

∫γzˉ dz=∫02πe−it⋅ieit dt=∫02πi dt=2πi\int_\gamma \bar{z}\, dz = \int_0^{2\pi} e^{-it} \cdot ie^{it}\, dt = \int_0^{2\pi} i\, dt = 2\pi i.

Note: Since zˉ\bar{z} is not analytic, this result is non-zero, as expected.

Problem. Evaluate ∫γdzz\int_\gamma \frac{dz}{z} where γ\gamma is the upper semicircle z=eiθz = e^{i\theta}, 0≤θ≤π0 \leq \theta \leq \pi.

∫0πieiθeiθ dθ=∫0πi dθ=iπ\int_0^\pi \frac{ie^{i\theta}}{e^{i\theta}}\, d\theta = \int_0^\pi i\, d\theta = i\pi.

Problem. Evaluate ∫γz dz\int_\gamma z\, dz where γ\gamma consists of the line segment from 00 to 11 followed by the line segment from 11 to 1+i1 + i.

γ1(t)=t\gamma_1(t) = t, 0≤t≤10 \leq t \leq 1: ∫01t⋅1 dt=12\int_0^1 t \cdot 1\, dt = \frac{1}{2}.

γ2(t)=1+it\gamma_2(t) = 1 + it, 0≤t≤10 \leq t \leq 1: ∫01(1+it)⋅i dt=∫01(i−t) dt=i−12\int_0^1 (1 + it) \cdot i\, dt = \int_0^1 (i - t)\, dt = i - \frac{1}{2}.

Total: 12+i−12=i\frac{1}{2} + i - \frac{1}{2} = i.

Check: Since zz is entire, the integral from 00 to 1+i1 + i is 12(1+i)2=i\frac{1}{2}(1+i)^2 = i. Consistent. ■\blacksquare

Solution

Problem. Use the ML inequality to show that lim⁡R→∞∫CReizz dz=0\lim_{R \to \infty} \int_{C_R} \frac{e^{iz}}{z}\, dz = 0 Where CRC_R is the upper semicircle ∣z∣=R|z| = R, Im(z)≥0\mathrm{Im}(z) \geq 0.

On CRC_R: z=Reiθz = Re^{i\theta}, 0≤θ≤π0 \leq \theta \leq \pi. ∣eiz∣=∣eiR(cos⁡θ+isin⁡θ)∣=e−Rsin⁡θ|e^{iz}| = |e^{iR(\cos\theta + i\sin\theta)}| = e^{-R\sin\theta}.

∣∫CReizz dz∣≤∫0πe−Rsin⁡θR⋅R dθ=∫0πe−Rsin⁡θ dθ\left|\int_{C_R} \frac{e^{iz}}{z}\, dz\right| \leq \int_0^\pi \frac{e^{-R\sin\theta}}{R} \cdot R\, d\theta = \int_0^\pi e^{-R\sin\theta}\, d\theta.

By Jordan’s inequality sin⁡θ≥2θπ\sin\theta \geq \frac{2\theta}{\pi} for θ∈[0,π/2]\theta \in [0, \pi/2]:

≤2∫0π/2e−2Rθ/π dθ=πR(1−e−R)→0\leq 2\int_0^{\pi/2} e^{-2R\theta/\pi}\, d\theta = \frac{\pi}{R}(1 - e^{-R}) \to 0 as R→∞R \to \infty. ■\blacksquare

Problem. Bound ∣∫γdzz2+4∣\left|\int_\gamma \frac{dz}{z^2 + 4}\right| where γ\gamma is ∣z∣=3|z| = 3.

On γ\gamma: ∣z2+4∣≥∣z∣2−4=9−4=5|z^2 + 4| \geq |z|^2 - 4 = 9 - 4 = 5 (reverse triangle inequality). So ∣1z2+4∣≤15\left|\frac{1}{z^2 + 4}\right| \leq \frac{1}{5}.

Length of γ\gamma: L=2π⋅3=6πL = 2\pi \cdot 3 = 6\pi.

∣∫γdzz2+4∣≤15⋅6π=6π5\left|\int_\gamma \frac{dz}{z^2 + 4}\right| \leq \frac{1}{5} \cdot 6\pi = \frac{6\pi}{5}.

When ff is analytic on a connected domain and has a known antiderivative FF with F′=fF' = f:

∫γf(z) dz=F(γ(b))−F(γ(a))\int_\gamma f(z)\, dz = F(\gamma(b)) - F(\gamma(a))

This follows from the fundamental theorem of calculus applied to F(γ(t))F(\gamma(t)).

Solution

Problem. Evaluate ∫γcos⁡z dz\int_\gamma \cos z\, dz where γ\gamma is any path from 00 to π+i\pi + i.

Since cos⁡z\cos z is entire with antiderivative sin⁡z\sin z:

∫γcos⁡z dz=sin⁡(π+i)−sin⁡(0)=sin⁡(π+i)\int_\gamma \cos z\, dz = \sin(\pi + i) - \sin(0) = \sin(\pi + i).

sin⁡(π+i)=sin⁡πcosh⁡1+icos⁡πsinh⁡1=−isinh⁡1\sin(\pi + i) = \sin\pi\cosh 1 + i\cos\pi\sinh 1 = -i\sinh 1.

So the integral equals −isinh⁡1-i\sinh 1.

Problem. Evaluate ∫γe2z dz\int_\gamma e^{2z}\, dz where γ\gamma is any path from 11 to ii.

Antiderivative: 12e2z\frac{1}{2}e^{2z}.

∫γe2z dz=12(e2i−e2)\int_\gamma e^{2z}\, dz = \frac{1}{2}(e^{2i} - e^{2}).

Theorem 5.1 (Cauchy’s Theorem). If ff is analytic on a connected domain DD and γ\gamma Is a simple closed contour in DDThen

∫γf(z) dz=0\int_\gamma f(z)\, dz = 0

Proof (for f′f' continuous). By Green’s theorem in the plane, writing f=u+ivf = u + iv:

∫γf dz=∫γ(u dx−v dy)+i∫γ(v dx+u dy)\int_\gamma f\, dz = \int_\gamma (u\, dx - v\, dy) + i\int_\gamma (v\, dx + u\, dy)

Applying Green’s theorem to each integral:

=∬D(−vx−uy) dA+i∬D(ux−vy) dA=0= \iint_D (-v_x - u_y)\, dA + i\iint_D (u_x - v_y)\, dA = 0

By the Cauchy-Riemann equations. ■\blacksquare

A domain D⊆CD \subseteq \mathbb{C} is ** connected** if every simple closed contour in DD can Be continuously shrunk to a point within DD.

Cauchy’s theorem may fail on multiply connected domains. For example, ∫γ1z dz=2πi\int_\gamma \frac{1}{z}\, dz = 2\pi i where γ\gamma is the unit circle (traversing a region that Excludes the singularity at z=0z = 0).

Corollary 5.2. If ff is analytic on a connected domain DDThen the integral ∫z0z1f(z) dz\int_{z_0}^{z_1} f(z)\, dz is independent of the path from z0z_0 to z1z_1 in DD.

Theorem 5.3. If ff is analytic on a connected domain DDThen ff has an antiderivative FF in DD (i.e., F′(z)=f(z)F'(z) = f(z)), and

∫γf(z) dz=F(z1)−F(z0)\int_\gamma f(z)\, dz = F(z_1) - F(z_0)

Where z0z_0 and z1z_1 are the endpoints of γ\gamma.

5.5 Cauchy’s Theorem for Multiply Connected Domains

Section titled “5.5 Cauchy’s Theorem for Multiply Connected Domains”

Theorem 5.4. If ff is analytic on a domain DD containing simple closed contours γ,γ1,…,γn\gamma, \gamma_1, \ldots, \gamma_n where γ1,…,γn\gamma_1, \ldots, \gamma_n Lie in the interior of γ\gamma and the region between γ\gamma and the γk\gamma_k is contained in DD And all contours are positively oriented, then

∫γf(z) dz=∑k=1n∫γkf(z) dz\int_\gamma f(z)\, dz = \sum_{k=1}^n \int_{\gamma_k} f(z)\, dz

Theorem 5.5 (Deformation of Contours). If ff is analytic on a domain containing two simple Closed contours γ1\gamma_1 and γ2\gamma_2 where one can be continuously deformed into the other Within the domain of analyticity of ffThen

∫γ1f(z) dz=∫γ2f(z) dz\int_{\gamma_1} f(z)\, dz = \int_{\gamma_2} f(z)\, dz

Proof. This follows directly from Theorem 5.4 applied to the region between γ1\gamma_1 and γ2\gamma_2. ■\blacksquare

Remark. This theorem is enormously useful: we can replace a complicated contour with a simpler one (a small circle around each singularity) without changing the value of the integral.

Solution

Problem. Evaluate ∫γdzz−2\int_\gamma \frac{dz}{z - 2} where γ\gamma is the ellipse x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1.

Since z=2z = 2 is inside the ellipse and 1z−2\frac{1}{z - 2} is analytic everywhere else, By deformation of contours we can replace γ\gamma with a small circle around z=2z = 2:

∫γdzz−2=2πi\int_\gamma \frac{dz}{z - 2} = 2\pi i.

Problem. Evaluate ∫γezz dz\int_\gamma \frac{e^z}{z}\, dz where γ\gamma is the square with vertices ±2±2i\pm 2 \pm 2i.

ezz\frac{e^z}{z} is analytic on and inside γ\gamma except at z=0z = 0. By deformation: ∫γezz dz=∫∣z∣=rezz dz=2πi⋅e0=2πi\int_\gamma \frac{e^z}{z}\, dz = \int_{|z|=r} \frac{e^z}{z}\, dz = 2\pi i \cdot e^0 = 2\pi i.

Problem. Evaluate ∫γdzz2−1\int_\gamma \frac{dz}{z^2 - 1} where γ\gamma is ∣z∣=2|z| = 2.

1z2−1=12(1z−1−1z+1)\frac{1}{z^2 - 1} = \frac{1}{2}\left(\frac{1}{z-1} - \frac{1}{z+1}\right).

Both z=±1z = \pm 1 are inside ∣z∣=2|z| = 2.

∫γdzz2−1=12(2πi−2πi)=0\int_\gamma \frac{dz}{z^2 - 1} = \frac{1}{2}(2\pi i - 2\pi i) = 0.

Theorem 6.1 (Cauchy’s Integral Formula). If ff is analytic on a connected domain Containing a simple closed positively oriented contour γ\gammaAnd z0z_0 is inside γ\gammaThen

f(z0)=12πi∫γf(z)z−z0 dzf(z_0) = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z - z_0}\, dz

Proof. Let γε\gamma_\varepsilon be a small circle of radius ε\varepsilon around z0z_0. Since f(z)z−z0\frac{f(z)}{z - z_0} is analytic on the region between γ\gamma and γε\gamma_\varepsilon

∫γf(z)z−z0 dz=∫γεf(z)z−z0 dz\int_\gamma \frac{f(z)}{z - z_0}\, dz = \int_{\gamma_\varepsilon} \frac{f(z)}{z - z_0}\, dz

On γε\gamma_\varepsilon: f(z)=f(z0)+(z−z0)f′(ζ)f(z) = f(z_0) + (z - z_0)f'(\zeta) for some ζ\zeta between zz and z0z_0.

