Coursera Complex Analysis Lesson 7.4 Finding Residues

Aug 19, 2016

IMPORTANT

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

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