## Slide 2 Restate Residue Theorem

• f has isolated singularity at $z_0$ and is analytic on the punctured disc 0 < | z- $z_0$| < r
• f has a unique Laurent series representation …
• The residue of f at $z_0$ is represented by Res(f, $z_0$) = $a_1$ the coefficient of $\frac{1}{z - z_0}$

Following slides are to calculate residues at removable singularities and poles (not essential singularities?)

## Slide 3 Residue at Removable singularity $z_0$ is a removable singularity of a Laurent series if there are no negative powers of the terms $(z -z_0)^k$

so $a_1$ = 0

so Res(f, $z_0$) = 0

## Slide 4 Residue at simple pole $f(z) = \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0)^1 + a_2(z-z_0)^2 + ...$ (*)

How to isolate $a_1$? Cross-multiply then take limit as z tends to $z_0$ $(z-z_0)f(z) ={a_{-1}} + a_0(z-z_0)^1 + a_1(z-z_0)^2 + a_2(z-z_0)^3 + ...$ $Res(f, z_0) = \underset{{z \to z_0}}{\lim}(z-z_0)f(z)$

We need to take limit because of the division by $z - z_0$ in the original Laurent series (*)

## Slide 5 Residue at simple pole example

(From previous slide) $Res(f, z_0) = \underset{{z \to z_0}}{\lim}(z-z_0)f(z)$

Consider $f(z) = \frac{1}{z^2 + 1}$ has simple poles at $z_0$= +i, -i $Res(\frac{1}{1+z^2}, i) = \underset{{z \to i}}{\lim}(z-i)\frac{1}{z^2 + 1}$ $= \underset{{z \to i}}{\lim}(z-i)\frac{1}{(z+i) (z-i)} = \underset{{z \to i}}{\lim}\frac{1}{(z+i)} = \frac{1}{2i}$

## Slide 6 Residue at double pole example

Laurent series: $f(z) = ...+ \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0)^1 + a_2(z-z_0)^2 + ...$ (*) $z_0$ is double pole when $a_{-2} \neq 0 \textrm{ and } a_{-k} = 0$ for k > 2.

To isolate $a_{-1}$. (i) Multiply by (z –  z_0)^2 and (ii) differentiate (iii) take limit. $f(z) = \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0)^1 + a_2(z-z_0)^2 + ...$ $(z-z_0)^2f(z) = {a_{-2}} + {a_{-1}}{(z-z_0)}^1 + a_0(z-z_0)^2 + a_1(z-z_0)^3 + a_2(z-z_0)^4 + ...$ $\frac{d}{dz}(z-z_0)^2f(z) = {a_{-1}} + 2a_0(z-z_0)+ 3a_1(z-z_0)^2 + ...$ $Res(f, z_0) = a_{-1}= \underset{{z \to z_0}}{\lim}\frac{d}{dz} (z-z_0)^2f(z)$

## Slide 7 Residue at double pole example $f(z) = \frac{1}{(z-1)^2(z-3)}$ double pole at z = 1, simple pole at z = 3. $Res(\frac{1}{(z-1)^2(z-3)}, 1) = \underset{{z \to 1}}{\lim}\frac{d}{dz} (z-1)^2\frac{1}{(z-1)^2(z-3)}$ $= \underset{{z \to 1}}{\lim}\frac{d}{dz} \frac{1}{(z-3)}$ $= \underset{{z \to 1}}{\lim} \frac{-1}{(z-3)^2} =-\frac{1}{4}$

## Slide 8 Residue at poles of order n

Use method above: (i) Multiply by (z –  z_0)^n and (ii) differentiate multiple times (iii) take limit.

Giving

Res(f,z_0) = $\frac{1}{(n-1)!}\underset{{z \to z_0}}{\lim}\frac{d^{n-1}}{dz^{n-1}}[(z - z_0)^n f(z)]$

## Slide 9 Residue of “quotient” functions

f(z) = $\frac{g(z)}{h(z)}$

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