=∫γεf(z0)z−z0 dz+∫γεf′(ζ) dz=f(z0)⋅2πi+0= \int_{\gamma_\varepsilon} \frac{f(z_0)}{z - z_0}\, dz + \int_{\gamma_\varepsilon} f'(\zeta)\, dz = f(z_0) \cdot 2\pi i + 0

Since ∫γεdzz−z0=2πi\int_{\gamma_\varepsilon} \frac{dz}{z - z_0} = 2\pi i (parameterize z=z0+εeiθz = z_0 + \varepsilon e^{i\theta}) and ∫γεf′(ζ) dz→0\int_{\gamma_\varepsilon} f'(\zeta)\, dz \to 0 as ε→0\varepsilon \to 0 by the ML inequality. ■\blacksquare

Theorem 6.2 (Cauchy’s Integral Formula for Derivatives). Under the same conditions,

f(n)(z0)=n!2πi∫γf(z)(z−z0)n+1 dzf^{(n)}(z_0) = \frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz

Proof. We proceed by induction. The base case n=0n = 0 is Theorem 6.1. For the inductive step, Assume the formula holds for nn. Using the difference quotient:

f(n+1)(z0)=lim⁡h→0f(n)(z0+h)−f(n)(z0)h=lim⁡h→0n!2πi∫γ1h[f(z)(z−z0−h)n+1−f(z)(z−z0)n+1]dzf^{(n+1)}(z_0) = \lim_{h \to 0} \frac{f^{(n)}(z_0 + h) - f^{(n)}(z_0)}{h} = \lim_{h \to 0} \frac{n!}{2\pi i}\int_\gamma \frac{1}{h}\left[\frac{f(z)}{(z - z_0 - h)^{n+1}} - \frac{f(z)}{(z - z_0)^{n+1}}\right] dz

=(n+1)!2πi∫γf(z)(z−z0)n+2 dz= \frac{(n+1)!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+2}}\, dz

Where we justified passing the limit inside the integral by uniform convergence of the integrand On compact subsets. ■\blacksquare

6.3 Consequences of Cauchy’s Integral Formula

Section titled “6.3 Consequences of Cauchy’s Integral Formula”

Corollary 6.3. If ff is analytic, then ff is infinitely differentiable.

This is remarkable: a single complex derivative implies the existence of all derivatives.

Corollary 6.4 (Cauchy’s Estimates). If ff is analytic on and inside a circle ∣z−z0∣=R|z - z_0| = R And ∣f(z)∣≤M|f(z)| \leq M on the circle, then

∣f(n)(z0)∣≤n!MRn|f^{(n)}(z_0)| \leq \frac{n!M}{R^n}

Proof. From the integral formula: ∣f(n)(z0)∣=n!2π∣∫∣z−z0∣=Rf(z)(z−z0)n+1 dz∣≤n!2π⋅MRn+1⋅2πR=n!MRn|f^{(n)}(z_0)| = \frac{n!}{2\pi}\left|\int_{|z-z_0|=R} \frac{f(z)}{(z-z_0)^{n+1}}\, dz\right| \leq \frac{n!}{2\pi} \cdot \frac{M}{R^{n+1}} \cdot 2\pi R = \frac{n!M}{R^n}. ■\blacksquare

Theorem 6.5 (Liouville’s Theorem). Every bounded entire function is constant.

Proof. If ∣f(z)∣≤M|f(z)| \leq M for all zzThen by Cauchy’s estimates with RR arbitrarily large: ∣f′(z0)∣≤MR→0|f'(z_0)| \leq \frac{M}{R} \to 0 as R→∞R \to \infty. So f′(z)=0f'(z) = 0 for all zzMeaning ff is Constant. ■\blacksquare

Corollary 6.6. If ff is entire and ∣f(z)∣≥M|f(z)| \geq M for all zz (bounded away from zero), then ff is constant.

Proof. 1/f1/f is entire and bounded by 1/M1/MSo constant by Liouville. ■\blacksquare

Theorem 6.7 (Fundamental Theorem of Algebra). Every non-constant polynomial p(z)∈C[z]p(z) \in \mathbb{C}[z] has a root in C\mathbb{C}.

Proof. Suppose p(z)p(z) has no root. Then f(z)=1/p(z)f(z) = 1/p(z) is entire. Since ∣p(z)∣→∞|p(z)| \to \infty as ∣z∣→∞|z| \to \infty, f(z)→0f(z) \to 0So ff is bounded. By Liouville’s theorem, ff is constant, so pp Is constant, a contradiction. ■\blacksquare

Corollary 6.8. Every polynomial of degree n≥1n \geq 1 has exactly nn roots in C\mathbb{C} Counting multiplicities.

6.6 Worked Examples: Cauchy’s Integral Formula

Section titled “6.6 Worked Examples: Cauchy’s Integral Formula”
Solution

Problem. Evaluate ∫γezz−1 dz\int_\gamma \frac{e^z}{z - 1}\, dz where γ\gamma is ∣z∣=2|z| = 2.

Solution. The function ezz−1\frac{e^z}{z - 1} has a singularity at z=1z = 1Which lies inside γ\gamma. By Cauchy’s integral formula with f(z)=ezf(z) = e^z and z0=1z_0 = 1:

∫γezz−1 dz=2πi⋅f(1)=2πi⋅e1=2πie\int_\gamma \frac{e^z}{z - 1}\, dz = 2\pi i \cdot f(1) = 2\pi i \cdot e^1 = 2\pi i e. ■\blacksquare

Problem. Evaluate ∫γz2+1(z−i)3 dz\int_\gamma \frac{z^2 + 1}{(z - i)^3}\, dz where γ\gamma is ∣z∣=2|z| = 2.

By Cauchy’s formula for derivatives with f(z)=z2+1f(z) = z^2 + 1 and z0=iz_0 = i:

∫γf(z)(z−i)3 dz=2πi2! f′′(i)\int_\gamma \frac{f(z)}{(z - i)^3}\, dz = \frac{2\pi i}{2!}\,f''(i).

f′(z)=2zf'(z) = 2z, f′′(z)=2f''(z) = 2. So f′′(i)=2f''(i) = 2.

∫γz2+1(z−i)3 dz=2πi2⋅2=2πi\int_\gamma \frac{z^2 + 1}{(z - i)^3}\, dz = \frac{2\pi i}{2} \cdot 2 = 2\pi i. ■\blacksquare

Problem. Evaluate ∫γsin⁡zz(z−π) dz\int_\gamma \frac{\sin z}{z(z - \pi)}\, dz where γ\gamma is ∣z∣=4|z| = 4.

Singularities inside γ\gamma: z=0z = 0 and z=πz = \pi.

sin⁡zz(z−π)=1π(sin⁡zz−π−sin⁡zz)\frac{\sin z}{z(z - \pi)} = \frac{1}{\pi}\left(\frac{\sin z}{z - \pi} - \frac{\sin z}{z}\right).

At z=0z = 0: by CIF, ∫γsin⁡zz dz=2πi⋅sin⁡(0)=0\int_\gamma \frac{\sin z}{z}\, dz = 2\pi i \cdot \sin(0) = 0. At z=πz = \pi: by CIF, ∫γsin⁡zz−π dz=2πi⋅sin⁡(π)=0\int_\gamma \frac{\sin z}{z - \pi}\, dz = 2\pi i \cdot \sin(\pi) = 0.

∫γsin⁡zz(z−π) dz=1π(0−0)=0\int_\gamma \frac{\sin z}{z(z - \pi)}\, dz = \frac{1}{\pi}(0 - 0) = 0.

Problem. Evaluate ∫γe2z(z−1)2(z+1) dz\int_\gamma \frac{e^{2z}}{(z - 1)^2(z + 1)}\, dz where γ\gamma is ∣z∣=3|z| = 3.

By partial fractions: 1(z−1)2(z+1)=1/4z+1−1/4z−1+1/2(z−1)2\frac{1}{(z-1)^2(z+1)} = \frac{1/4}{z+1} - \frac{1/4}{z-1} + \frac{1/2}{(z-1)^2}.

∫γe2z(z−1)2(z+1) dz=14⋅2πi⋅e−2−14⋅2πi⋅e2+12⋅2πi1!⋅2e2\int_\gamma \frac{e^{2z}}{(z-1)^2(z+1)}\, dz = \frac{1}{4} \cdot 2\pi i \cdot e^{-2} - \frac{1}{4} \cdot 2\pi i \cdot e^2 + \frac{1}{2} \cdot \frac{2\pi i}{1!} \cdot 2e^2

=πie−22−πie22+2πie2=πie−22+3πie22= \frac{\pi i e^{-2}}{2} - \frac{\pi i e^2}{2} + 2\pi i e^2 = \frac{\pi i e^{-2}}{2} + \frac{3\pi i e^2}{2}.

Theorem 7.1. If ff is analytic on ∣z−z0∣<R|z - z_0| \lt RThen

f(z)=∑n=0∞f(n)(z0)n!(z−z0)nf(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z - z_0)^n

And the series converges uniformly on compact subsets of ∣z−z0∣<R|z - z_0| \lt R.

Proof. For ∣z−z0∣<r<R|z - z_0| \lt r \lt RApply Cauchy’s integral formula on ∣ζ−z0∣=r|\zeta - z_0| = r:

f(z)=12πi∫∣ζ−z0∣=rf(ζ)ζ−z dζf(z) = \frac{1}{2\pi i}\int_{|\zeta - z_0| = r} \frac{f(\zeta)}{\zeta - z}\, d\zeta

Write 1ζ−z=1(ζ−z0)−(z−z0)=1ζ−z0⋅11−(z−z0)/(ζ−z0)\frac{1}{\zeta - z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0} \cdot \frac{1}{1 - (z - z_0)/(\zeta - z_0)} =∑n=0∞(z−z0)n(ζ−z0)n+1= \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}} (geometric series, convergent since ∣z−z0∣/∣ζ−z0∣<1|z - z_0|/|\zeta - z_0| \lt 1).

Substituting and integrating term by term gives the Taylor series. ■\blacksquare

Remark. The radius of convergence RR is the distance from z0z_0 to the nearest singularity of ff.

ez=∑n=0∞znn!=1+z+z22!+⋯e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \cdots

sin⁡z=∑n=0∞(−1)nz2n+1(2n+1)!\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}

cos⁡z=∑n=0∞(−1)nz2n(2n)!\cos z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}

11−z=∑n=0∞zn,∣z∣<1\frac{1}{1 - z} = \sum_{n=0}^{\infty} z^n, \quad |z| \lt 1

ln⁡(1+z)=∑n=1∞(−1)n+1znn,∣z∣<1\ln(1 + z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} z^n}{n}, \quad |z| \lt 1

Solution

Problem. Find the Taylor series of f(z)=1zf(z) = \frac{1}{z} centered at z0=1z_0 = 1.

1z=11+(z−1)=∑n=0∞(−1)n(z−1)n\frac{1}{z} = \frac{1}{1 + (z - 1)} = \sum_{n=0}^{\infty} (-1)^n (z - 1)^n for ∣z−1∣<1|z - 1| \lt 1.

