MATH3401

(1) \(\,f(z)=\ds\frac{\sin z}{z}\): \(z=0\) is a removable singularity.

\[g(z)= \left\{\array { \frac{\sin z}{z} & z\neq 0 \cr 1 & z=0 } \right.\] is entire and \(g(z) = \ds \sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n+1)!}\). Also note \(\underset{z=0}{\res} g(z) = 0\).

(2) \(\,f(z)=\ds\frac{1}{z^4}\): analytic on \(\C_*\), isolated singularity at \(z=0\).

This is a pole of order 4, (largest negative power occuring).

Here \(\underset{z=0}{\res} f(z) = 0\) (coefficient of the \(z^{-1}\) term).

(3) \(\,h(z)=\ds\frac{\sinh z}{z^4}\): analytic on \(\C_*\), isolated singularity at \(z=0\).

On \(\C_*\): \[ h(z) = \frac{1}{z^4}\left( z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots \right) \] \[\;=\frac{1}{z^3}+\frac{1}{3!z}+\frac{z}{5!}+\cdots\]

Pole of order 3 at $z=0$; here \(\underset{z=0}{\res} h(z) = \frac{1}{3!}=\frac{1}{6}\).

(4) \(\,f(z)=e^{1/z}\): analytic on \(\C_*\), isolated singularity at \(z=0\).

On \(\C_*\): \[ f(z) =\ds \sum_{n=0}^{\infty}\frac{1}{n!z^n} \] Essential singularity at $z=0$. Here \(\underset{z=0}{\res} f(z) = 1\).

(5) \(\,f(z)=\ds\frac{1}{1-\frac{1}{z}}\), \(z\neq 0\): isolated singularity at \(z=0,1\).

For \(z\neq 0,1\): \( \quad f(z) =\ds \frac{z}{z-1} \) \(=\ds \frac{-z}{1-z}\) \[=-z\left(1+z+z^2+\cdots \right) \qquad\quad\;\; \] \[=-z-z^2-z^3-z^4-\cdots \qquad\quad\;\; \] Removable singularity at $z=0$. Taking \(f(0)=0\) extends \(f\) to an anlaytic function on \(\{z:|z|\lt 1\}\). What happens at \(1\)? 📝

Theorem 1: An isolated singularity \(z_0\) of \(f\) is a pole of order \(m\ge 1\) at \(z_0\), iff \(f\) can be written near \(z_0\) as \[ f(z) = \frac{\phi (z) }{(z-z_0)^m} \] where \(\phi\) is analytic on some \(B_{R}(z_0)\) and \(\phi(z_0) \neq 0\).

Theorem 2: In this case \[ \underset{z=z_0}{\res} f =\frac{\phi^{(m-1)}(z_0)}{(m-1)!}. \quad(*) \] In particular, for \(m=1\) (simple pole): \[ \underset{z=z_0}{\res} f =\frac{\phi^{(0)}(z_0)}{0!} = \phi(z_0). \quad(**) \]