Complex Analysis

1. Complex Numbers Review

1.1 Definition and Arithmetic

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

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

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

$z + w = w + z$ and $zw = wz$ (commutativity)
$(z + w) + u = z + (w + u)$ and $(zw)u = z(wu)$ (associativity)
$z(w + u) = zw + zu$ (distributivity)
There exist additive identity $0$ and multiplicative identity $1$ .
Every $z \neq 0$ has a multiplicative inverse $\frac{1}{z} = \frac{\bar{z}}{|z|^2}$ .

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

1.2 The Complex Conjugate and Modulus

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

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

$\overline{z + w} = \bar{z} + \bar{w}$ and $\overline{zw} = \bar{z}\bar{w}$
$z\bar{z} = |z|^2$
$z + \bar{z} = 2\,\mathrm{Re}(z)$ and $z - \bar{z} = 2i\,\mathrm{Im}(z)$
$\bar{\bar{z}} = z$

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

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

$|z| \geq 0$ with equality iff $z = 0$
$|zw| = |z||w|$
$|z + w| \leq |z| + |w|$ (triangle inequality)
$\bigl||z| - |w|\bigr| \leq |z - w|$ (reverse triangle inequality)

Proof of (3). $|z + w|^2 = (z + w)(\bar{z} + \bar{w}) = |z|^2 + z\bar{w} + \bar{z}w + |w|^2 = |z|^2 + 2\,\mathrm{Re}(z\bar{w}) + |w|^2 \leq |z|^2 + 2|z||w| + |w|^2 = (|z| + |w|)^2$ . The inequality follows from $\mathrm{Re}(z\bar{w}) \leq |z\bar{w}| = |z||w|$ . $\blacksquare$

1.3 Polar Form

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

$z = r(\cos\theta + i\sin\theta) = re^{i\theta}$

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

Definition. The principal argument $\mathrm{Arg}(z)$ is the unique $\theta \in (-\pi, \pi]$ Such that $z = |z|e^{i\theta}$ . The argument $\arg(z)$ is multi-valued: $\arg(z) = \mathrm{Arg}(z) + 2\pi k$ for $k \in \mathbb{Z}$ .

Proposition 1.4. If $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$ Then $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$ and $z_1/z_2 = (r_1/r_2)\, e^{i(\theta_1 - \theta_2)}$ .

Worked Examples: Polar Form Conversions

Solution

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

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

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

$\theta = \arctan\!\left(\frac{\sqrt{3}}{-1}\right) = \frac{2\pi}{3}$ (adjusting to second quadrant).

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

All arguments: $\arg(z) = \frac{2\pi}{3} + 2\pi k$ for $k \in \mathbb{Z}$ .

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

$z = 3\left(\cos\!\left(-\frac{\pi}{4}\right) + i\sin\!\left(-\frac{\pi}{4}\right)\right) = 3\left(\frac{\sqrt{2}}{2} - i\,\frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2}\,i$ .

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

$|z| = \sqrt{9 + 16} = 5$ .

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

$\theta = \arctan(4/3) + \pi = \pi + \arctan(4/3)$ .

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

1.4 Euler”s Formula and De Moivre’s Theorem

Euler’s formula: $e^{i\theta} = \cos\theta + i\sin\theta$ .

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

$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$

Proposition 1.5. De Moivre’s theorem holds for all integers $n$ Including negative values.

Proof. For $n \geq 0$ It follows by induction from the multiplication law $e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)}$ . For $n \lt 0$ Write $n = -m$ with $m \gt 0$ : $(\cos\theta + i\sin\theta)^n = \frac{1}{(\cos\theta + i\sin\theta)^m} = \frac{1}{\cos(m\theta) + i\sin(m\theta)} = \cos(-m\theta) + i\sin(-m\theta) = \cos(n\theta) + i\sin(n\theta)$ . $\blacksquare$

Applications of De Moivre’s Theorem

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

$1 + i = \sqrt{2}\,e^{i\pi/4}$ So $(1 + i)^{20} = (\sqrt{2})^{20}\, e^{20\pi i/4} = 2^{10}\, e^{5\pi i} = 1024\,e^{\pi i} = -1024$ .

Solution

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

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

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

$\cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$ .

Using $\sin^2\theta = 1 - \cos^2\theta$ :

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

Problem. Show that $\sum_{k=0}^{n-1} \cos(k\theta) = \frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos\!\left(\frac{(n-1)\theta}{2}\right)$ For $\theta \notin 2\pi\mathbb{Z}$ .

Consider $S = \sum_{k=0}^{n-1} e^{ik\theta} = \frac{1 - e^{in\theta}}{1 - e^{i\theta}}$ (geometric series with $r = e^{i\theta} \neq 1$ ).

$S = \frac{e^{in\theta/2}(e^{-in\theta/2} - e^{in\theta/2})}{e^{i\theta/2}(e^{-i\theta/2} - e^{i\theta/2})} = e^{i(n-1)\theta/2} \cdot \frac{\sin(n\theta/2)}{\sin(\theta/2)}$ .

Taking real parts gives the result.

1.5 Roots of Complex Numbers

Definition. An $n$ -th root of $w \in \mathbb{C}$ is a complex number $z$ such that $z^n = w$ .

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

$z_k = \rho^{1/n}\, e^{i(\phi + 2\pi k)/n}, \quad k = 0, 1, \ldots, n - 1$

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

Proof. If $z^n = w$ Write $z = r\,e^{i\theta}$ . Then $r^n e^{in\theta} = \rho\, e^{i\phi}$ So $r = \rho^{1/n}$ and $n\theta = \phi + 2\pi k$ . For $k = 0, 1, \ldots, n-1$ these give distinct Values of $\theta$ ; for $k \geq n$ they repeat. $\blacksquare$

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

1.6 Roots of Unity

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

$z_k = e^{2\pi i k / n}, \quad k = 0, 1, \ldots, n - 1$

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

Proposition 1.7. If $\omega = e^{2\pi i/n}$ is a primitive $n$ -th root of unity, then $\sum_{k=0}^{n-1} \omega^k = 0$ and $\sum_{k=0}^{n-1} \omega^{jk} = 0$ for any $j$ not divisible by $n$ .

Proof. The sum $\sum_{k=0}^{n-1} \omega^k = \frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0$ Provided $\omega \neq 1$ . For $j$ not divisible by $n$ , $\omega^j$ is a non-trivial root of unity, So the same argument applies. $\blacksquare$

Solution

Problem. Find all cube roots of $-8$ .

$-8 = 8\,e^{i\pi}$ . The cube roots are: $z_k = 8^{1/3}\, e^{i(\pi + 2\pi k)/3} = 2\, e^{i(\pi + 2\pi k)/3}$ for $k = 0, 1, 2$ .

$z_0 = 2\,e^{i\pi/3} = 2\left(\frac{1}{2} + i\,\frac{\sqrt{3}}{2}\right) = 1 + i\sqrt{3}$ . $z_1 = 2\,e^{i\pi} = -2$ . $z_2 = 2\,e^{i5\pi/3} = 2\left(\frac{1}{2} - i\,\frac{\sqrt{3}}{2}\right) = 1 - i\sqrt{3}$ .

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

$16i = 16\,e^{i\pi/2}$ . The fourth roots are: $z_k = 16^{1/4}\, e^{i(\pi/2 + 2\pi k)/4} = 2\, e^{i(\pi/8 + \pi k/2)}$ for $k = 0, 1, 2, 3$ .

$z_0 = 2\,e^{i\pi/8}$ , $z_1 = 2\,e^{i5\pi/8}$ , $z_2 = 2\,e^{i9\pi/8}$ , $z_3 = 2\,e^{i13\pi/8}$ .

Problem. Show that the $n$ -th roots of any non-zero $w$ are in geometric progression.

The roots are $z_k = \rho^{1/n}\, e^{i(\phi + 2\pi k)/n} = z_0 \cdot \left(e^{2\pi i/n}\right)^k = z_0 \cdot \omega^k$ Where $\omega = e^{2\pi i/n}$ is a primitive $n$ -th root of unity. This is a geometric sequence With ratio $\omega$ .

2. Complex Functions and Analyticity

2.1 Complex Functions

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

Example. $f(z) = z^2 = (x + iy)^2 = (x^2 - y^2) + i(2xy)$ . Here $u = x^2 - y^2$ and $v = 2xy$ .

Example. $f(z) = \bar{z} = x - iy$ . Here $u = x$ and $v = -y$ .

Example. $f(z) = |z|^2 = x^2 + y^2$ . Here $u = x^2 + y^2$ and $v = 0$ .

2.2 Limits and Continuity

The limit $\lim_{z \to z_0} f(z) = L$ means: for every $\varepsilon \gt 0$ There exists $\delta \gt 0$ Such that $0 \lt |z - z_0| \lt \delta$ implies $|f(z) - L| \lt \varepsilon$ .