Radius of convergence: distance from z0=1z_0 = 1 to the singularity at z=0z = 0Which is 11.

Problem. Find the Taylor series of f(z)=1(1−z)2f(z) = \frac{1}{(1 - z)^2} centered at z0=0z_0 = 0.

1(1−z)2=ddz[11−z]=ddz∑n=0∞zn=∑n=1∞nzn−1=∑n=0∞(n+1)zn\frac{1}{(1-z)^2} = \frac{d}{dz}\left[\frac{1}{1 - z}\right] = \frac{d}{dz}\sum_{n=0}^{\infty} z^n = \sum_{n=1}^{\infty} nz^{n-1} = \sum_{n=0}^{\infty} (n+1)z^n for ∣z∣<1|z| \lt 1.

Problem. Find the Taylor series of f(z)=ezsin⁡zf(z) = e^z \sin z up to the z4z^4 term.

ez=1+z+z2/2+z3/6+z4/24+⋯e^z = 1 + z + z^2/2 + z^3/6 + z^4/24 + \cdots sin⁡z=z−z3/6+z5/120−⋯\sin z = z - z^3/6 + z^5/120 - \cdots

ezsin⁡z=(1+z+z2/2+z3/6+z4/24+⋯ )(z−z3/6+⋯ )e^z \sin z = (1 + z + z^2/2 + z^3/6 + z^4/24 + \cdots)(z - z^3/6 + \cdots)

=z+z2+z3/2+z4/6+⋯−z3/6−z4/6+⋯= z + z^2 + z^3/2 + z^4/6 + \cdots - z^3/6 - z^4/6 + \cdots =z+z2+z3/3−z4/30+⋯= z + z^2 + z^3/3 - z^4/30 + \cdots

Theorem 7.2 (Laurent Series). If ff is analytic on the annulus r<∣z−z0∣<Rr \lt |z - z_0| \lt R Then

f(z)=∑n=−∞∞an(z−z0)n=⋯+a−2(z−z0)2+a−1z−z0+a0+a1(z−z0)+⋯f(z) = \sum_{n=-\infty}^{\infty} a_n(z - z_0)^n = \cdots + \frac{a_{-2}}{(z - z_0)^2} + \frac{a_{-1}}{z - z_0} + a_0 + a_1(z - z_0) + \cdots

Where

an=12πi∫γf(z)(z−z0)n+1 dza_n = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz

For any simple closed contour γ\gamma in the annulus encircling z0z_0.

The principal part is ∑n=−∞−1an(z−z0)n\sum_{n=-\infty}^{-1} a_n(z - z_0)^n (negative powers). The analytic Part is ∑n=0∞an(z−z0)n\sum_{n=0}^{\infty} a_n(z - z_0)^n (non-negative powers).

The Laurent series expansion depends on the annulus of convergence. A function may have different Laurent expansions in different annuli.

Proposition 7.3. The Laurent series expansion of ff in an annulus is unique.

Solution

Problem. Find the Laurent series of f(z)=1z(z−1)f(z) = \frac{1}{z(z-1)} in 0<∣z∣<10 \lt |z| \lt 1.

Solution. Using partial fractions: 1z(z−1)=1z−1−1z\frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}.

In ∣z∣<1|z| \lt 1: 1z−1=−11−z=−∑n=0∞zn\frac{1}{z - 1} = -\frac{1}{1 - z} = -\sum_{n=0}^{\infty} z^n.

So f(z)=−∑n=0∞zn−1z=⋯−z2−z−1−1zf(z) = -\sum_{n=0}^{\infty} z^n - \frac{1}{z} = \cdots - z^2 - z - 1 - \frac{1}{z}.

The principal part is −1/z-1/zSo z=0z = 0 is a simple pole. ■\blacksquare

Problem. Find the Laurent series of f(z)=1z(z−1)f(z) = \frac{1}{z(z-1)} in 1<∣z∣<∞1 \lt |z| \lt \infty.

In ∣z∣>1|z| \gt 1: 1z−1=1z⋅11−1/z=∑n=2∞z−n\frac{1}{z - 1} = \frac{1}{z} \cdot \frac{1}{1 - 1/z} = \sum_{n=2}^{\infty} z^{-n}.

f(z)=∑n=2∞z−n−1z=1z2+1z3+⋯f(z) = \sum_{n=2}^{\infty} z^{-n} - \frac{1}{z} = \frac{1}{z^2} + \frac{1}{z^3} + \cdots

Problem. Find the Laurent series of f(z)=ezz2f(z) = \frac{e^z}{z^2} in 0<∣z∣<∞0 \lt |z| \lt \infty.

ez=∑n=0∞znn!e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}So f(z)=∑n=0∞zn−2n!=1z2+1z+12+z6+⋯f(z) = \sum_{n=0}^{\infty} \frac{z^{n-2}}{n!} = \frac{1}{z^2} + \frac{1}{z} + \frac{1}{2} + \frac{z}{6} + \cdots

Residue at z=0z = 0: a−1=1a_{-1} = 1.

Problem. Find the Laurent series of f(z)=1z2(z−3)f(z) = \frac{1}{z^2(z - 3)} in 0<∣z∣<30 \lt |z| \lt 3.

1z−3=−13∑n=0∞zn3n\frac{1}{z - 3} = -\frac{1}{3}\sum_{n=0}^{\infty} \frac{z^n}{3^n}.

f(z)=−∑n=0∞zn−23n+1=−13z2−19z−127−z81−⋯f(z) = -\sum_{n=0}^{\infty} \frac{z^{n-2}}{3^{n+1}} = -\frac{1}{3z^2} - \frac{1}{9z} - \frac{1}{27} - \frac{z}{81} - \cdots

Residue at z=0z = 0: a−1=−19a_{-1} = -\frac{1}{9}.

Definition. The residue at infinity of ff is defined as

Res(f,∞)=−12πi∫∣z∣=Rf(z) dz\mathrm{Res}(f, \infty) = -\frac{1}{2\pi i}\int_{|z|=R} f(z)\, dz

For sufficiently large RR (enclosing all finite singularities).

Proposition 7.4. For a function ff with finitely many singularities in C\mathbb{C}:

∑all finite zkRes(f,zk)+Res(f,∞)=0\sum_{\mathrm{all\ finite\ } z_k} \mathrm{Res}(f, z_k) + \mathrm{Res}(f, \infty) = 0

Proof. By the residue theorem applied to ∣z∣=R|z| = R enclosing all finite singularities:

∫∣z∣=Rf dz=2πi∑finiteRes(f,zk)\int_{|z|=R} f\, dz = 2\pi i \sum_{\mathrm{finite} \mathrm{Res}(f, z_k)}.

But Res(f,∞)=−12πi∫∣z∣=Rf dz\mathrm{Res}(f, \infty) = -\frac{1}{2\pi i}\int_{|z|=R} f\, dzSo the sum is zero. ■\blacksquare

Let z0z_0 be an isolated singularity of ff (i.e., ff is analytic in a punctured neighbourhood of z0z_0).

Classification by Laurent series:

  1. Removable singularity: an=0a_n = 0 for all n<0n \lt 0. Can be removed by redefining f(z0)=a0f(z_0) = a_0.
  2. Pole of order mm: a−m≠0a_{-m} \neq 0 and an=0a_n = 0 for n<−mn \lt -m. The principal part is finite.
  3. Essential singularity: infinitely many non-zero ana_n with n<0n \lt 0.

Proposition 8.1 (Riemann’s Removable Singularity Theorem). If ff is bounded near z0z_0Then z0z_0 is a removable singularity.

Proposition 8.2. z0z_0 is a pole of order mm if and only if lim⁡z→z0(z−z0)mf(z)\lim_{z \to z_0} (z - z_0)^m f(z) Exists and is non-zero.

Theorem 8.3 (Casorati-Weierstrass). If z0z_0 is an essential singularity of ffThen ff takes Values arbitrarily close to any complex number in every neighbourhood of z0z_0.

Solution

Problem. Classify the singularities of f(z)=sin⁡zzf(z) = \frac{\sin z}{z}.

z=0z = 0: sin⁡z=z−z3/6+⋯\sin z = z - z^3/6 + \cdotsSo f(z)=1−z2/6+⋯f(z) = 1 - z^2/6 + \cdots. No negative powers, so z=0z = 0 is a removable singularity. f(0)=1f(0) = 1 by continuity.

Problem. Classify the singularities of f(z)=ez−1z2f(z) = \frac{e^z - 1}{z^2}.

z=0z = 0: ez−1=z+z2/2+⋯e^z - 1 = z + z^2/2 + \cdotsSo f(z)=1z+12+⋯f(z) = \frac{1}{z} + \frac{1}{2} + \cdots. Principal part is 1/z1/zSo z=0z = 0 is a simple pole with residue 11.

Problem. Classify the singularity of f(z)=e1/zf(z) = e^{1/z} at z=0z = 0.

e1/z=∑n=0∞1n! zn=1+1z+12z2+⋯e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n!\, z^n} = 1 + \frac{1}{z} + \frac{1}{2z^2} + \cdots

Infinitely many negative powers ⇒\Rightarrow z=0z = 0 is an essential singularity.

Problem. Classify the singularities of f(z)=z+1z3(z2+1)f(z) = \frac{z + 1}{z^3(z^2 + 1)}.

z=0z = 0: pole of order 33. z=iz = i: simple pole. z=−iz = -i: simple pole.

Problem. Determine the type of singularity of f(z)=zsin⁡zf(z) = \frac{z}{\sin z} at z=0z = 0.

sin⁡z=z−z3/6+⋯\sin z = z - z^3/6 + \cdotsSo f(z)=11−z2/6+⋯=1+z26+⋯f(z) = \frac{1}{1 - z^2/6 + \cdots} = 1 + \frac{z^2}{6} + \cdots.

No negative powers, so z=0z = 0 is a removable singularity with f(0)=1f(0) = 1.

Definition. The residue of ff at an isolated singularity z0z_0 is the coefficient a−1a_{-1} In the Laurent expansion:

Res(f,z0)=a−1=12πi∫γf(z) dz\mathrm{Res}(f, z_0) = a_{-1} = \frac{1}{2\pi i}\int_\gamma f(z)\, dz

Where γ\gamma is a small positively oriented circle around z0z_0.

For a simple pole at z0z_0:

Res(f,z0)=lim⁡z→z0(z−z0)f(z)\mathrm{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)

If f=g/hf = g/h where g(z0)≠0g(z_0) \neq 0, h(z0)=0h(z_0) = 0, h′(z0)≠0h'(z_0) \neq 0:

Res(f,z0)=g(z0)h′(z0)\mathrm{Res}(f, z_0) = \frac{g(z_0)}{h'(z_0)}

For a pole of order mm at z0z_0:

Res(f,z0)=1(m−1)!lim⁡z→z0dm−1dzm−1[(z−z0)mf(z)]\mathrm{Res}(f, z_0) = \frac{1}{(m-1)!}\lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}}\left[(z - z_0)^m f(z)\right]

Solution

Problem. Find the residue of f(z)=zz2+4z+3f(z) = \frac{z}{z^2 + 4z + 3} at each pole.

z2+4z+3=(z+1)(z+3)z^2 + 4z + 3 = (z + 1)(z + 3)So simple poles at z=−1z = -1 and z=−3z = -3.

