Complex Analysis: Zeros, Poles, and Residues

This flashcard set covers the identification of zeros, poles (simple, second order, and order 6), removable singularities, essential singularities, and residue calculations for various complex functions based on the lecture notes by Steffen Solgren.

Zeros of $\tan^3(\frac{z}{2})$

Located at z = n\text{\pi} where $n \in \mathbb{Z}$, specifically with an order of $3$.

Simple pole of $f(z) = \frac{(z+2i)^2}{z-i}$

A singularity occurring at $z = i$ where the denominator $z - i = 0$.

Second order poles of $f(z)$ (per lecture example)

Singularities located where $z + 2i = 0 \Rightarrow z = -2i$ and $z - i = 0 \Rightarrow z = i$.

Removable singularity of $f(z) = \frac{\sin(z)}{z-\pi}$

Located at $z = \pi$, because the expression $\frac{\sin(z)}{z-\pi} \rightarrow -1$ as $z \rightarrow \pi$.

Pole of order 6 at $z=0$

The classification of the singularity for the function $f(z) = \frac{\sin(2z)}{z^6}$ at the origin.

Essential singularity of $f(z) = e^{\frac{1}{1-z}}$

A singularity located at $z = 1$, where the function cannot be defined by a pole of any finite order.

Residue of $f(z) = e^{\frac{1}{1-z}}$ at $z = 1$

The coefficient $b_{-1}$ in the Laurent series expansion, calculated to be $-1$.

Simple poles of $f(z) = \frac{z-23}{z^2-4z-5}$

Located at $z = -1$ and $z = 5$, derived from factoring the denominator as $(z+1)(z-5)$.

Residue of $f(z) = \frac{z-23}{z^2-4z-5}$ at $z = -1$

The value calculated using the limit or derivative formula, which equals $4$.

Residue of $f(z) = \frac{z-23}{z^2-4z-5}$ at $z = 5$

The value calculated using the limit or derivative formula, which equals $-3$.

Residue of $f(z) = \frac{z \cosh(\pi z)}{z^4 + 13z^2 + 36}$ at $z = \pi i$

The coefficient of the $\frac{1}{z}$ term in the series expansion, identified as $b_{-1} = \frac{1}{12}$.

Upper half unit circle poles for $f(z) = \frac{1}{z^4+1}$

The simple poles located at $z = e^{i\frac{\pi}{4}}$ and $z = e^{i\frac{3\pi}{4}}$.

Real poles of $f(x) = \frac{x+5}{x(x+1)(x-1)}$

The singularities located on the real axis at $x = 0$, $x = -1$, and $x = 1$.

Residue of $f(x) = \frac{x+5}{x(x+1)(x-1)}$ at $x = 0$

Calculated by evaluating the remaining parts of the function at the pole, resulting in $-5$.