Problem Set

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

$f(z) = (x + iy)^2 + (x - iy)^2 = 2(x^2 - y^2)$. 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).

$f(z) = 1/(z^2 + 1)$ 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).

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

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).

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).

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).

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

$z = 0$: $e^{1/z}$ 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).

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).

$f(z) = \frac{1}{(z^2 + 1)(z^2 + 4)}$. 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).

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

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).

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

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).

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).

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

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).

$\frac{z^3}{z^2 + 1}$ 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).

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).

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

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

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).

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).

$f(z) = x^2 + y^2 + 2x - 2iy$. 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).

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).