Unlike the real case, $z$ can approach $z_0$ from any direction in $\mathbb{C}$ . This makes limits More restrictive.

Proposition 2.1. $\lim_{z \to z_0} f(z) = L$ if and only if $\lim_{(x,y) \to (x_0, y_0)} u(x, y) = a$ And $\lim_{(x,y) \to (x_0, y_0)} v(x, y) = b$ where $L = a + bi$ .

Definition. $f$ is continuous at $z_0$ if $\lim_{z \to z_0} f(z) = f(z_0)$ .

Solution

Problem. Show that $\lim_{z \to 0} \frac{\bar{z}}{z}$ does not exist.

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

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

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

2.3 The Derivative

Definition. $f$ is differentiable at $z_0$ if

$f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$

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

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

2.4 Analytic Functions

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

Examples of entire functions: $z^n$ , $e^z$ , $\sin z$ , $\cos z$ Polynomials.

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

Solution

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

$f(z) = x^2 + y^2$ So $u = x^2 + y^2$ and $v = 0$ . $u_x = 2x$ , $u_y = 2y$ , $v_x = 0$ , $v_y = 0$ . The Cauchy-Riemann equations require $2x = 0$ and $2y = 0$ So $x = y = 0$ . Thus $f$ satisfies CR only at $z = 0$ .

At $z = 0$ : $f'(0) = \lim_{h \to 0} \frac{|h|^2}{h} = \lim_{h \to 0} \bar{h} = 0$ So $f$ is Differentiable at $0$ but not analytic anywhere (no neighbourhood of $0$ is analytic).

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

$f(z) = |z|^2 + z = (x^2 + y^2 + x) + iy$ . $u_x = 2x + 1$ , $u_y = 2y$ , $v_x = 0$ , $v_y = 1$ . CR equations: $2x + 1 = 1 \Rightarrow x = 0$ And $2y = 0 \Rightarrow y = 0$ . At $(0, 0)$ : $f'(0) = \lim_{h \to 0} \frac{h\bar{h} + h}{h} = \lim_{h \to 0} (\bar{h} + 1) = 1$ . So $f$ is differentiable at $z = 0$ only, hence nowhere analytic.

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 $f$ is a single-valued analytic function $g$ Defined on a domain $D$ such that $g(z) \in f(z)$ for all $z \in D$ .

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

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

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

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

Complex Powers. For $z, \alpha \in \mathbb{C}$ with $z \neq 0$ :

$z^\alpha = e^{\alpha \log z}$

This is multi-valued . When $\alpha$ is rational with reduced form $p/q$ There are exactly $q$ distinct values.

Solution

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

$(-1)^i = e^{i \log(-1)} = e^{i(i\pi + 2\pi i k)} = e^{-\pi - 2\pi k}$ for $k \in \mathbb{Z}$ .

These are all positive real numbers: $\ldots, e^{3\pi}, e^{\pi}, e^{-\pi}, e^{-3\pi}, \ldots$ . The principal value (using the principal branch) is $e^{-\pi}$ .

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

$i^{1/2} = e^{(1/2)\log i} = e^{(1/2)(i\pi/2 + 2\pi i k)} = e^{i\pi/4 + i\pi k}$ .

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

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

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

$z^2 + 1 \leq 0$ when $z^2 \leq -1$ I.e., $z \in [-i, 0] \cup [0, i]$ (the imaginary axis Segment from $-i$ to $i$ ). Also $z^2 + 1 = 0$ at $z = \pm i$ .

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

3. The Cauchy-Riemann Equations

3.1 Statement

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

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

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

$f'(z) = \lim_{h \to 0} \frac{u(x+h, y) - u(x, y)}{h} + i\lim_{h \to 0} \frac{v(x+h, y) - v(x, y)}{h} = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}$

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

$f'(z) = \lim_{k \to 0} \frac{u(x, y+k) - u(x, y)}{ik} + i\lim_{k \to 0} \frac{v(x, y+k) - v(x, y)}{ik} = -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}$

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

3.2 Sufficiency Condition

Theorem 3.2. If $u$ and $v$ have continuous first partial derivatives on an open set $U$ and Satisfy the Cauchy-Riemann equations on $U$ Then $f = u + iv$ is analytic on $U$ .

Proof. Since $u_x, u_y, v_x, v_y$ are continuous on $U$ , $u$ and $v$ are (real) differentiable. Let $\Delta z = \Delta x + i\Delta y$ . By real differentiability:

$u(x + \Delta x, y + \Delta y) - u(x, y) = u_x\,\Delta x + u_y\,\Delta y + \varepsilon_1$ $v(x + \Delta x, y + \Delta y) - v(x, y) = v_x\,\Delta x + v_y\,\Delta y + \varepsilon_2$

Where $\varepsilon_1, \varepsilon_2 = o(|\Delta z|)$ . Therefore

$\frac{f(z + \Delta z) - f(z)}{\Delta z} = \frac{(u_x + iv_x)\Delta x + (u_y + iv_y)\Delta y + \varepsilon_1 + i\varepsilon_2}{\Delta x + i\Delta y}$

By CR: $u_y + iv_y = -v_x + iu_x = i(u_x + iv_x)$ . Substituting:

$= (u_x + iv_x)\frac{\Delta x + i\Delta y}{\Delta x + i\Delta y} + \frac{o(|\Delta z|)}{\Delta z} \to u_x + iv_x$

As $\Delta z \to 0$ . $\blacksquare$

3.3 The Derivative in Terms of Partial Derivatives

When the Cauchy-Riemann equations hold:

$f'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i\frac{\partial u}{\partial y}$

3.4 Harmonic Functions

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

Proposition 3.3. If $f = u + iv$ is analytic, then $u$ and $v$ are harmonic.

Proof. From the Cauchy-Riemann equations: $u_x = v_y$ and $u_y = -v_x$ . Differentiating: $u_{xx} = v_{yx}$ and $u_{yy} = -v_{xy}$ . By equality of mixed partials, $u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0$ . Similarly for $v$ . $\blacksquare$

Definition. If $u$ and $v$ are harmonic on $U$ and satisfy the Cauchy-Riemann equations, then $v$ is the harmonic conjugate of $u$ .

Proposition 3.4. If $U$ is a connected domain and $u$ is harmonic on $U$ Then $u$ has A harmonic conjugate on $U$ Unique up to an additive constant.

Proof. Define $v(x, y) = \int_{(x_0, y_0)}^{(x, y)} (-u_y\, dx + u_x\, dy)$ . The integrand is closed (since $(-u_y)_y = -u_{yy} = u_{xx} = (u_x)_x$ ) and since $U$ is Connected, $v$ is well-defined (path-independent) by Green’s theorem. Then $v_x = -u_y$ and $v_y = u_x$ Which are the CR equations. $\blacksquare$

Solution

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

Verify $u$ is harmonic: $u_{xx} = 6x$ , $u_{yy} = -6x$ So $u_{xx} + u_{yy} = 0$ . $\checkmark$

By CR: $v_y = u_x = 3x^2 - 3y^2$ So $v = 3x^2 y - y^3 + g(x)$ . Also $v_x = -u_y = 6xy$ So $6xy = 6xy + g'(x)$ Giving $g'(x) = 0$ So $g(x) = C$ .

Harmonic conjugate: $v(x, y) = 3x^2 y - y^3 + C$ .

Note: $f(z) = u + iv = x^3 - 3xy^2 + i(3x^2 y - y^3) = (x + iy)^3 = z^3$ .

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

$u_x = \frac{2x}{x^2 + y^2}$ , $u_{xx} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2}$ . $u_y = \frac{2y}{x^2 + y^2}$ , $u_{yy} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2}$ . $\Delta u = 0$ . $\checkmark$

However, $\oint_{|z|=1} (-u_y\, dx + u_x\, dy) = \oint_{|z|=1} \frac{-y\, dx + x\, dy}{x^2 + y^2} = \int_0^{2\pi} 1\, d\theta = 2\pi \neq 0$ .

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

3.5 Worked Examples: Verifying CR Equations

Solution

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

Solution. $e^z = e^{x+iy} = e^x(\cos y + i\sin y)$ . So $u = e^x \cos y$ and $v = e^x \sin y$ .

$u_x = e^x \cos y$ , $u_y = -e^x \sin y$ , $v_x = e^x \sin y$ , $v_y = e^x \cos y$ .

Cauchy-Riemann: $u_x = e^x \cos y = v_y$ and $u_y = -e^x \sin y = -v_x$ . Both satisfied.

$f'(z) = u_x + iv_x = e^x \cos y + ie^x \sin y = e^z$ . $\blacksquare$

Problem. Verify CR for $f(z) = \sin z$ and find $f'(z)$ .

$\sin z = \sin(x + iy) = \sin x \cosh y + i\cos x \sinh y$ .

