Finding Residue using Cauchy's Integral & Residue Thms

[brandones]

Homework Statement

Calculate

$$\stackrel{Res}{z=0}(\frac{z^5}{(z^2-4)(z-4))}$$

Homework Equations

I've learned enough Latex for one day, thank you very much. Wikipedia:

http://en.wikipedia.org/wiki/Residue_theorem

http://en.wikipedia.org/wiki/Cauchy's_integral_formula

The Attempt at a Solution

$$\stackrel{Res}{z=0}(\frac{z^5}{(z^2-4)(z-4))}$$

via Cauchy's Residue Theorem:
= $$\frac{1}{2πi}\int \frac{z^5}{(z-2)(z+2)(z-4)} dz$$

which, via Cauchy's Integral Formula:
= $$(\frac{z^5}{(z+2)(z-4)})$$ solved at z=2

= $$\frac{32}{-8}$$

=-4

Is this legit? This seems way too convenient, and weird that there isn't an explicit theorem saying I can do this.

 

[brandones]

Oh wait, I just went a really roundabout way of defining the residue, didn't I?

Well, now it seems totally obvious.

Hooray!

 

[ideasrule]

Homework Statement

Calculate

$$\stackrel{Res}{z=0}(\frac{z^5}{(z^2-4)(z-4))}$$

You mean at z=2, right?

Oh wait, I just went a really roundabout way of defining the residue, didn't I?

Yeah, for simple poles at z=c, the residue is just the limit of (z-c)*f(z).