At z=−1z = -1: Res=lim⁡z→−1zz+3=−12\mathrm{Res} = \lim_{z \to -1} \frac{z}{z + 3} = \frac{-1}{2}. At z=−3z = -3: Res=lim⁡z→−3zz+1=−3−2=32\mathrm{Res} = \lim_{z \to -3} \frac{z}{z + 1} = \frac{-3}{-2} = \frac{3}{2}.

Problem. Find the residue of f(z)=ez(z−1)2(z−2)f(z) = \frac{e^z}{(z - 1)^2(z - 2)} at each pole.

At z=1z = 1 (pole of order 22): Res=ddz[ezz−2]z=1=ez(z−2)−ez(z−2)2∣z=1=−e−e1=−2e\mathrm{Res} = \frac{d}{dz}\left[\frac{e^z}{z - 2}\right]_{z=1} = \frac{e^z(z - 2) - e^z}{(z-2)^2}\Big|_{z=1} = \frac{-e - e}{1} = -2e.

At z=2z = 2 (simple pole): Res=e2(2−1)2=e2\mathrm{Res} = \frac{e^2}{(2-1)^2} = e^2.

Theorem 8.4 (Residue Theorem). If ff is analytic inside and on a simple closed positively Oriented contour γ\gamma except for isolated singularities z1,z2,…,znz_1, z_2, \ldots, z_n inside γ\gamma Then

∫γf(z) dz=2πi∑k=1nRes(f,zk)\int_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^{n} \mathrm{Res}(f, z_k)

Proof. For each singularity zkz_kDraw a small circle γk\gamma_k around it. By Cauchy’s theorem Applied to the multiply connected region between γ\gamma and the γk\gamma_k:

∫γf dz=∑k=1n∫γkf dz=∑k=1n2πi⋅Res(f,zk)\int_\gamma f\, dz = \sum_{k=1}^n \int_{\gamma_k} f\, dz = \sum_{k=1}^n 2\pi i \cdot \mathrm{Res}(f, z_k). ■\blacksquare

Solution

Problem 1. Evaluate ∫γezz(z−1)2 dz\int_\gamma \frac{e^z}{z(z-1)^2}\, dz where γ\gamma is ∣z∣=2|z| = 2.

Solution. Singularities inside γ\gamma: z=0z = 0 (simple pole) and z=1z = 1 (pole of order 22).

At z=0z = 0: Res=lim⁡z→0ez(z−1)2=1(−1)2=1\mathrm{Res} = \lim_{z \to 0} \frac{e^z}{(z-1)^2} = \frac{1}{(-1)^2} = 1.

At z=1z = 1: Res(f,1)=ddz[(z−1)2⋅ezz(z−1)2]z=1=ddz[ezz]z=1=ez⋅z−ezz2∣z=1=e−e1=0\mathrm{Res}(f, 1) = \frac{d}{dz}\left[(z-1)^2 \cdot \frac{e^z}{z(z-1)^2}\right]_{z=1} = \frac{d}{dz}\left[\frac{e^z}{z}\right]_{z=1} = \frac{e^z \cdot z - e^z}{z^2}\Big|_{z=1} = \frac{e - e}{1} = 0.

∫γf dz=2πi(1+0)=2πi\int_\gamma f\, dz = 2\pi i(1 + 0) = 2\pi i. ■\blacksquare

Problem 2. Evaluate ∫γ1z4+1 dz\int_\gamma \frac{1}{z^4 + 1}\, dz where γ\gamma is ∣z∣=2|z| = 2.

Solution. The poles are the fourth roots of −1-1: zk=eiπ/4+ikπ/2z_k = e^{i\pi/4 + ik\pi/2} for k=0,1,2,3k = 0, 1, 2, 3. All four lie inside ∣z∣=2|z| = 2.

Each is a simple pole with Res(f,zk)=14zk3\mathrm{Res}(f, z_k) = \frac{1}{4z_k^3}. Since zk4=−1z_k^4 = -1: zk−3=−zkz_k^{-3} = -z_kSo the sum equals −14∑zk=0-\frac{1}{4}\sum z_k = 0.

∫γdzz4+1=2πi⋅0=0\int_\gamma \frac{dz}{z^4 + 1} = 2\pi i \cdot 0 = 0. ■\blacksquare

Contour integration is a powerful tool for evaluating definite integrals.

9.2 Integrals of Rational Functions over the Real Line

Section titled “9.2 Integrals of Rational Functions over the Real Line”

Theorem 9.1. If f(x)=P(x)/Q(x)f(x) = P(x)/Q(x) where deg⁡(Q)≥deg⁡(P)+2\deg(Q) \geq \deg(P) + 2 and QQ has no real roots, Then

∫−∞∞f(x) dx=2πi∑Im(zk)>0Res(f,zk)\int_{-\infty}^{\infty} f(x)\, dx = 2\pi i \sum_{\mathrm{Im}(z_k) > 0} \mathrm{Res}(f, z_k)

Where the sum is over poles in the upper half-plane.

Proof. Integrate f(z)f(z) over the semicircular contour γR\gamma_R consisting of [−R,R][-R, R] on the Real axis and the semicircle ∣z∣=R|z| = R in the upper half-plane. As R→∞R \to \inftyThe integral over The semicircle vanishes (since ∣f(z)∣≤M/R2|f(z)| \leq M/R^2 and the length is πR\pi R). ■\blacksquare

Problem. Evaluate ∫−∞∞dxx2+1\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1}.

Solution. f(z)=1z2+1f(z) = \frac{1}{z^2 + 1} has simple poles at z=±iz = \pm i.

Only z=iz = i is in the upper half-plane.

Res(1z2+1,i)=12z∣z=i=12i\mathrm{Res}\left(\frac{1}{z^2 + 1}, i\right) = \frac{1}{2z}\Big|_{z = i} = \frac{1}{2i}.

∫−∞∞dxx2+1=2πi⋅12i=π\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1} = 2\pi i \cdot \frac{1}{2i} = \pi. ■\blacksquare

9.4 Integrals Involving Trigonometric Functions

Section titled “9.4 Integrals Involving Trigonometric Functions”

For integrals of the form ∫02πR(cos⁡θ,sin⁡θ) dθ\int_0^{2\pi} R(\cos\theta, \sin\theta)\, d\thetaSubstitute z=eiθz = e^{i\theta}So dz=iz dθdz = iz\, d\theta, cos⁡θ=z+z−12\cos\theta = \frac{z + z^{-1}}{2} sin⁡θ=z−z−12i\sin\theta = \frac{z - z^{-1}}{2i}.

The integral becomes ∫∣z∣=1f(z) dz\int_{|z|=1} f(z)\, dz where f(z)f(z) is a rational function.

Problem. Evaluate ∫02πdθ2+cos⁡θ\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta}.

Solution. Substitute z=eiθz = e^{i\theta}: dθ=dzizd\theta = \frac{dz}{iz} cos⁡θ=z+1/z2\cos\theta = \frac{z + 1/z}{2}.

∫∣z∣=1dziz(2+z+1/z2)=∫∣z∣=12 dzi(z2+4z+1)\int_{|z|=1} \frac{dz}{iz\left(2 + \frac{z + 1/z}{2}\right)} = \int_{|z|=1} \frac{2\, dz}{i(z^2 + 4z + 1)}

Poles: z2+4z+1=0⇒z=−2±3z^2 + 4z + 1 = 0 \Rightarrow z = -2 \pm \sqrt{3}.

∣z1∣=∣−2+3∣=2−3<1|z_1| = |-2 + \sqrt{3}| = 2 - \sqrt{3} \lt 1 (inside). ∣z2∣=∣−2−3∣=2+3>1|z_2| = |-2 - \sqrt{3}| = 2 + \sqrt{3} \gt 1 (outside).

Res(1z2+4z+1,z1)=123\mathrm{Res}\left(\frac{1}{z^2 + 4z + 1}, z_1\right) = \frac{1}{2\sqrt{3}}.

∫02πdθ2+cos⁡θ=2i⋅2πi⋅123=2π3\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta} = \frac{2}{i} \cdot 2\pi i \cdot \frac{1}{2\sqrt{3}} = \frac{2\pi}{\sqrt{3}}. ■\blacksquare

Theorem 9.2 (Jordan’s Lemma). If f(z)→0f(z) \to 0 uniformly as ∣z∣→∞|z| \to \infty in the upper Half-plane and a>0a \gt 0Then

lim⁡R→∞∫CReiazf(z) dz=0\lim_{R \to \infty} \int_{C_R} e^{iaz}f(z)\, dz = 0

Where CRC_R is the upper semicircle ∣z∣=R|z| = R, Im(z)≥0\mathrm{Im}(z) \geq 0.

This allows evaluation of integrals of the form ∫−∞∞f(x)cos⁡(ax) dx\int_{-\infty}^{\infty} f(x)\cos(ax)\, dx and ∫−∞∞f(x)sin⁡(ax) dx\int_{-\infty}^{\infty} f(x)\sin(ax)\, dx.

Solution

Problem. Evaluate ∫−∞∞cos⁡xx2+1 dx\int_{-\infty}^{\infty} \frac{\cos x}{x^2 + 1}\, dx.

Consider ∫−∞∞eixx2+1 dx=2πi⋅Res ⁣(eizz2+1,i)\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2 + 1}\, dx = 2\pi i \cdot \mathrm{Res}\!\left(\frac{e^{iz}}{z^2+1}, i\right).

Res ⁣(eizz2+1,i)=ei⋅i2i=e−12i\mathrm{Res}\!\left(\frac{e^{iz}}{z^2+1}, i\right) = \frac{e^{i \cdot i}}{2i} = \frac{e^{-1}}{2i}.

∫−∞∞eixx2+1 dx=2πi⋅e−12i=πe\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2 + 1}\, dx = 2\pi i \cdot \frac{e^{-1}}{2i} = \frac{\pi}{e}.

Taking real parts: ∫−∞∞cos⁡xx2+1 dx=πe\int_{-\infty}^{\infty} \frac{\cos x}{x^2 + 1}\, dx = \frac{\pi}{e}.

Problem. Evaluate ∫−∞∞xsin⁡xx2+a2 dx\int_{-\infty}^{\infty} \frac{x \sin x}{x^2 + a^2}\, dx for a>0a \gt 0.

Consider ∫−∞∞z eizz2+a2 dz\int_{-\infty}^{\infty} \frac{z\, e^{iz}}{z^2 + a^2}\, dz. Only z=iaz = ia is in the upper half-plane.

Res ⁣(zeizz2+a2,ia)=ia⋅ei⋅ia2ia=e−a2\mathrm{Res}\!\left(\frac{ze^{iz}}{z^2 + a^2}, ia\right) = \frac{ia \cdot e^{i \cdot ia}}{2ia} = \frac{e^{-a}}{2}.