$u = \sin x \cosh y$ , $v = \cos x \sinh y$ .

$u_x = \cos x \cosh y$ , $u_y = \sin x \sinh y$ . $v_x = -\sin x \sinh y$ , $v_y = \cos x \cosh y$ .

CR: $u_x = \cos x \cosh y = v_y$ $\checkmark$ and $u_y = \sin x \sinh y = -v_x$ $\checkmark$ .

$f'(z) = u_x + iv_x = \cos x \cosh y - i\sin x \sinh y = \cos z$ . $\blacksquare$

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

$\frac{1}{z} = \frac{\bar{z}}{|z|^2} = \frac{x - iy}{x^2 + y^2}$ .

$u = \frac{x}{x^2 + y^2}$ , $v = \frac{-y}{x^2 + y^2}$ .

$u_x = \frac{y^2 - x^2}{(x^2 + y^2)^2}$ , $v_y = \frac{y^2 - x^2}{(x^2 + y^2)^2}$ . So $u_x = v_y$ . $\checkmark$

$u_y = \frac{-2xy}{(x^2 + y^2)^2}$ , $v_x = \frac{2xy}{(x^2 + y^2)^2}$ . So $u_y = -v_x$ . $\checkmark$

$f'(z) = u_x + iv_x = \frac{-(x^2 - y^2 + 2ixy)}{(x^2 + y^2)^2} = \frac{-1}{z^2}$ . $\blacksquare$

4. Complex Integration

4.1 Contours

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

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

4.2 The Complex Integral

Definition. For a contour $\gamma$ and a continuous function $f$ on $\gamma$ :

$\int_{\gamma} f(z)\, dz = \int_a^b f(\gamma(t))\gamma'(t)\, dt$

4.3 Basic Properties

Proposition 4.1. The complex integral is linear:

$\int_\gamma (af + bg)\, dz = a\int_\gamma f\, dz + b\int_\gamma g\, dz$

Proposition 4.2. Reversing orientation changes the sign:

$\int_{-\gamma} f\, dz = -\int_\gamma f\, dz$

Proposition 4.3. Additivity over contours:

$\int_{\gamma_1 + \gamma_2} f\, dz = \int_{\gamma_1} f\, dz + \int_{\gamma_2} f\, dz$

4.4 ML Inequality

Proposition 4.4 (ML Inequality). If $|f(z)| \leq M$ for all $z$ on a contour $\gamma$ of length $L$ Then

$\left|\int_\gamma f(z)\, dz\right| \leq ML$

Proof. $\left|\int_a^b f(\gamma(t))\gamma'(t)\, dt\right| \leq \int_a^b |f(\gamma(t))||\gamma'(t)|\, dt \leq M \int_a^b |\gamma'(t)|\, dt = ML$ . $\blacksquare$

4.5 Worked Examples: Contour Integrals

Solution

Problem. Evaluate $\int_\gamma z^2\, dz$ where $\gamma$ is the line segment from $0$ to $1 + i$ .

Solution. Parameterize $\gamma(t) = t(1 + i)$ for $0 \leq t \leq 1$ . Then $\gamma'(t) = 1 + i$ .

$\int_\gamma z^2\, dz = \int_0^1 (t(1+i))^2 (1+i)\, dt = (1+i)^3 \int_0^1 t^2\, dt = (1+i)^3 \cdot \frac{1}{3}$

$(1+i)^3 = (1+i)(1+i)^2 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i$ .

$\int_\gamma z^2\, dz = \frac{-2 + 2i}{3}$ . $\blacksquare$

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

$\gamma(t) = e^{it}$ , $0 \leq t \leq 2\pi$ , $\gamma'(t) = ie^{it}$ . $\bar{z} = e^{-it}$ on $\gamma$ .

$\int_\gamma \bar{z}\, dz = \int_0^{2\pi} e^{-it} \cdot ie^{it}\, dt = \int_0^{2\pi} i\, dt = 2\pi i$ .

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

Problem. Evaluate $\int_\gamma \frac{dz}{z}$ where $\gamma$ is the upper semicircle $z = e^{i\theta}$ , $0 \leq \theta \leq \pi$ .

$\int_0^\pi \frac{ie^{i\theta}}{e^{i\theta}}\, d\theta = \int_0^\pi i\, d\theta = i\pi$ .

Problem. Evaluate $\int_\gamma z\, dz$ where $\gamma$ consists of the line segment from $0$ to $1$ followed by the line segment from $1$ to $1 + i$ .

$\gamma_1(t) = t$ , $0 \leq t \leq 1$ : $\int_0^1 t \cdot 1\, dt = \frac{1}{2}$ .

$\gamma_2(t) = 1 + it$ , $0 \leq t \leq 1$ : $\int_0^1 (1 + it) \cdot i\, dt = \int_0^1 (i - t)\, dt = i - \frac{1}{2}$ .

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

Check: Since $z$ is entire, the integral from $0$ to $1 + i$ is $\frac{1}{2}(1+i)^2 = i$ . Consistent. $\blacksquare$

4.6 ML Inequality Applications

Solution

Problem. Use the ML inequality to show that $\lim_{R \to \infty} \int_{C_R} \frac{e^{iz}}{z}\, dz = 0$ Where $C_R$ is the upper semicircle $|z| = R$ , $\mathrm{Im}(z) \geq 0$ .

On $C_R$ : $z = Re^{i\theta}$ , $0 \leq \theta \leq \pi$ . $|e^{iz}| = |e^{iR(\cos\theta + i\sin\theta)}| = e^{-R\sin\theta}$ .

$\left|\int_{C_R} \frac{e^{iz}}{z}\, dz\right| \leq \int_0^\pi \frac{e^{-R\sin\theta}}{R} \cdot R\, d\theta = \int_0^\pi e^{-R\sin\theta}\, d\theta$ .

By Jordan’s inequality $\sin\theta \geq \frac{2\theta}{\pi}$ for $\theta \in [0, \pi/2]$ :

$\leq 2\int_0^{\pi/2} e^{-2R\theta/\pi}\, d\theta = \frac{\pi}{R}(1 - e^{-R}) \to 0$ as $R \to \infty$ . $\blacksquare$

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

On $\gamma$ : $|z^2 + 4| \geq |z|^2 - 4 = 9 - 4 = 5$ (reverse triangle inequality). So $\left|\frac{1}{z^2 + 4}\right| \leq \frac{1}{5}$ .

Length of $\gamma$ : $L = 2\pi \cdot 3 = 6\pi$ .

$\left|\int_\gamma \frac{dz}{z^2 + 4}\right| \leq \frac{1}{5} \cdot 6\pi = \frac{6\pi}{5}$ .

4.7 Antiderivative Method

When $f$ is analytic on a connected domain and has a known antiderivative $F$ with $F' = f$ :

$\int_\gamma f(z)\, dz = F(\gamma(b)) - F(\gamma(a))$

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

Solution

Problem. Evaluate $\int_\gamma \cos z\, dz$ where $\gamma$ is any path from $0$ to $\pi + i$ .

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

$\int_\gamma \cos z\, dz = \sin(\pi + i) - \sin(0) = \sin(\pi + i)$ .

$\sin(\pi + i) = \sin\pi\cosh 1 + i\cos\pi\sinh 1 = -i\sinh 1$ .

So the integral equals $-i\sinh 1$ .

Problem. Evaluate $\int_\gamma e^{2z}\, dz$ where $\gamma$ is any path from $1$ to $i$ .

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

$\int_\gamma e^{2z}\, dz = \frac{1}{2}(e^{2i} - e^{2})$ .

5. Cauchy’s Theorem

5.1 Statement

Theorem 5.1 (Cauchy’s Theorem). If $f$ is analytic on a connected domain $D$ and $\gamma$ Is a simple closed contour in $D$ Then

$\int_\gamma f(z)\, dz = 0$

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

$\int_\gamma f\, dz = \int_\gamma (u\, dx - v\, dy) + i\int_\gamma (v\, dx + u\, dy)$

Applying Green’s theorem to each integral:

$= \iint_D (-v_x - u_y)\, dA + i\iint_D (u_x - v_y)\, dA = 0$

By the Cauchy-Riemann equations. $\blacksquare$

5.2 Connected Domains

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

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

5.3 Path Independence

Corollary 5.2. If $f$ is analytic on a connected domain $D$ Then the integral $\int_{z_0}^{z_1} f(z)\, dz$ is independent of the path from $z_0$ to $z_1$ in $D$ .

5.4 Antiderivatives

Theorem 5.3. If $f$ is analytic on a connected domain $D$ Then $f$ has an antiderivative $F$ in $D$ (i.e., $F'(z) = f(z)$ ), and

$\int_\gamma f(z)\, dz = F(z_1) - F(z_0)$

Where $z_0$ and $z_1$ are the endpoints of $\gamma$ .

