Integrate (z^2 - 4)/(z^2 + 4) Around |z - i|=2: -4π

[geft]

Homework Statement

Integrate (z^2 - 4)/(z^2 + 4) counterclockwise around the circle |z - i| = 2.

Homework Equations

Cauchy's integral formula

The Attempt at a Solution

|z - i| = 2
|z - 2| = i

z0 = 2

(z^2 - 4)/(z^2 + 4)
= ((z + 2)(z - 2))/(z^2 + 4)
= (z + 2)/(z^2 + 4) * i

f(z) = (z + 2)/(z^2 + 4)

2i*pi*f(z0) = 2i*pi*(1/2) = i*pi

The answer is -4*pi. Please tell me what I'm doing wrong.

 

[jackmell]

I can give you a very good piece of advice: you're got to make that look nicer so that it's easier to "see":

$$\mathop\oint\limits_{|z-i|=2} \frac{z^2-4}{z^2+4}dz=\mathop\oint\limits_{|z-i|=2} \frac{z^2-4}{(z+2i)(z-2i)}dz=\mathop\oint\limits_{|z-i|=2} \frac{z^2-4}{z+2i}\frac{1}{z-2i}dz$$

Now we in the big house. Can you now just apply Cauchy's Integral formula with:

$$f(z)=\frac{z^2-4}{z+2i}$$

 

[geft]

But isn't the factor z - i? How can you factor z - 2i?

Also, 2*pi*i*f(i) = 2*pi*i*(-5/3i) = -10*pi/3