∫−∞∞x eixx2+a2 dx=2πi⋅e−a2=πi e−a\int_{-\infty}^{\infty} \frac{x\, e^{ix}}{x^2 + a^2}\, dx = 2\pi i \cdot \frac{e^{-a}}{2} = \pi i\, e^{-a}.

Taking imaginary parts: ∫−∞∞xsin⁡xx2+a2 dx=π e−a\int_{-\infty}^{\infty} \frac{x \sin x}{x^2 + a^2}\, dx = \pi\, e^{-a}.

Problem. Evaluate ∫02πcos⁡2θ5+4cos⁡θ dθ\int_0^{2\pi} \frac{\cos 2\theta}{5 + 4\cos\theta}\, d\theta.

Substitute z=eiθz = e^{i\theta}: cos⁡θ=(z+z−1)/2\cos\theta = (z + z^{-1})/2, cos⁡2θ=(z2+z−2)/2\cos 2\theta = (z^2 + z^{-2})/2.

I=12i∫∣z∣=1z4+1z2(2z+1)(z+2) dzI = \frac{1}{2i}\int_{|z|=1} \frac{z^4 + 1}{z^2(2z + 1)(z + 2)}\, dz.

Poles inside ∣z∣=1|z| = 1: z=0z = 0 (order 22) and z=−1/2z = -1/2 (simple).

At z=0z = 0: Res=ddz[z4+1(2z+1)(z+2)]z=0=−54\mathrm{Res} = \frac{d}{dz}\left[\frac{z^4 + 1}{(2z+1)(z+2)}\right]_{z=0} = -\frac{5}{4}.

At z=−1/2z = -1/2: Res=17/163/4=1712\mathrm{Res} = \frac{17/16}{3/4} = \frac{17}{12}.

I=12i⋅2πi(−54+1712)=π6I = \frac{1}{2i} \cdot 2\pi i \left(-\frac{5}{4} + \frac{17}{12}\right) = \frac{\pi}{6}.

9.8 Improper Integrals and Principal Value

Section titled “9.8 Improper Integrals and Principal Value”

For integrals where the integrand has poles on the real axis, we use the Cauchy principal value:

PV ⁣∫−∞∞f(x) dx=lim⁡ε→0+(∫−∞a−εf(x) dx+∫a+ε∞f(x) dx)\mathrm{PV}\!\int_{-\infty}^{\infty} f(x)\, dx = \lim_{\varepsilon \to 0^+} \left(\int_{-\infty}^{a-\varepsilon} f(x)\, dx + \int_{a+\varepsilon}^{\infty} f(x)\, dx\right)

Solution

Problem. Evaluate PV ⁣∫−∞∞sin⁡xx dx\mathrm{PV}\!\int_{-\infty}^{\infty} \frac{\sin x}{x}\, dx.

Consider ∮γeizz dz\oint_\gamma \frac{e^{iz}}{z}\, dz where γ\gamma consists of [−R,−ε][-R, -\varepsilon] [ε,R][\varepsilon, R] on the real axis, small upper semicircle CεC_\varepsilon around 00And large Upper semicircle CRC_R.

No poles inside the contour, so the integral is 00.

On CRC_R: vanishes as R→∞R \to \infty by Jordan’s lemma. On CεC_\varepsilon (indenting above): ∫Cεeizz dz→−iπ\int_{C_\varepsilon} \frac{e^{iz}}{z}\, dz \to -i\pi as ε→0\varepsilon \to 0 (half residue contribution).

0=PV ⁣∫−∞∞eixx dx+(−iπ)0 = \mathrm{PV}\!\int_{-\infty}^{\infty} \frac{e^{ix}}{x}\, dx + (-i\pi).

PV ⁣∫−∞∞eixx dx=iπ\mathrm{PV}\!\int_{-\infty}^{\infty} \frac{e^{ix}}{x}\, dx = i\pi.

Taking imaginary parts: PV ⁣∫−∞∞sin⁡xx dx=π\mathrm{PV}\!\int_{-\infty}^{\infty} \frac{\sin x}{x}\, dx = \pi.

Definition. An analytic function ff is conformal at z0z_0 if f′(z0)≠0f'(z_0) \neq 0. A conformal Mapping preserves angles (both magnitude and orientation) between curves.

If f′(z0)=reiθf'(z_0) = re^{i\theta}Then near z0z_0 the mapping ff acts as a rotation by θ\theta followed By a scaling by rr. The Jacobian determinant is ∣f′(z0)∣2>0|f'(z_0)|^2 \gt 0So orientation is preserved.

MappingEffect
w=az+bw = az + b (a≠0a \neq 0)Translation, rotation, scaling
w=1/zw = 1/zInversion in the unit circle
w=z2w = z^2Squaring (doubles angles)
w=ezw = e^zExponential (maps strips to sectors)
w=z−a1−aˉzw = \frac{z - a}{1 - \bar{a}z}Möbius (maps disk to disk)

A Möbius transformation (or linear fractional transformation) is

T(z)=az+bcz+d,ad−bc≠0T(z) = \frac{az + b}{cz + d}, \quad ad - bc \neq 0

Proposition 10.1. Möbius transformations are conformal (where defined) and map circles and lines To circles and lines.

Proposition 10.2. Three points determine a unique Möbius transformation: T(z1)=w1T(z_1) = w_1 T(z2)=w2T(z_2) = w_2, T(z3)=w3T(z_3) = w_3.

Definition. The cross-ratio of four distinct points z1,z2,z3,z4z_1, z_2, z_3, z_4 is

(z1,z2,z3,z4)=(z1−z3)(z2−z4)(z1−z4)(z2−z3)(z_1, z_2, z_3, z_4) = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)}

Proposition 10.3. The cross-ratio is invariant under Möbius transformations: (Tz1,Tz2,Tz3,Tz4)=(z1,z2,z3,z4)(Tz_1, Tz_2, Tz_3, Tz_4) = (z_1, z_2, z_3, z_4).

Proposition 10.4. The unique Möbius transformation sending z1↦0z_1 \mapsto 0, z2↦1z_2 \mapsto 1 z3↦∞z_3 \mapsto \infty is

T(z)=(z−z1)(z2−z3)(z−z3)(z2−z1)T(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}

10.6 Classification of Möbius Transformations

Section titled “10.6 Classification of Möbius Transformations”

A Möbius transformation T(z)=az+bcz+dT(z) = \frac{az + b}{cz + d} is classified by its fixed points (solutions of T(z)=zT(z) = z).

  1. Parabolic: Exactly one fixed point. Conjugate to w=z+kw = z + k.
  2. Elliptic: Two fixed points, ∣T′(z0)∣=1|T'(z_0)| = 1. Conjugate to a rotation w=eiθzw = e^{i\theta} z.
  3. Hyperbolic: Two fixed points, T′(z0)∈R+T'(z_0) \in \mathbb{R}^+, T′(z0)≠1T'(z_0) \neq 1. Conjugate to w=kzw = kz.
  4. Loxodromic: Two fixed points, T′(z0)∉R∪{z:∣z∣=1}T'(z_0) \notin \mathbb{R} \cup \{z : |z| = 1\}. Conjugate to w=keiθzw = ke^{i\theta}z.
Solution

Problem. Find the Möbius transformation mapping 0↦i0 \mapsto i, 1↦01 \mapsto 0, ∞↦−i\infty \mapsto -i.

T(z)=az+bcz+dT(z) = \frac{az + b}{cz + d} with T(0)=i⇒b/d=i⇒b=idT(0) = i \Rightarrow b/d = i \Rightarrow b = id. T(1)=0⇒a=−b=−idT(1) = 0 \Rightarrow a = -b = -id. T(∞)=−i⇒a/c=−i⇒c=dT(\infty) = -i \Rightarrow a/c = -i \Rightarrow c = d.

T(z)=−idz+iddz+d=i(1−z)z+1T(z) = \frac{-idz + id}{dz + d} = \frac{i(1 - z)}{z + 1}.

Problem. Show that T(z)=z−1z+1T(z) = \frac{z - 1}{z + 1} maps the right half-plane to the unit disk.

If Re(z)>0\mathrm{Re}(z) \gt 0Then ∣z−1∣<∣z+1∣|z - 1| \lt |z + 1|So ∣T(z)∣<1|T(z)| \lt 1.

Check boundary: T(i)=i−1i+1=(i−1)(−i+1)(i+1)(−i+1)=22=1T(i) = \frac{i - 1}{i + 1} = \frac{(i-1)(-i+1)}{(i+1)(-i+1)} = \frac{2}{2} = 1. ∣T(i)∣=1|T(i)| = 1. ✓\checkmark

Problem. Classify T(z)=2z+1z+2T(z) = \frac{2z + 1}{z + 2}.

Fixed points: z=2z+1z+2⇒z2=1⇒z=±1z = \frac{2z + 1}{z + 2} \Rightarrow z^2 = 1 \Rightarrow z = \pm 1.

T′(z)=3(z+2)2T'(z) = \frac{3}{(z + 2)^2}. T′(1)=1/3T'(1) = 1/3, T′(−1)=3T'(-1) = 3.

Both multipliers are real and positive (not equal to 11), so TT is hyperbolic.

Theorem 10.5 (Riemann Mapping Theorem). Let UU be a connected open proper subset of C\mathbb{C}. Then there exists a bijective conformal map from UU onto the unit disk D={z:∣z∣<1}\mathbb{D} = \{z : |z| \lt 1\}.

This is one of the most profound results in complex analysis, establishing that all connected Domains (other than C\mathbb{C} itself) are conformally equivalent.

Remark. The Riemann mapping theorem is an existence theorem; it does not provide an explicit Formula for the conformal map .

11. Liouville’s Theorem and the Maximum Modulus Principle

Section titled “11. Liouville’s Theorem and the Maximum Modulus Principle”

Theorem 11.1 (Liouville’s Theorem). Every bounded entire function is constant.

Proof. If ∣f(z)∣≤M|f(z)| \leq M for all zzThen by Cauchy’s estimates with RR arbitrarily large: ∣f′(z0)∣≤MR→0|f'(z_0)| \leq \frac{M}{R} \to 0 as R→∞R \to \infty. So f′(z)=0f'(z) = 0 for all zzMeaning ff is Constant. ■\blacksquare

Theorem 11.2 (Fundamental Theorem of Algebra). Every non-constant polynomial p(z)∈C[z]p(z) \in \mathbb{C}[z] has a root in C\mathbb{C}.

Proof. Suppose p(z)p(z) has no root. Then f(z)=1/p(z)f(z) = 1/p(z) is entire. Since ∣p(z)∣→∞|p(z)| \to \infty as ∣z∣→∞|z| \to \infty, f(z)→0f(z) \to 0So ff is bounded. By Liouville’s theorem, ff is constant, so pp Is constant, a contradiction. ■\blacksquare

Theorem 11.3 (Maximum Modulus Principle). If ff is analytic and non-constant on a domain DD Then ∣f∣|f| has no local maximum in DD.

Corollary 11.4. If ff is analytic on a bounded domain DD and continuous on Dˉ=D∪∂D\bar{D} = D \cup \partial DThen ∣f∣|f| attains its maximum on ∂D\partial D.

Theorem 11.5 (Minimum Modulus Principle). If ff is analytic and non-zero on a bounded domain DD And continuous on Dˉ\bar{D}Then ∣f∣|f| attains its minimum on ∂D\partial D.