5.5 Cauchy’s Theorem for Multiply Connected Domains

Theorem 5.4. If $f$ is analytic on a domain $D$ containing simple closed contours $\gamma, \gamma_1, \ldots, \gamma_n$ where $\gamma_1, \ldots, \gamma_n$ Lie in the interior of $\gamma$ and the region between $\gamma$ and the $\gamma_k$ is contained in $D$ And all contours are positively oriented, then

$\int_\gamma f(z)\, dz = \sum_{k=1}^n \int_{\gamma_k} f(z)\, dz$

5.6 Deformation of Contours

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

$\int_{\gamma_1} f(z)\, dz = \int_{\gamma_2} f(z)\, dz$

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

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 $\int_\gamma \frac{dz}{z - 2}$ where $\gamma$ is the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ .

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

$\int_\gamma \frac{dz}{z - 2} = 2\pi i$ .

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

$\frac{e^z}{z}$ is analytic on and inside $\gamma$ except at $z = 0$ . By deformation: $\int_\gamma \frac{e^z}{z}\, dz = \int_{|z|=r} \frac{e^z}{z}\, dz = 2\pi i \cdot e^0 = 2\pi i$ .

Problem. Evaluate $\int_\gamma \frac{dz}{z^2 - 1}$ where $\gamma$ is $|z| = 2$ .

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

Both $z = \pm 1$ are inside $|z| = 2$ .

$\int_\gamma \frac{dz}{z^2 - 1} = \frac{1}{2}(2\pi i - 2\pi i) = 0$ .

6. Cauchy’s Integral Formula

6.1 Statement

Theorem 6.1 (Cauchy’s Integral Formula). If $f$ is analytic on a connected domain Containing a simple closed positively oriented contour $\gamma$ And $z_0$ is inside $\gamma$ Then

$f(z_0) = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z - z_0}\, dz$

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

$\int_\gamma \frac{f(z)}{z - z_0}\, dz = \int_{\gamma_\varepsilon} \frac{f(z)}{z - z_0}\, dz$

On $\gamma_\varepsilon$ : $f(z) = f(z_0) + (z - z_0)f'(\zeta)$ for some $\zeta$ between $z$ and $z_0$ .

$= \int_{\gamma_\varepsilon} \frac{f(z_0)}{z - z_0}\, dz + \int_{\gamma_\varepsilon} f'(\zeta)\, dz = f(z_0) \cdot 2\pi i + 0$

Since $\int_{\gamma_\varepsilon} \frac{dz}{z - z_0} = 2\pi i$ (parameterize $z = z_0 + \varepsilon e^{i\theta}$ ) and $\int_{\gamma_\varepsilon} f'(\zeta)\, dz \to 0$ as $\varepsilon \to 0$ by the ML inequality. $\blacksquare$

6.2 Derivatives

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

$f^{(n)}(z_0) = \frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz$

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

$f^{(n+1)}(z_0) = \lim_{h \to 0} \frac{f^{(n)}(z_0 + h) - f^{(n)}(z_0)}{h} = \lim_{h \to 0} \frac{n!}{2\pi i}\int_\gamma \frac{1}{h}\left[\frac{f(z)}{(z - z_0 - h)^{n+1}} - \frac{f(z)}{(z - z_0)^{n+1}}\right] dz$

$= \frac{(n+1)!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+2}}\, dz$

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

Corollary 6.3. If $f$ is analytic, then $f$ is infinitely differentiable.

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

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

$|f^{(n)}(z_0)| \leq \frac{n!M}{R^n}$

Proof. From the integral formula: $|f^{(n)}(z_0)| = \frac{n!}{2\pi}\left|\int_{|z-z_0|=R} \frac{f(z)}{(z-z_0)^{n+1}}\, dz\right| \leq \frac{n!}{2\pi} \cdot \frac{M}{R^{n+1}} \cdot 2\pi R = \frac{n!M}{R^n}$ . $\blacksquare$

6.4 Liouville’s Theorem

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

Proof. If $|f(z)| \leq M$ for all $z$ Then by Cauchy’s estimates with $R$ arbitrarily large: $|f'(z_0)| \leq \frac{M}{R} \to 0$ as $R \to \infty$ . So $f'(z) = 0$ for all $z$ Meaning $f$ is Constant. $\blacksquare$

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

Proof. $1/f$ is entire and bounded by $1/M$ So constant by Liouville. $\blacksquare$

6.5 Fundamental Theorem of Algebra

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

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

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

6.6 Worked Examples: Cauchy’s Integral Formula

Solution

Problem. Evaluate $\int_\gamma \frac{e^z}{z - 1}\, dz$ where $\gamma$ is $|z| = 2$ .

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

$\int_\gamma \frac{e^z}{z - 1}\, dz = 2\pi i \cdot f(1) = 2\pi i \cdot e^1 = 2\pi i e$ . $\blacksquare$

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

By Cauchy’s formula for derivatives with $f(z) = z^2 + 1$ and $z_0 = i$ :

$\int_\gamma \frac{f(z)}{(z - i)^3}\, dz = \frac{2\pi i}{2!}\,f''(i)$ .

$f'(z) = 2z$ , $f''(z) = 2$ . So $f''(i) = 2$ .

$\int_\gamma \frac{z^2 + 1}{(z - i)^3}\, dz = \frac{2\pi i}{2} \cdot 2 = 2\pi i$ . $\blacksquare$

Problem. Evaluate $\int_\gamma \frac{\sin z}{z(z - \pi)}\, dz$ where $\gamma$ is $|z| = 4$ .

Singularities inside $\gamma$ : $z = 0$ and $z = \pi$ .

$\frac{\sin z}{z(z - \pi)} = \frac{1}{\pi}\left(\frac{\sin z}{z - \pi} - \frac{\sin z}{z}\right)$ .

At $z = 0$ : by CIF, $\int_\gamma \frac{\sin z}{z}\, dz = 2\pi i \cdot \sin(0) = 0$ . At $z = \pi$ : by CIF, $\int_\gamma \frac{\sin z}{z - \pi}\, dz = 2\pi i \cdot \sin(\pi) = 0$ .

$\int_\gamma \frac{\sin z}{z(z - \pi)}\, dz = \frac{1}{\pi}(0 - 0) = 0$ .

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

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

$\int_\gamma \frac{e^{2z}}{(z-1)^2(z+1)}\, dz = \frac{1}{4} \cdot 2\pi i \cdot e^{-2} - \frac{1}{4} \cdot 2\pi i \cdot e^2 + \frac{1}{2} \cdot \frac{2\pi i}{1!} \cdot 2e^2$

$= \frac{\pi i e^{-2}}{2} - \frac{\pi i e^2}{2} + 2\pi i e^2 = \frac{\pi i e^{-2}}{2} + \frac{3\pi i e^2}{2}$ .

7. Taylor and Laurent Series

7.1 Taylor Series

Theorem 7.1. If $f$ is analytic on $|z - z_0| \lt R$ Then

$f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z - z_0)^n$

And the series converges uniformly on compact subsets of $|z - z_0| \lt R$ .

Proof. For $|z - z_0| \lt r \lt R$ Apply Cauchy’s integral formula on $|\zeta - z_0| = r$ :

$f(z) = \frac{1}{2\pi i}\int_{|\zeta - z_0| = r} \frac{f(\zeta)}{\zeta - z}\, d\zeta$

Write $\frac{1}{\zeta - z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0} \cdot \frac{1}{1 - (z - z_0)/(\zeta - z_0)}$ $= \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}}$ (geometric series, convergent since $|z - z_0|/|\zeta - z_0| \lt 1$ ).

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

Remark. The radius of convergence $R$ is the distance from $z_0$ to the nearest singularity of $f$ .

7.2 Common Taylor Series

$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \cdots$

$\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}$

$\cos z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$

$\frac{1}{1 - z} = \sum_{n=0}^{\infty} z^n, \quad |z| \lt 1$

$\ln(1 + z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} z^n}{n}, \quad |z| \lt 1$

7.3 Worked Examples: Taylor Series

Solution

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

$\frac{1}{z} = \frac{1}{1 + (z - 1)} = \sum_{n=0}^{\infty} (-1)^n (z - 1)^n$ for $|z - 1| \lt 1$ .

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

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

$\frac{1}{(1-z)^2} = \frac{d}{dz}\left[\frac{1}{1 - z}\right] = \frac{d}{dz}\sum_{n=0}^{\infty} z^n = \sum_{n=1}^{\infty} nz^{n-1} = \sum_{n=0}^{\infty} (n+1)z^n$ for $|z| \lt 1$ .