Remark. If ff has zeros in DDThen ∣f∣|f| attains its minimum of 00 at those zeros. The minimum modulus principle requires the non-vanishing hypothesis.

Theorem 11.6 (Schwarz Lemma). If f:D→Df : \mathbb{D} \to \mathbb{D} is analytic with f(0)=0f(0) = 0 Then

∣f(z)∣≤∣z∣for all z∈D|f(z)| \leq |z| \quad \mathrm{for\ all\ } z \in \mathbb{D}

And ∣f′(0)∣≤1|f'(0)| \leq 1. Equality in either case implies f(z)=eiθzf(z) = e^{i\theta} z for some real θ\theta.

Proof. Define g(z)=f(z)/zg(z) = f(z)/z for z≠0z \neq 0 and g(0)=f′(0)g(0) = f'(0). Then gg is analytic on D\mathbb{D}. For ∣z∣=r<1|z| = r \lt 1: ∣g(z)∣=∣f(z)∣/∣z∣≤1/r|g(z)| = |f(z)|/|z| \leq 1/r. By the maximum modulus Principle, ∣g(z)∣≤1/r|g(z)| \leq 1/r for ∣z∣≤r|z| \leq r. Letting r→1r \to 1: ∣g(z)∣≤1|g(z)| \leq 1So ∣f(z)∣≤∣z∣|f(z)| \leq |z|. Also ∣f′(0)∣=∣g(0)∣≤1|f'(0)| = |g(0)| \leq 1. If ∣f′(0)∣=1|f'(0)| = 1Then ∣g∣|g| attains its maximum At an interior point, so gg is constant: g(z)=eiθg(z) = e^{i\theta}. ■\blacksquare

12. Argument Principle and Rouché’s Theorem

Section titled “12. Argument Principle and Rouché’s Theorem”

Theorem 12.1 (Argument Principle). If ff is meromorphic inside and on a simple closed contour γ\gamma with no zeros or poles on γ\gammaThen

12πi∫γf′(z)f(z) dz=N−P\frac{1}{2\pi i}\int_\gamma \frac{f'(z)}{f(z)}\, dz = N - P

Where NN is the number of zeros and PP is the number of poles of ff inside γ\gamma (counting Multiplicities).

Theorem 12.2 (Rouché’s Theorem). If ff and gg are analytic inside and on a simple closed Contour γ\gammaAnd ∣f(z)∣>∣g(z)∣|f(z)| \gt |g(z)| on γ\gammaThen ff and f+gf + g have the same number of Zeros inside γ\gamma.

Proof. On γ\gamma: ∣g(z)/f(z)∣<1|g(z)/f(z)| \lt 1. The function h(z)=1+g(z)/f(z)h(z) = 1 + g(z)/f(z) satisfies ∣h(z)−1∣<1|h(z) - 1| \lt 1 on γ\gammaSo h(γ)h(\gamma) does not wind around 00. By the argument principle Applied to hh: 0=Nh−Ph0 = N_h - P_hMeaning hh has the same number of zeros and poles inside γ\gamma. But h=(f+g)/fh = (f + g)/fSo zeros of hh are zeros of f+gf + g and poles of hh are zeros of ff. Therefore ff and f+gf + g have the same number of zeros. ■\blacksquare

Problem. Show that z4+6z+3z^4 + 6z + 3 has exactly one root in ∣z∣<1|z| \lt 1.

Solution. On ∣z∣=1|z| = 1: ∣6z∣=6>∣z4+3∣≤∣z∣4+3=4|6z| = 6 \gt |z^4 + 3| \leq |z|^4 + 3 = 4. By Rouché’s theorem with f(z)=6zf(z) = 6z and g(z)=z4+3g(z) = z^4 + 3: f+g=z4+6z+3f + g = z^4 + 6z + 3 has the same number of zeros in ∣z∣<1|z| \lt 1 as f(z)=6zf(z) = 6zWhich has exactly one zero (at z=0z = 0). ■\blacksquare

Solution

Problem. Show that all roots of z4+z+1=0z^4 + z + 1 = 0 satisfy ∣z∣<2|z| \lt 2.

On ∣z∣=2|z| = 2: ∣z4∣=16>∣z+1∣≤3|z^4| = 16 \gt |z + 1| \leq 3. By Rouché with f(z)=z4f(z) = z^4 and g(z)=z+1g(z) = z + 1: z4+z+1z^4 + z + 1 has 44 zeros in ∣z∣<2|z| \lt 2 (same as z4z^4).

Problem. Show that z5+3z2+1z^5 + 3z^2 + 1 has exactly two roots in ∣z∣<1|z| \lt 1.

On ∣z∣=1|z| = 1: ∣3z2+1∣≥∣3z2∣−∣1∣=2>∣z5∣=1|3z^2 + 1| \geq |3z^2| - |1| = 2 \gt |z^5| = 1. By Rouché with f(z)=3z2+1f(z) = 3z^2 + 1 and g(z)=z5g(z) = z^5: z5+3z2+1z^5 + 3z^2 + 1 has the same number of zeros as 3z2+13z^2 + 1 in ∣z∣<1|z| \lt 1. 3z2+1=0⇒z=±i/33z^2 + 1 = 0 \Rightarrow z = \pm i/\sqrt{3}Both in ∣z∣<1|z| \lt 1. So 22 zeros.

Definition. If f1f_1 is analytic on D1D_1 and f2f_2 is analytic on D2D_2 with D1∩D2≠∅D_1 \cap D_2 \neq \emptyset and f1=f2f_1 = f_2 on D1∩D2D_1 \cap D_2Then f2f_2 is an analytic Continuation of f1f_1.

Theorem 13.1 (Identity Theorem). If ff and gg are analytic on a domain DD and agree on a set With a limit point in DDThen f=gf = g on all of DD.

Proof. Let E={z∈D:f(n)(z)=g(n)(z) for all n≥0}E = \{z \in D : f^{(n)}(z) = g^{(n)}(z) \mathrm{\ for\ all\ } n \geq 0\}. EE is Non-empty (it contains the limit point by continuity of derivatives). EE is closed (by continuity). If z0∈Ez_0 \in EThe Taylor series of ff and gg at z0z_0 coincide, so f=gf = g in a neighbourhood of z0z_0Giving EE open. Since DD is connected, E=DE = D. ■\blacksquare

:::caution Common Pitfall The Cauchy-Riemann equations are necessary but not sufficient for Differentiability. The partial derivatives must also be continuous. For example, f(z)=exp⁡(−1/z4)f(z) = \exp(-1/z^4) extended by f(0)=0f(0) = 0 satisfies the Cauchy-Riemann equations at the origin But is not differentiable there. :::

:::caution Common Pitfall Cauchy’s theorem requires a connected domain. On a multiply Connected domain, the integral of an analytic function around a closed contour may be non-zero. The Classic example is ∮∣z∣=1dz/z=2πi\oint_{|z|=1} dz/z = 2\pi i. :::

:::caution Common Pitfall When computing residues at poles of order m≥2m \geq 2The formula involves Differentiation. A common error is forgetting the (m−1)!(m-1)! in the denominator or differentiating (z−z0)mf(z)(z - z_0)^m f(z) the wrong number of times. :::

:::caution Common Pitfall The residue at infinity is Res(f,∞)=−Res(1/z2⋅f(1/z),0)\mathrm{Res}(f, \infty) = -\mathrm{Res}(1/z^2 \cdot f(1/z), 0). It is NOT f(∞)f(\infty). For A function that is analytic everywhere in the finite plane except for finitely many singularities, The sum of all residues (including the residue at infinity) is zero. :::

:::caution Common Pitfall A conformal mapping preserves angles but not necessarily distances. The Mapping w=z2w = z^2 is conformal at every z≠0z \neq 0But it doubles the angle between curves at each Point. At z=0z = 0It is not conformal because f′(0)=0f'(0) = 0. :::

:::caution Common Pitfall The maximum modulus principle says that ∣f∣|f| has no local maximum in the Interior, but the minimum can occur in the interior (e.g., f(z)=zf(z) = z on the unit disk has minimum ∣f∣=0|f| = 0 at z=0z = 0). For the minimum principle, one needs the additional hypothesis that ff has No zeros in the domain. :::

:::caution Common Pitfall The complex logarithm is multi-valued. When a problem asks for “logarithm” without specifying a branch, you must either compute all values or explicitly state which Branch you are using. The principal branch Log z\mathrm{Log}\, z has a branch cut along (−∞,0](-\infty, 0] And is undefined on this cut. :::

:::caution Common Pitfall When applying the ML inequality, make sure MM is a valid upper bound for ∣f(z)∣|f(z)| on the entire contour. A common error is bounding ∣f∣|f| on only part of the contour. Also, LL must be the arc length of the contour, not a diameter or radius. :::

Express z=−3+iz = -\sqrt{3} + i in polar form and find all values of z1/3z^{1/3}.

Solution

∣z∣=3+1=2|z| = \sqrt{3 + 1} = 2. Since Re(z)<0\mathrm{Re}(z) \lt 0 and Im(z)>0\mathrm{Im}(z) \gt 0: arg⁡(z)=π−π/6=5π/6\arg(z) = \pi - \pi/6 = 5\pi/6.

z=2 e5πi/6z = 2\,e^{5\pi i/6}.

z1/3=21/3 e(5π/6+2πk)/3z^{1/3} = 2^{1/3}\, e^{(5\pi/6 + 2\pi k)/3} for k=0,1,2k = 0, 1, 2.

z0=21/3 e5πi/18z_0 = 2^{1/3}\, e^{5\pi i/18}, z1=21/3 e17πi/18z_1 = 2^{1/3}\, e^{17\pi i/18}, z2=21/3 e29πi/18z_2 = 2^{1/3}\, e^{29\pi i/18}.

If you get this wrong, revise: Section 1.5 (Roots of Complex Numbers).

Let f(z)=z2+zˉ2f(z) = z^2 + \bar{z}^2. Find where ff is differentiable and where it is analytic.

Solution

f(z)=(x+iy)2+(x−iy)2=2(x2−y2)f(z) = (x + iy)^2 + (x - iy)^2 = 2(x^2 - y^2). So u=2(x2−y2)u = 2(x^2 - y^2), v=0v = 0.

ux=4xu_x = 4x, uy=−4yu_y = -4y, vx=0v_x = 0, vy=0v_y = 0.

CR: 4x=0⇒x=04x = 0 \Rightarrow x = 0, −4y=0⇒y=0-4y = 0 \Rightarrow y = 0.

ff is differentiable only at z=0z = 0 and analytic nowhere.

f′(0)=0f'(0) = 0 (verified by direct computation).

If you get this wrong, revise: Sections 2.4 and 3.1 (Analyticity and Cauchy-Riemann).

Verify that f(z)=1z2+1f(z) = \frac{1}{z^2 + 1} satisfies the Cauchy-Riemann equations on its domain and Find f′(z)f'(z).

Solution

f(z)=1/(z2+1)f(z) = 1/(z^2 + 1) is a rational function with denominator non-zero away from ±i\pm iSo ff Is analytic on C∖{i,−i}\mathbb{C} \setminus \{i, -i\}.