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

$e^z = 1 + z + z^2/2 + z^3/6 + z^4/24 + \cdots$ $\sin z = z - z^3/6 + z^5/120 - \cdots$

$e^z \sin z = (1 + z + z^2/2 + z^3/6 + z^4/24 + \cdots)(z - z^3/6 + \cdots)$

$= z + z^2 + z^3/2 + z^4/6 + \cdots - z^3/6 - z^4/6 + \cdots$ $= z + z^2 + z^3/3 - z^4/30 + \cdots$

7.4 Laurent Series

Theorem 7.2 (Laurent Series). If $f$ is analytic on the annulus $r \lt |z - z_0| \lt R$ Then

$f(z) = \sum_{n=-\infty}^{\infty} a_n(z - z_0)^n = \cdots + \frac{a_{-2}}{(z - z_0)^2} + \frac{a_{-1}}{z - z_0} + a_0 + a_1(z - z_0) + \cdots$

Where

$a_n = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz$

For any simple closed contour $\gamma$ in the annulus encircling $z_0$ .

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

7.5 Classification of Laurent Series

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 $f$ in an annulus is unique.

7.6 Worked Examples: Laurent Series

Solution

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

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

In $|z| \lt 1$ : $\frac{1}{z - 1} = -\frac{1}{1 - z} = -\sum_{n=0}^{\infty} z^n$ .

So $f(z) = -\sum_{n=0}^{\infty} z^n - \frac{1}{z} = \cdots - z^2 - z - 1 - \frac{1}{z}$ .

The principal part is $-1/z$ So $z = 0$ is a simple pole. $\blacksquare$

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

In $|z| \gt 1$ : $\frac{1}{z - 1} = \frac{1}{z} \cdot \frac{1}{1 - 1/z} = \sum_{n=2}^{\infty} z^{-n}$ .

$f(z) = \sum_{n=2}^{\infty} z^{-n} - \frac{1}{z} = \frac{1}{z^2} + \frac{1}{z^3} + \cdots$

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

$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$ So $f(z) = \sum_{n=0}^{\infty} \frac{z^{n-2}}{n!} = \frac{1}{z^2} + \frac{1}{z} + \frac{1}{2} + \frac{z}{6} + \cdots$

Residue at $z = 0$ : $a_{-1} = 1$ .

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

$\frac{1}{z - 3} = -\frac{1}{3}\sum_{n=0}^{\infty} \frac{z^n}{3^n}$ .

$f(z) = -\sum_{n=0}^{\infty} \frac{z^{n-2}}{3^{n+1}} = -\frac{1}{3z^2} - \frac{1}{9z} - \frac{1}{27} - \frac{z}{81} - \cdots$

Residue at $z = 0$ : $a_{-1} = -\frac{1}{9}$ .

7.7 Residue at Infinity

Definition. The residue at infinity of $f$ is defined as

$\mathrm{Res}(f, \infty) = -\frac{1}{2\pi i}\int_{|z|=R} f(z)\, dz$

For sufficiently large $R$ (enclosing all finite singularities).

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

$\sum_{\mathrm{all\ finite\ } z_k} \mathrm{Res}(f, z_k) + \mathrm{Res}(f, \infty) = 0$

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

$\int_{|z|=R} f\, dz = 2\pi i \sum_{\mathrm{finite} \mathrm{Res}(f, z_k)}$ .

But $\mathrm{Res}(f, \infty) = -\frac{1}{2\pi i}\int_{|z|=R} f\, dz$ So the sum is zero. $\blacksquare$

8. Singularities and Residue Theory

8.1 Isolated Singularities

Let $z_0$ be an isolated singularity of $f$ (i.e., $f$ is analytic in a punctured neighbourhood of $z_0$ ).

Classification by Laurent series:

Removable singularity: $a_n = 0$ for all $n \lt 0$ . Can be removed by redefining $f(z_0) = a_0$ .
Pole of order $m$ : $a_{-m} \neq 0$ and $a_n = 0$ for $n \lt -m$ . The principal part is finite.
Essential singularity: infinitely many non-zero $a_n$ with $n \lt 0$ .

Proposition 8.1 (Riemann’s Removable Singularity Theorem). If $f$ is bounded near $z_0$ Then $z_0$ is a removable singularity.

Proposition 8.2. $z_0$ is a pole of order $m$ if and only if $\lim_{z \to z_0} (z - z_0)^m f(z)$ Exists and is non-zero.

Theorem 8.3 (Casorati-Weierstrass). If $z_0$ is an essential singularity of $f$ Then $f$ takes Values arbitrarily close to any complex number in every neighbourhood of $z_0$ .

8.2 Classification with Worked Examples

Solution

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

$z = 0$ : $\sin z = z - z^3/6 + \cdots$ So $f(z) = 1 - z^2/6 + \cdots$ . No negative powers, so $z = 0$ is a removable singularity. $f(0) = 1$ by continuity.

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

$z = 0$ : $e^z - 1 = z + z^2/2 + \cdots$ So $f(z) = \frac{1}{z} + \frac{1}{2} + \cdots$ . Principal part is $1/z$ So $z = 0$ is a simple pole with residue $1$ .

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

$e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n!\, z^n} = 1 + \frac{1}{z} + \frac{1}{2z^2} + \cdots$

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

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

$z = 0$ : pole of order $3$ . $z = i$ : simple pole. $z = -i$ : simple pole.

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

$\sin z = z - z^3/6 + \cdots$ So $f(z) = \frac{1}{1 - z^2/6 + \cdots} = 1 + \frac{z^2}{6} + \cdots$ .

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

8.3 Definition of the Residue

Definition. The residue of $f$ at an isolated singularity $z_0$ is the coefficient $a_{-1}$ In the Laurent expansion:

$\mathrm{Res}(f, z_0) = a_{-1} = \frac{1}{2\pi i}\int_\gamma f(z)\, dz$

Where $\gamma$ is a small positively oriented circle around $z_0$ .

8.4 Computing Residues

For a simple pole at $z_0$ :

$\mathrm{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)$

If $f = g/h$ where $g(z_0) \neq 0$ , $h(z_0) = 0$ , $h'(z_0) \neq 0$ :

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

For a pole of order $m$ at $z_0$ :

$\mathrm{Res}(f, z_0) = \frac{1}{(m-1)!}\lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}}\left[(z - z_0)^m f(z)\right]$

Solution

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

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

At $z = -1$ : $\mathrm{Res} = \lim_{z \to -1} \frac{z}{z + 3} = \frac{-1}{2}$ . At $z = -3$ : $\mathrm{Res} = \lim_{z \to -3} \frac{z}{z + 1} = \frac{-3}{-2} = \frac{3}{2}$ .

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

At $z = 1$ (pole of order $2$ ): $\mathrm{Res} = \frac{d}{dz}\left[\frac{e^z}{z - 2}\right]_{z=1} = \frac{e^z(z - 2) - e^z}{(z-2)^2}\Big|_{z=1} = \frac{-e - e}{1} = -2e$ .

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

8.5 The Residue Theorem

Theorem 8.4 (Residue Theorem). If $f$ is analytic inside and on a simple closed positively Oriented contour $\gamma$ except for isolated singularities $z_1, z_2, \ldots, z_n$ inside $\gamma$ Then

$\int_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^{n} \mathrm{Res}(f, z_k)$

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

$\int_\gamma f\, dz = \sum_{k=1}^n \int_{\gamma_k} f\, dz = \sum_{k=1}^n 2\pi i \cdot \mathrm{Res}(f, z_k)$ . $\blacksquare$

8.6 Worked Examples: Residue Theorem

Solution

Problem 1. Evaluate $\int_\gamma \frac{e^z}{z(z-1)^2}\, dz$ where $\gamma$ is $|z| = 2$ .

Solution. Singularities inside $\gamma$ : $z = 0$ (simple pole) and $z = 1$ (pole of order $2$ ).

At $z = 0$ : $\mathrm{Res} = \lim_{z \to 0} \frac{e^z}{(z-1)^2} = \frac{1}{(-1)^2} = 1$ .

At $z = 1$ : $\mathrm{Res}(f, 1) = \frac{d}{dz}\left[(z-1)^2 \cdot \frac{e^z}{z(z-1)^2}\right]_{z=1} = \frac{d}{dz}\left[\frac{e^z}{z}\right]_{z=1} = \frac{e^z \cdot z - e^z}{z^2}\Big|_{z=1} = \frac{e - e}{1} = 0$ .

$\int_\gamma f\, dz = 2\pi i(1 + 0) = 2\pi i$ . $\blacksquare$

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

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

Each is a simple pole with $\mathrm{Res}(f, z_k) = \frac{1}{4z_k^3}$ . Since $z_k^4 = -1$ : $z_k^{-3} = -z_k$ So the sum equals $-\frac{1}{4}\sum z_k = 0$ .

$\int_\gamma \frac{dz}{z^4 + 1} = 2\pi i \cdot 0 = 0$ . $\blacksquare$

9. Applications of Contour Integration

9.1 Evaluation of Real Integrals

Contour integration is a powerful tool for evaluating definite integrals.

9.2 Integrals of Rational Functions over the Real Line

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

$\int_{-\infty}^{\infty} f(x)\, dx = 2\pi i \sum_{\mathrm{Im}(z_k) > 0} \mathrm{Res}(f, z_k)$

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

Proof. Integrate $f(z)$ over the semicircular contour $\gamma_R$ consisting of $[-R, R]$ on the Real axis and the semicircle $|z| = R$ in the upper half-plane. As $R \to \infty$ The integral over The semicircle vanishes (since $|f(z)| \leq M/R^2$ and the length is $\pi R$ ). $\blacksquare$

9.3 Worked Example

Problem. Evaluate $\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1}$ .

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

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

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

$\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1} = 2\pi i \cdot \frac{1}{2i} = \pi$ . $\blacksquare$

9.4 Integrals Involving Trigonometric Functions

For integrals of the form $\int_0^{2\pi} R(\cos\theta, \sin\theta)\, d\theta$ Substitute $z = e^{i\theta}$ So $dz = iz\, d\theta$ , $\cos\theta = \frac{z + z^{-1}}{2}$ $\sin\theta = \frac{z - z^{-1}}{2i}$ .

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

9.5 Worked Example

Problem. Evaluate $\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta}$ .

Solution. Substitute $z = e^{i\theta}$ : $d\theta = \frac{dz}{iz}$ $\cos\theta = \frac{z + 1/z}{2}$ .

$\int_{|z|=1} \frac{dz}{iz\left(2 + \frac{z + 1/z}{2}\right)} = \int_{|z|=1} \frac{2\, dz}{i(z^2 + 4z + 1)}$

Poles: $z^2 + 4z + 1 = 0 \Rightarrow z = -2 \pm \sqrt{3}$ .

$|z_1| = |-2 + \sqrt{3}| = 2 - \sqrt{3} \lt 1$ (inside). $|z_2| = |-2 - \sqrt{3}| = 2 + \sqrt{3} \gt 1$ (outside).

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

$\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta} = \frac{2}{i} \cdot 2\pi i \cdot \frac{1}{2\sqrt{3}} = \frac{2\pi}{\sqrt{3}}$ . $\blacksquare$

9.6 Jordan’s Lemma

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

$\lim_{R \to \infty} \int_{C_R} e^{iaz}f(z)\, dz = 0$

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

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

9.7 Fourier-Type Integrals

Solution

Problem. Evaluate $\int_{-\infty}^{\infty} \frac{\cos x}{x^2 + 1}\, dx$ .

Consider $\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2 + 1}\, dx = 2\pi i \cdot \mathrm{Res}\!\left(\frac{e^{iz}}{z^2+1}, i\right)$ .

$\mathrm{Res}\!\left(\frac{e^{iz}}{z^2+1}, i\right) = \frac{e^{i \cdot i}}{2i} = \frac{e^{-1}}{2i}$ .

$\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2 + 1}\, dx = 2\pi i \cdot \frac{e^{-1}}{2i} = \frac{\pi}{e}$ .

Taking real parts: $\int_{-\infty}^{\infty} \frac{\cos x}{x^2 + 1}\, dx = \frac{\pi}{e}$ .

Problem. Evaluate $\int_{-\infty}^{\infty} \frac{x \sin x}{x^2 + a^2}\, dx$ for $a \gt 0$ .

Consider $\int_{-\infty}^{\infty} \frac{z\, e^{iz}}{z^2 + a^2}\, dz$ . Only $z = ia$ is in the upper half-plane.

$\mathrm{Res}\!\left(\frac{ze^{iz}}{z^2 + a^2}, ia\right) = \frac{ia \cdot e^{i \cdot ia}}{2ia} = \frac{e^{-a}}{2}$ .

$\int_{-\infty}^{\infty} \frac{x\, e^{ix}}{x^2 + a^2}\, dx = 2\pi i \cdot \frac{e^{-a}}{2} = \pi i\, e^{-a}$ .

Taking imaginary parts: $\int_{-\infty}^{\infty} \frac{x \sin x}{x^2 + a^2}\, dx = \pi\, e^{-a}$ .

Problem. Evaluate $\int_0^{2\pi} \frac{\cos 2\theta}{5 + 4\cos\theta}\, d\theta$ .

Substitute $z = e^{i\theta}$ : $\cos\theta = (z + z^{-1})/2$ , $\cos 2\theta = (z^2 + z^{-2})/2$ .

$I = \frac{1}{2i}\int_{|z|=1} \frac{z^4 + 1}{z^2(2z + 1)(z + 2)}\, dz$ .

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

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

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

$I = \frac{1}{2i} \cdot 2\pi i \left(-\frac{5}{4} + \frac{17}{12}\right) = \frac{\pi}{6}$ .

9.8 Improper Integrals and Principal Value

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

$\mathrm{PV}\!\int_{-\infty}^{\infty} f(x)\, dx = \lim_{\varepsilon \to 0^+} \left(\int_{-\infty}^{a-\varepsilon} f(x)\, dx + \int_{a+\varepsilon}^{\infty} f(x)\, dx\right)$

Solution

Problem. Evaluate $\mathrm{PV}\!\int_{-\infty}^{\infty} \frac{\sin x}{x}\, dx$ .

Consider $\oint_\gamma \frac{e^{iz}}{z}\, dz$ where $\gamma$ consists of $[-R, -\varepsilon]$ $[\varepsilon, R]$ on the real axis, small upper semicircle $C_\varepsilon$ around $0$ And large Upper semicircle $C_R$ .

No poles inside the contour, so the integral is $0$ .

On $C_R$ : vanishes as $R \to \infty$ by Jordan’s lemma. On $C_\varepsilon$ (indenting above): $\int_{C_\varepsilon} \frac{e^{iz}}{z}\, dz \to -i\pi$ as $\varepsilon \to 0$ (half residue contribution).

$0 = \mathrm{PV}\!\int_{-\infty}^{\infty} \frac{e^{ix}}{x}\, dx + (-i\pi)$ .

$\mathrm{PV}\!\int_{-\infty}^{\infty} \frac{e^{ix}}{x}\, dx = i\pi$ .

Taking imaginary parts: $\mathrm{PV}\!\int_{-\infty}^{\infty} \frac{\sin x}{x}\, dx = \pi$ .

10. Conformal Mappings

10.1 Definition

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

10.2 Geometric Interpretation

If $f'(z_0) = re^{i\theta}$ Then near $z_0$ the mapping $f$ acts as a rotation by $\theta$ followed By a scaling by $r$ . The Jacobian determinant is $|f'(z_0)|^2 \gt 0$ So orientation is preserved.

10.3 Common Conformal Mappings

Mapping	Effect
$w = az + b$ ( $a \neq 0$ )	Translation, rotation, scaling
$w = 1/z$	Inversion in the unit circle
$w = z^2$	Squaring (doubles angles)
$w = e^z$	Exponential (maps strips to sectors)
$w = \frac{z - a}{1 - \bar{a}z}$	Möbius (maps disk to disk)

10.4 Möbius Transformations

A Möbius transformation (or linear fractional transformation) is

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

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: $T(z_1) = w_1$ $T(z_2) = w_2$ , $T(z_3) = w_3$ .

10.5 Cross-Ratio

Definition. The cross-ratio of four distinct points $z_1, z_2, z_3, z_4$ is

$(z_1, z_2, z_3, z_4) = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)}$

Proposition 10.3. The cross-ratio is invariant under Möbius transformations: $(Tz_1, Tz_2, Tz_3, Tz_4) = (z_1, z_2, z_3, z_4)$ .

Proposition 10.4. The unique Möbius transformation sending $z_1 \mapsto 0$ , $z_2 \mapsto 1$ $z_3 \mapsto \infty$ is

$T(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}$

10.6 Classification of Möbius Transformations

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

Parabolic: Exactly one fixed point. Conjugate to $w = z + k$ .
Elliptic: Two fixed points, $|T'(z_0)| = 1$ . Conjugate to a rotation $w = e^{i\theta} z$ .
Hyperbolic: Two fixed points, $T'(z_0) \in \mathbb{R}^+$ , $T'(z_0) \neq 1$ . Conjugate to $w = kz$ .
Loxodromic: Two fixed points, $T'(z_0) \notin \mathbb{R} \cup \{z : |z| = 1\}$ . Conjugate to $w = ke^{i\theta}z$ .

Solution

Problem. Find the Möbius transformation mapping $0 \mapsto i$ , $1 \mapsto 0$ , $\infty \mapsto -i$ .

$T(z) = \frac{az + b}{cz + d}$ with $T(0) = i \Rightarrow b/d = i \Rightarrow b = id$ . $T(1) = 0 \Rightarrow a = -b = -id$ . $T(\infty) = -i \Rightarrow a/c = -i \Rightarrow c = d$ .