By the quotient rule: f′(z)=−2z(z2+1)2f'(z) = \frac{-2z}{(z^2 + 1)^2}.

Verify via CR at z=1z = 1: u=x2−y2+1(x2−y2+1)2+4x2y2u = \frac{x^2 - y^2 + 1}{(x^2 - y^2 + 1)^2 + 4x^2y^2} v=−2xy(x2−y2+1)2+4x2y2v = \frac{-2xy}{(x^2 - y^2 + 1)^2 + 4x^2y^2}.

ux(1,0)=−1/2=f′(1)u_x(1, 0) = -1/2 = f'(1). ✓\checkmark

If you get this wrong, revise: Sections 3.1 and 3.3 (CR Equations).

Show that u(x,y)=x3−3xy2+3x2−3y2u(x, y) = x^3 - 3xy^2 + 3x^2 - 3y^2 is harmonic and find its harmonic conjugate.

Solution

uxx=6x+6u_{xx} = 6x + 6, uyy=−6x−6u_{yy} = -6x - 6. Δu=0\Delta u = 0. ✓\checkmark

By CR: vy=ux=3x2−3y2+6xv_y = u_x = 3x^2 - 3y^2 + 6x. v=3x2y−y3+6xy+g(x)v = 3x^2 y - y^3 + 6xy + g(x).

vx=−uy=6xy+6yv_x = -u_y = 6xy + 6y. 6xy+6y=6xy+6y+g′(x)⇒g′(x)=0⇒g(x)=C6xy + 6y = 6xy + 6y + g'(x) \Rightarrow g'(x) = 0 \Rightarrow g(x) = C.

Harmonic conjugate: v(x,y)=3x2y−y3+6xy+Cv(x, y) = 3x^2 y - y^3 + 6xy + C.

f(z)=u+iv=z3+3z2f(z) = u + iv = z^3 + 3z^2.

If you get this wrong, revise: Section 3.4 (Harmonic Functions).

Evaluate ∫γ(z2+2z) dz\int_\gamma (z^2 + 2z)\, dz where γ\gamma is the upper half of the unit circle from z=1z = 1 to z=−1z = -1.

Solution

Since z2+2zz^2 + 2z is entire, the integral is path-independent. Let F(z)=z3/3+z2F(z) = z^3/3 + z^2.

∫γ(z2+2z) dz=F(−1)−F(1)=23−43=−23\int_\gamma (z^2 + 2z)\, dz = F(-1) - F(1) = \frac{2}{3} - \frac{4}{3} = -\frac{2}{3}.

If you get this wrong, revise: Sections 4.5 and 4.7 (Contour Integrals).

Use the ML inequality to bound ∣∫γezz−2 dz∣\left|\int_\gamma \frac{e^z}{z - 2}\, dz\right| where γ\gamma Is the circle ∣z∣=1|z| = 1.

Solution

On γ\gamma: ∣z∣=1|z| = 1So ∣ez∣≤e|e^z| \leq e and ∣z−2∣≥1|z - 2| \geq 1.

∣ezz−2∣≤e\left|\frac{e^z}{z - 2}\right| \leq e. L=2πL = 2\pi.

∣∫γezz−2 dz∣≤2πe\left|\int_\gamma \frac{e^z}{z - 2}\, dz\right| \leq 2\pi e.

If you get this wrong, revise: Section 4.6 (ML Inequality).

Evaluate ∮γz+1z2−z dz\oint_\gamma \frac{z + 1}{z^2 - z}\, dz where γ\gamma is ∣z∣=2|z| = 2.

Solution

z+1z2−z=z+1z(z−1)\frac{z + 1}{z^2 - z} = \frac{z + 1}{z(z - 1)}. Simple poles at z=0z = 0 and z=1z = 1Both inside ∣z∣=2|z| = 2.

At z=0z = 0: Res=lim⁡z→0z+1z−1=−1\mathrm{Res} = \lim_{z \to 0} \frac{z + 1}{z - 1} = -1. At z=1z = 1: Res=lim⁡z→1z+1z=2\mathrm{Res} = \lim_{z \to 1} \frac{z + 1}{z} = 2.

∮γz+1z2−z dz=2πi(−1+2)=2πi\oint_\gamma \frac{z + 1}{z^2 - z}\, dz = 2\pi i(-1 + 2) = 2\pi i.

If you get this wrong, revise: Sections 8.4 and 8.5 (Residues).

Classify all singularities of f(z)=e1/zz2+1f(z) = \frac{e^{1/z}}{z^2 + 1} and find all residues.

Solution

z=0z = 0: e1/ze^{1/z} has an essential singularity at 00So z=0z = 0 is an essential singularity of ff. z=iz = i: simple pole. z=−iz = -i: simple pole.

At z=iz = i: Res=e1/i2i=e−i2i\mathrm{Res} = \frac{e^{1/i}}{2i} = \frac{e^{-i}}{2i}. At z=−iz = -i: Res=e1/(−i)−2i=ei−2i\mathrm{Res} = \frac{e^{1/(-i)}}{-2i} = \frac{e^{i}}{-2i}.

At z=0z = 0: find the coefficient of 1/z1/z in e1/zz2+1\frac{e^{1/z}}{z^2 + 1}. 1z2+1=1−z2+z4−⋯\frac{1}{z^2 + 1} = 1 - z^2 + z^4 - \cdots near z=0z = 0. e1/z=1+1/z+1/(2z2)+⋯e^{1/z} = 1 + 1/z + 1/(2z^2) + \cdots. The 1/z1/z coefficient in the product: from 1⋅1/z=1/z1 \cdot 1/z = 1/zGiving residue 11.

If you get this wrong, revise: Sections 8.1 and 8.4 (Singularities and Residues).

Evaluate ∫02πcos⁡θ5+4cos⁡θ dθ\int_0^{2\pi} \frac{\cos\theta}{5 + 4\cos\theta}\, d\theta.

Solution

Substitute z=eiθz = e^{i\theta}:

I=∫∣z∣=1(z+z−1)/25+4(z+z−1)/2⋅dziz=12i∫∣z∣=1z2+1z(2z2+5z+2) dz=12i∫∣z∣=1z2+1z(2z+1)(z+2) dzI = \int_{|z|=1} \frac{(z + z^{-1})/2}{5 + 4(z + z^{-1})/2} \cdot \frac{dz}{iz} = \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 1}{z(2z^2 + 5z + 2)}\, dz = \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 1}{z(2z + 1)(z + 2)}\, dz.

Poles inside ∣z∣=1|z| = 1: z=0z = 0 (simple) and z=−1/2z = -1/2 (simple).

At z=0z = 0: Res=1(2⋅0+1)(0+2)=12\mathrm{Res} = \frac{1}{(2 \cdot 0 + 1)(0 + 2)} = \frac{1}{2}. At z=−1/2z = -1/2: Res=1/4+1(−1/2)(−1+2)=5/4−1/2=−52\mathrm{Res} = \frac{1/4 + 1}{(-1/2)(-1 + 2)} = \frac{5/4}{-1/2} = -\frac{5}{2}.

I=12i⋅2πi(12−52)=π(−2)=−π3I = \frac{1}{2i} \cdot 2\pi i\left(\frac{1}{2} - \frac{5}{2}\right) = \pi(-2) = -\frac{\pi}{3}.

If you get this wrong, revise: Section 9.4 (Trigonometric Integrals).

Evaluate ∫−∞∞dx(x2+1)(x2+4)\int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)(x^2 + 4)}.

Solution

f(z)=1(z2+1)(z2+4)f(z) = \frac{1}{(z^2 + 1)(z^2 + 4)}. Poles in upper half-plane: z=iz = i (simple) and z=2iz = 2i (simple).

At z=iz = i: Res=1(2i)(i2+4)=12i⋅3=16i\mathrm{Res} = \frac{1}{(2i)(i^2 + 4)} = \frac{1}{2i \cdot 3} = \frac{1}{6i}. At z=2iz = 2i: Res=1(4i−1)(4i)=14i(−3)=−112i\mathrm{Res} = \frac{1}{(4i - 1)(4i)} = \frac{1}{4i(-3)} = -\frac{1}{12i}.

∫−∞∞f(x) dx=2πi(16i−112i)=2πi⋅112i=π6\int_{-\infty}^{\infty} f(x)\, dx = 2\pi i\left(\frac{1}{6i} - \frac{1}{12i}\right) = 2\pi i \cdot \frac{1}{12i} = \frac{\pi}{6}.

If you get this wrong, revise: Section 9.2 (Rational Function Integrals).

Find the Taylor series of f(z)=zz2+4f(z) = \frac{z}{z^2 + 4} centered at z0=0z_0 = 0 and state the radius Of convergence.

Solution

zz2+4=z4⋅11+z2/4=z4∑n=0∞(−1)nz2n4n=∑n=0∞(−1)nz2n+14n+1\frac{z}{z^2 + 4} = \frac{z}{4} \cdot \frac{1}{1 + z^2/4} = \frac{z}{4}\sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{4^n} = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{4^{n+1}}

For ∣z∣<2|z| \lt 2. Radius of convergence: distance from 00 to nearest singularity (±2i\pm 2i), which is 22.

If you get this wrong, revise: Section 7.1 (Taylor Series).

Find the Laurent series of f(z)=1(z−1)(z−2)f(z) = \frac{1}{(z - 1)(z - 2)} in the annulus 1<∣z∣<21 \lt |z| \lt 2.

Solution

1(z−1)(z−2)=1z−2−1z−1\frac{1}{(z-1)(z-2)} = \frac{1}{z - 2} - \frac{1}{z - 1}.

For ∣z∣>1|z| \gt 1: 1z−1=1z⋅11−1/z=∑n=0∞z−n−1\frac{1}{z - 1} = \frac{1}{z} \cdot \frac{1}{1 - 1/z} = \sum_{n=0}^{\infty} z^{-n-1}.

For ∣z∣<2|z| \lt 2: 1z−2=−12⋅11−z/2=−∑n=0∞zn2n+1\frac{1}{z - 2} = -\frac{1}{2} \cdot \frac{1}{1 - z/2} = -\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}.

f(z)=−∑n=0∞zn2n+1−∑n=0∞z−n−1f(z) = -\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}} - \sum_{n=0}^{\infty} z^{-n-1}.

If you get this wrong, revise: Section 7.4 (Laurent Series).

Using Rouché’s theorem, determine the number of roots of z5−5z+1=0z^5 - 5z + 1 = 0 in ∣z∣<1|z| \lt 1.

Solution

On ∣z∣=1|z| = 1: ∣−5z∣=5>∣z5+1∣≤2|-5z| = 5 \gt |z^5 + 1| \leq 2.

By Rouché with f(z)=−5zf(z) = -5z and g(z)=z5+1g(z) = z^5 + 1: z5−5z+1z^5 - 5z + 1 has the same number of zeros In ∣z∣<1|z| \lt 1 as −5z-5zWhich has exactly one zero (at z=0z = 0).

So exactly one root in ∣z∣<1|z| \lt 1.

If you get this wrong, revise: Section 12.2 (Rouché’s Theorem).