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

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

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

Check boundary: $T(i) = \frac{i - 1}{i + 1} = \frac{(i-1)(-i+1)}{(i+1)(-i+1)} = \frac{2}{2} = 1$ . $|T(i)| = 1$ . $\checkmark$

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

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

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

Both multipliers are real and positive (not equal to $1$ ), so $T$ is hyperbolic.

10.7 The Riemann Mapping Theorem

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

This is one of the most profound results in complex analysis, establishing that all connected Domains (other than $\mathbb{C}$ 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

11.1 Liouville’s Theorem

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

11.2 The Fundamental Theorem of Algebra

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

11.3 The Maximum Modulus Principle

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

Corollary 11.4. If $f$ is analytic on a bounded domain $D$ and continuous on $\bar{D} = D \cup \partial D$ Then $|f|$ attains its maximum on $\partial D$ .

11.4 Minimum Modulus Principle

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

Remark. If $f$ has zeros in $D$ Then $|f|$ attains its minimum of $0$ at those zeros. The minimum modulus principle requires the non-vanishing hypothesis.

11.5 Schwarz Lemma

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

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

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

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

12. Argument Principle and Rouché’s Theorem

12.1 The Argument Principle

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

$\frac{1}{2\pi i}\int_\gamma \frac{f'(z)}{f(z)}\, dz = N - P$

Where $N$ is the number of zeros and $P$ is the number of poles of $f$ inside $\gamma$ (counting Multiplicities).

12.2 Rouché’s Theorem

Theorem 12.2 (Rouché’s Theorem). If $f$ and $g$ are analytic inside and on a simple closed Contour $\gamma$ And $|f(z)| \gt |g(z)|$ on $\gamma$ Then $f$ and $f + g$ have the same number of Zeros inside $\gamma$ .

Proof. On $\gamma$ : $|g(z)/f(z)| \lt 1$ . The function $h(z) = 1 + g(z)/f(z)$ satisfies $|h(z) - 1| \lt 1$ on $\gamma$ So $h(\gamma)$ does not wind around $0$ . By the argument principle Applied to $h$ : $0 = N_h - P_h$ Meaning $h$ has the same number of zeros and poles inside $\gamma$ . But $h = (f + g)/f$ So zeros of $h$ are zeros of $f + g$ and poles of $h$ are zeros of $f$ . Therefore $f$ and $f + g$ have the same number of zeros. $\blacksquare$

12.3 Worked Example

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

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

Solution

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

On $|z| = 2$ : $|z^4| = 16 \gt |z + 1| \leq 3$ . By Rouché with $f(z) = z^4$ and $g(z) = z + 1$ : $z^4 + z + 1$ has $4$ zeros in $|z| \lt 2$ (same as $z^4$ ).

Problem. Show that $z^5 + 3z^2 + 1$ has exactly two roots in $|z| \lt 1$ .

On $|z| = 1$ : $|3z^2 + 1| \geq |3z^2| - |1| = 2 \gt |z^5| = 1$ . By Rouché with $f(z) = 3z^2 + 1$ and $g(z) = z^5$ : $z^5 + 3z^2 + 1$ has the same number of zeros as $3z^2 + 1$ in $|z| \lt 1$ . $3z^2 + 1 = 0 \Rightarrow z = \pm i/\sqrt{3}$ Both in $|z| \lt 1$ . So $2$ zeros.

13. Analytic Continuation

13.1 Definition

Definition. If $f_1$ is analytic on $D_1$ and $f_2$ is analytic on $D_2$ with $D_1 \cap D_2 \neq \emptyset$ and $f_1 = f_2$ on $D_1 \cap D_2$ Then $f_2$ is an analytic Continuation of $f_1$ .

13.2 Identity Theorem

Theorem 13.1 (Identity Theorem). If $f$ and $g$ are analytic on a domain $D$ and agree on a set With a limit point in $D$ Then $f = g$ on all of $D$ .

Proof. Let $E = \{z \in D : f^{(n)}(z) = g^{(n)}(z) \mathrm{\ for\ all\ } n \geq 0\}$ . $E$ is Non-empty (it contains the limit point by continuity of derivatives). $E$ is closed (by continuity). If $z_0 \in E$ The Taylor series of $f$ and $g$ at $z_0$ coincide, so $f = g$ in a neighbourhood of $z_0$ Giving $E$ open. Since $D$ is connected, $E = D$ . $\blacksquare$

14. Common Pitfalls

:::caution Common Pitfall The Cauchy-Riemann equations are necessary but not sufficient for Differentiability. The partial derivatives must also be continuous. For example, $f(z) = \exp(-1/z^4)$ extended by $f(0) = 0$ 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 $\oint_{|z|=1} dz/z = 2\pi i$ . :::

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

:::caution Common Pitfall The residue at infinity is $\mathrm{Res}(f, \infty) = -\mathrm{Res}(1/z^2 \cdot f(1/z), 0)$ . It is NOT $f(\infty)$ . 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 $w = z^2$ is conformal at every $z \neq 0$ But it doubles the angle between curves at each Point. At $z = 0$ It is not conformal because $f'(0) = 0$ . :::

:::caution Common Pitfall The maximum modulus principle says that $|f|$ has no local maximum in the Interior, but the minimum can occur in the interior (e.g., $f(z) = z$ on the unit disk has minimum $|f| = 0$ at $z = 0$ ). For the minimum principle, one needs the additional hypothesis that $f$ 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 $\mathrm{Log}\, z$ has a branch cut along $(-\infty, 0]$ And is undefined on this cut. :::

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

15. Problem Set

Problem 1

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

Solution

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

$z = 2\,e^{5\pi i/6}$ .

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

$z_0 = 2^{1/3}\, e^{5\pi i/18}$ , $z_1 = 2^{1/3}\, e^{17\pi i/18}$ , $z_2 = 2^{1/3}\, e^{29\pi i/18}$ .

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

Problem 2

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

Solution

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

$u_x = 4x$ , $u_y = -4y$ , $v_x = 0$ , $v_y = 0$ .

CR: $4x = 0 \Rightarrow x = 0$ , $-4y = 0 \Rightarrow y = 0$ .

$f$ is differentiable only at $z = 0$ and analytic nowhere.

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

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

Problem 3

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

Solution

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

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

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

$u_x(1, 0) = -1/2 = f'(1)$ . $\checkmark$

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

Problem 4

Show that $u(x, y) = x^3 - 3xy^2 + 3x^2 - 3y^2$ is harmonic and find its harmonic conjugate.

Solution

$u_{xx} = 6x + 6$ , $u_{yy} = -6x - 6$ . $\Delta u = 0$ . $\checkmark$

By CR: $v_y = u_x = 3x^2 - 3y^2 + 6x$ . $v = 3x^2 y - y^3 + 6xy + g(x)$ .

$v_x = -u_y = 6xy + 6y$ . $6xy + 6y = 6xy + 6y + g'(x) \Rightarrow g'(x) = 0 \Rightarrow g(x) = C$ .

Harmonic conjugate: $v(x, y) = 3x^2 y - y^3 + 6xy + C$ .

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

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

Problem 5

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

Solution

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

$\int_\gamma (z^2 + 2z)\, dz = F(-1) - F(1) = \frac{2}{3} - \frac{4}{3} = -\frac{2}{3}$ .

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

Problem 6

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

Solution

On $\gamma$ : $|z| = 1$ So $|e^z| \leq e$ and $|z - 2| \geq 1$ .

$\left|\frac{e^z}{z - 2}\right| \leq e$ . $L = 2\pi$ .

$\left|\int_\gamma \frac{e^z}{z - 2}\, dz\right| \leq 2\pi e$ .

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

Problem 7

Evaluate $\oint_\gamma \frac{z + 1}{z^2 - z}\, dz$ where $\gamma$ is $|z| = 2$ .

Solution

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

At $z = 0$ : $\mathrm{Res} = \lim_{z \to 0} \frac{z + 1}{z - 1} = -1$ . At $z = 1$ : $\mathrm{Res} = \lim_{z \to 1} \frac{z + 1}{z} = 2$ .

$\oint_\gamma \frac{z + 1}{z^2 - z}\, dz = 2\pi i(-1 + 2) = 2\pi i$ .

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

Problem 8

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

Solution

$z = 0$ : $e^{1/z}$ has an essential singularity at $0$ So $z = 0$ is an essential singularity of $f$ . $z = i$ : simple pole. $z = -i$ : simple pole.

At $z = i$ : $\mathrm{Res} = \frac{e^{1/i}}{2i} = \frac{e^{-i}}{2i}$ . At $z = -i$ : $\mathrm{Res} = \frac{e^{1/(-i)}}{-2i} = \frac{e^{i}}{-2i}$ .