Find the Möbius transformation that maps 1↦01 \mapsto 0, i↦1i \mapsto 1, −1↦∞-1 \mapsto \infty.

Solution

T(z)=(z−1)(i−(−1))(z−(−1))(i−1)=(z−1)(i+1)(z+1)(i−1)T(z) = \frac{(z - 1)(i - (-1))}{(z - (-1))(i - 1)} = \frac{(z - 1)(i + 1)}{(z + 1)(i - 1)}.

Simplify: i+1i−1=(i+1)(−i−1)(i−1)(−i−1)=−i2−2i−1−i2+1=−2i2=−i\frac{i + 1}{i - 1} = \frac{(i+1)(-i-1)}{(i-1)(-i-1)} = \frac{-i^2 - 2i - 1}{-i^2 + 1} = \frac{-2i}{2} = -i.

T(z)=−i⋅z−1z+1T(z) = -i \cdot \frac{z - 1}{z + 1}.

Verify: T(1)=0T(1) = 0 ✓\checkmark, T(i)=−i⋅i−1i+1=−i⋅(−i)=−1T(i) = -i \cdot \frac{i-1}{i+1} = -i \cdot (-i) = -1.

That gives −1-1Not 11. Let me recompute.

T(z)=(z−z1)(z2−z3)(z−z3)(z2−z1)T(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} with z1=1z_1 = 1, z2=iz_2 = i, z3=−1z_3 = -1.

T(z)=(z−1)(i+1)(z+1)(i−1)T(z) = \frac{(z - 1)(i + 1)}{(z + 1)(i - 1)}.

T(i)=(i−1)(i+1)(i+1)(i−1)=1T(i) = \frac{(i - 1)(i + 1)}{(i + 1)(i - 1)} = 1. ✓\checkmark

T(1)=0T(1) = 0. ✓\checkmark. T(−1)=∞T(-1) = \infty. ✓\checkmark.

So T(z)=(z−1)(i+1)(z+1)(i−1)T(z) = \frac{(z - 1)(i + 1)}{(z + 1)(i - 1)}.

If you get this wrong, revise: Section 10.5 (Cross-Ratio).

Evaluate ∫γz3z2+1 dz\int_\gamma \frac{z^3}{z^2 + 1}\, dz where γ\gamma is ∣z∣=2|z| = 2.

Solution

z3z2+1\frac{z^3}{z^2 + 1} has simple poles at z=±iz = \pm iBoth inside ∣z∣=2|z| = 2.

At z=iz = i: Res=i32i=−i2i=−12\mathrm{Res} = \frac{i^3}{2i} = \frac{-i}{2i} = -\frac{1}{2}. At z=−iz = -i: Res=(−i)3−2i=i−2i=−12\mathrm{Res} = \frac{(-i)^3}{-2i} = \frac{i}{-2i} = -\frac{1}{2}.

∫γz3z2+1 dz=2πi(−12−12)=−2πi\int_\gamma \frac{z^3}{z^2 + 1}\, dz = 2\pi i\left(-\frac{1}{2} - \frac{1}{2}\right) = -2\pi i.

Alternatively: z3z2+1=z−zz2+1\frac{z^3}{z^2 + 1} = z - \frac{z}{z^2 + 1}. ∫γz dz=0\int_\gamma z\, dz = 0 (entire), and ∫γzz2+1 dz=2πi(1/2+1/2)=2πi\int_\gamma \frac{z}{z^2 + 1}\, dz = 2\pi i(1/2 + 1/2) = 2\pi i. So the integral equals 0−2πi=−2πi0 - 2\pi i = -2\pi i. ✓\checkmark

If you get this wrong, revise: Sections 8.4 and 8.5 (Residues).

Show that ∫−∞∞cos⁡2xx2+1 dx=πe2\int_{-\infty}^{\infty} \frac{\cos 2x}{x^2 + 1}\, dx = \frac{\pi}{e^2}.

Solution

Consider ∫−∞∞e2ixx2+1 dx\int_{-\infty}^{\infty} \frac{e^{2ix}}{x^2 + 1}\, dx.

f(z)=e2izz2+1f(z) = \frac{e^{2iz}}{z^2 + 1} has a simple pole at z=iz = i in the upper half-plane.

Res ⁣(e2izz2+1,i)=e2i⋅i2i=e−22i\mathrm{Res}\!\left(\frac{e^{2iz}}{z^2 + 1}, i\right) = \frac{e^{2i \cdot i}}{2i} = \frac{e^{-2}}{2i}.

∫−∞∞e2ixx2+1 dx=2πi⋅e−22i=πe2\int_{-\infty}^{\infty} \frac{e^{2ix}}{x^2 + 1}\, dx = 2\pi i \cdot \frac{e^{-2}}{2i} = \frac{\pi}{e^2}.

Taking real parts: ∫−∞∞cos⁡2xx2+1 dx=πe2\int_{-\infty}^{\infty} \frac{\cos 2x}{x^2 + 1}\, dx = \frac{\pi}{e^2}.

If you get this wrong, revise: Section 9.7 (Fourier-Type Integrals).

Find the residue of f(z)=sin⁡zz4f(z) = \frac{\sin z}{z^4} at z=0z = 0.

Solution

sin⁡z=z−z3/6+z5/120−⋯\sin z = z - z^3/6 + z^5/120 - \cdots

f(z)=z−z3/6+z5/120−⋯z4=1z3−16z+z120−⋯f(z) = \frac{z - z^3/6 + z^5/120 - \cdots}{z^4} = \frac{1}{z^3} - \frac{1}{6z} + \frac{z}{120} - \cdots

The coefficient of 1/z1/z is −1/6-1/6So Res(f,0)=−16\mathrm{Res}(f, 0) = -\frac{1}{6}.

If you get this wrong, revise: Section 8.4 (Computing Residues).

Evaluate ∫γdz(z−1)2(z−2)\int_\gamma \frac{dz}{(z - 1)^2(z - 2)} where γ\gamma is ∣z−1∣=1/2|z - 1| = 1/2.

Solution

Only z=1z = 1 is inside γ\gamma (a pole of order 22). z=2z = 2 is outside.

Res(f,1)=ddz[1z−2]z=1=−1(z−2)2∣z=1=−1\mathrm{Res}(f, 1) = \frac{d}{dz}\left[\frac{1}{z - 2}\right]_{z=1} = -\frac{1}{(z-2)^2}\Big|_{z=1} = -1.

∫γf dz=2πi⋅(−1)=−2πi\int_\gamma f\, dz = 2\pi i \cdot (-1) = -2\pi i.

If you get this wrong, revise: Section 6.2 (CIF for Derivatives) and 8.4 (Residues).

Use the Cauchy-Riemann equations to show that f(z)=∣z∣2+2zˉf(z) = |z|^2 + 2\bar{z} is differentiable at Exactly one point and find f′(z)f'(z) there.

Solution

f(z)=x2+y2+2x−2iyf(z) = x^2 + y^2 + 2x - 2iy. So u=x2+y2+2xu = x^2 + y^2 + 2x, v=−2yv = -2y.

ux=2x+2u_x = 2x + 2, uy=2yu_y = 2y, vx=0v_x = 0, vy=−2v_y = -2.

CR: 2x+2=−2⇒x=−22x + 2 = -2 \Rightarrow x = -2And 2y=0⇒y=02y = 0 \Rightarrow y = 0.

ff is differentiable only at z=−2z = -2.

f′(−2)=ux(−2,0)+ivx(−2,0)=(2(−2)+2)+0=−2f'(-2) = u_x(-2, 0) + iv_x(-2, 0) = (2(-2) + 2) + 0 = -2.

If you get this wrong, revise: Section 3.1 (Cauchy-Riemann Equations).

Evaluate ∫γezsin⁡z(z−π)3 dz\int_\gamma \frac{e^z \sin z}{(z - \pi)^3}\, dz where γ\gamma is ∣z∣=4|z| = 4.

Solution

Only z=πz = \pi is inside γ\gamma (a pole of order 33).

By CIF for derivatives: ∫γf(z)(z−π)3 dz=2πi2! f′′(π)\int_\gamma \frac{f(z)}{(z - \pi)^3}\, dz = \frac{2\pi i}{2!}\,f''(\pi) Where f(z)=ezsin⁡zf(z) = e^z \sin z.

f′(z)=ezsin⁡z+ezcos⁡z=ez(sin⁡z+cos⁡z)f'(z) = e^z \sin z + e^z \cos z = e^z(\sin z + \cos z). f′′(z)=ez(sin⁡z+cos⁡z)+ez(cos⁡z−sin⁡z)=2ezcos⁡zf''(z) = e^z(\sin z + \cos z) + e^z(\cos z - \sin z) = 2e^z \cos z.

f′′(π)=2eπcos⁡π=−2eπf''(\pi) = 2e^\pi \cos\pi = -2e^\pi.

∫γezsin⁡z(z−π)3 dz=πi⋅(−2eπ)=−2πi eπ\int_\gamma \frac{e^z \sin z}{(z - \pi)^3}\, dz = \pi i \cdot (-2e^\pi) = -2\pi i\, e^\pi.

If you get this wrong, revise: Section 6.2 (Cauchy’s Integral Formula for Derivatives).

Example 1: Definite integration

Evaluate ∫02(3x2+2x) dx\displaystyle\int_0^2 (3x^2 + 2x)\,dx.

Solution:

∫(3x2+2x) dx=x3+x2+c\int (3x^2 + 2x)\,dx = x^3 + x^2 + c

[x3+x2]02=(8+4)−(0)=12\left[x^3 + x^2\right]_0^2 = (8 + 4) - (0) = 12

Example 2: Integration by parts

Find ∫xe2x dx\displaystyle\int x e^{2x}\,dx.

Solution:

Let u=x  ⟹  dudx=1u = x \implies \frac{du}{dx} = 1 and dvdx=e2x  ⟹  v=12e2x\frac{dv}{dx} = e^{2x} \implies v = \frac{1}{2}e^{2x}.

∫xe2x dx=x⋅12e2x−∫12e2x dx=xe2x2−e2x4+c=e2x(2x−1)4+c\int x e^{2x}\,dx = x \cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x}\,dx = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + c = \frac{e^{2x}(2x - 1)}{4} + c

  • Holomorphic functions: complex differentiable on an open set; Cauchy-Riemann equations ux=vyu_x = v_y, uy=−vxu_y = -v_x are necessary conditions.
  • Cauchy’s integral theorem: ∮γf(z) dz=0\oint_\gamma f(z)\,dz = 0 for ff holomorphic inside and on γ\gamma; Cauchy’s integral formula evaluates f(a)f(a) and all derivatives via contour integrals.
  • Residue theorem: ∮γf(z) dz=2πi∑Res(f,zk)\oint_\gamma f(z)\,dz = 2\pi i \sum \text{Res}(f, z_k); residues computed via Laurent series coefficients.
  • Laurent series: generalises Taylor series to annular regions; classifies singularities as removable, poles, or essential.
  • Conformal mappings: holomorphic functions with non-zero derivative preserve angles; applications in fluid dynamics and electrostatics.
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Real AnalysisWyattsNotesView
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Complex Analysis — MIT 18.04MIT OCWView