At $z = 0$ : find the coefficient of $1/z$ in $\frac{e^{1/z}}{z^2 + 1}$ . $\frac{1}{z^2 + 1} = 1 - z^2 + z^4 - \cdots$ near $z = 0$ . $e^{1/z} = 1 + 1/z + 1/(2z^2) + \cdots$ . The $1/z$ coefficient in the product: from $1 \cdot 1/z = 1/z$ Giving residue $1$ .

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

Problem 9

Evaluate $\int_0^{2\pi} \frac{\cos\theta}{5 + 4\cos\theta}\, d\theta$ .

Solution

Substitute $z = e^{i\theta}$ :

$I = \int_{|z|=1} \frac{(z + z^{-1})/2}{5 + 4(z + z^{-1})/2} \cdot \frac{dz}{iz} = \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 1}{z(2z^2 + 5z + 2)}\, dz = \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 1}{z(2z + 1)(z + 2)}\, dz$ .

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

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

$I = \frac{1}{2i} \cdot 2\pi i\left(\frac{1}{2} - \frac{5}{2}\right) = \pi(-2) = -\frac{\pi}{3}$ .

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

Problem 10

Evaluate $\int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)(x^2 + 4)}$ .

Solution

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

At $z = i$ : $\mathrm{Res} = \frac{1}{(2i)(i^2 + 4)} = \frac{1}{2i \cdot 3} = \frac{1}{6i}$ . At $z = 2i$ : $\mathrm{Res} = \frac{1}{(4i - 1)(4i)} = \frac{1}{4i(-3)} = -\frac{1}{12i}$ .

$\int_{-\infty}^{\infty} f(x)\, dx = 2\pi i\left(\frac{1}{6i} - \frac{1}{12i}\right) = 2\pi i \cdot \frac{1}{12i} = \frac{\pi}{6}$ .

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

Problem 11

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

Solution

$\frac{z}{z^2 + 4} = \frac{z}{4} \cdot \frac{1}{1 + z^2/4} = \frac{z}{4}\sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{4^n} = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{4^{n+1}}$

For $|z| \lt 2$ . Radius of convergence: distance from $0$ to nearest singularity ( $\pm 2i$ ), which is $2$ .

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

Problem 12

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

Solution

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

For $|z| \gt 1$ : $\frac{1}{z - 1} = \frac{1}{z} \cdot \frac{1}{1 - 1/z} = \sum_{n=0}^{\infty} z^{-n-1}$ .

For $|z| \lt 2$ : $\frac{1}{z - 2} = -\frac{1}{2} \cdot \frac{1}{1 - z/2} = -\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$ .

$f(z) = -\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}} - \sum_{n=0}^{\infty} z^{-n-1}$ .

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

Problem 13

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

Solution

On $|z| = 1$ : $|-5z| = 5 \gt |z^5 + 1| \leq 2$ .

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

So exactly one root in $|z| \lt 1$ .

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

Problem 14

Find the Möbius transformation that maps $1 \mapsto 0$ , $i \mapsto 1$ , $-1 \mapsto \infty$ .

Solution

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

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

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

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

That gives $-1$ Not $1$ . Let me recompute.

$T(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}$ with $z_1 = 1$ , $z_2 = i$ , $z_3 = -1$ .

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

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

$T(1) = 0$ . $\checkmark$ . $T(-1) = \infty$ . $\checkmark$ .

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

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

Problem 15

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

Solution

$\frac{z^3}{z^2 + 1}$ has simple poles at $z = \pm i$ Both inside $|z| = 2$ .

At $z = i$ : $\mathrm{Res} = \frac{i^3}{2i} = \frac{-i}{2i} = -\frac{1}{2}$ . At $z = -i$ : $\mathrm{Res} = \frac{(-i)^3}{-2i} = \frac{i}{-2i} = -\frac{1}{2}$ .

$\int_\gamma \frac{z^3}{z^2 + 1}\, dz = 2\pi i\left(-\frac{1}{2} - \frac{1}{2}\right) = -2\pi i$ .

Alternatively: $\frac{z^3}{z^2 + 1} = z - \frac{z}{z^2 + 1}$ . $\int_\gamma z\, dz = 0$ (entire), and $\int_\gamma \frac{z}{z^2 + 1}\, dz = 2\pi i(1/2 + 1/2) = 2\pi i$ . So the integral equals $0 - 2\pi i = -2\pi i$ . $\checkmark$

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

Problem 16

Show that $\int_{-\infty}^{\infty} \frac{\cos 2x}{x^2 + 1}\, dx = \frac{\pi}{e^2}$ .

Solution

Consider $\int_{-\infty}^{\infty} \frac{e^{2ix}}{x^2 + 1}\, dx$ .

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

$\mathrm{Res}\!\left(\frac{e^{2iz}}{z^2 + 1}, i\right) = \frac{e^{2i \cdot i}}{2i} = \frac{e^{-2}}{2i}$ .

$\int_{-\infty}^{\infty} \frac{e^{2ix}}{x^2 + 1}\, dx = 2\pi i \cdot \frac{e^{-2}}{2i} = \frac{\pi}{e^2}$ .

Taking real parts: $\int_{-\infty}^{\infty} \frac{\cos 2x}{x^2 + 1}\, dx = \frac{\pi}{e^2}$ .

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

Problem 17

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

Solution

$\sin z = z - z^3/6 + z^5/120 - \cdots$

$f(z) = \frac{z - z^3/6 + z^5/120 - \cdots}{z^4} = \frac{1}{z^3} - \frac{1}{6z} + \frac{z}{120} - \cdots$

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

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

Problem 18

Evaluate $\int_\gamma \frac{dz}{(z - 1)^2(z - 2)}$ where $\gamma$ is $|z - 1| = 1/2$ .

Solution

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

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

$\int_\gamma f\, dz = 2\pi i \cdot (-1) = -2\pi i$ .

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

Problem 19

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

Solution

$f(z) = x^2 + y^2 + 2x - 2iy$ . So $u = x^2 + y^2 + 2x$ , $v = -2y$ .

$u_x = 2x + 2$ , $u_y = 2y$ , $v_x = 0$ , $v_y = -2$ .

CR: $2x + 2 = -2 \Rightarrow x = -2$ And $2y = 0 \Rightarrow y = 0$ .

$f$ is differentiable only at $z = -2$ .

$f'(-2) = u_x(-2, 0) + iv_x(-2, 0) = (2(-2) + 2) + 0 = -2$ .

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

Problem 20

Evaluate $\int_\gamma \frac{e^z \sin z}{(z - \pi)^3}\, dz$ where $\gamma$ is $|z| = 4$ .

Solution

Only $z = \pi$ is inside $\gamma$ (a pole of order $3$ ).

By CIF for derivatives: $\int_\gamma \frac{f(z)}{(z - \pi)^3}\, dz = \frac{2\pi i}{2!}\,f''(\pi)$ Where $f(z) = e^z \sin z$ .

$f'(z) = e^z \sin z + e^z \cos z = e^z(\sin z + \cos z)$ . $f''(z) = e^z(\sin z + \cos z) + e^z(\cos z - \sin z) = 2e^z \cos z$ .

$f''(\pi) = 2e^\pi \cos\pi = -2e^\pi$ .

$\int_\gamma \frac{e^z \sin z}{(z - \pi)^3}\, dz = \pi i \cdot (-2e^\pi) = -2\pi i\, e^\pi$ .

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

Worked Examples

Example 1: Definite integration

Evaluate $\displaystyle\int_0^2 (3x^2 + 2x)\,dx$ .

Solution:

$\int (3x^2 + 2x)\,dx = x^3 + x^2 + c$

$\left[x^3 + x^2\right]_0^2 = (8 + 4) - (0) = 12$

Example 2: Integration by parts

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

Solution:

Let $u = x \implies \frac{du}{dx} = 1$ and $\frac{dv}{dx} = e^{2x} \implies v = \frac{1}{2}e^{2x}$ .

$\int x e^{2x}\,dx = x \cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x}\,dx = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + c = \frac{e^{2x}(2x - 1)}{4} + c$

Summary

Holomorphic functions: complex differentiable on an open set; Cauchy-Riemann equations $u_x = v_y$ , $u_y = -v_x$ are necessary conditions.
Cauchy’s integral theorem: $\oint_\gamma f(z)\,dz = 0$ for $f$ holomorphic inside and on $\gamma$ ; Cauchy’s integral formula evaluates $f(a)$ and all derivatives via contour integrals.
Residue theorem: $\oint_\gamma f(z)\,dz = 2\pi i \sum \text{Res}(f, z_k)$ ; 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.

Cross-References

Topic	Site	Link
Real Analysis	WyattsNotes	View
Multivariable Calculus	WyattsNotes	View
Differential Equations	WyattsNotes	View
Complex Analysis — MIT 18.04	MIT OCW	View

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