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1. [SOLVED] Application of Cauchy's Integral Formula

Here's what I got (since $\frac{1}{2 - z} = \frac{-1}{z-2}$).

The book says the answer is $2\pi i$, not $-2\pi i$. Who is right?

2. The book is right. Check you have got the right singularity.

With "right" i mean the one who is inside your curve and apply the theorem as you just did.

3. Originally Posted by mabruka
The book is right. Check you have got the right singularity.

With "right" i mean the one who is inside your curve and apply the theorem as you just did.
Oh, because the curve is |z+1|= 2, a = -2 is inside the curve, and a = 2 is not. Therefore, I get $\int \! \frac{z^2 (\frac{1}{2 - z}) \, dz}{z - (-2)}$

$= f(-2) = 2\pi i(-2)^2(\frac{1}{2 - (-2)}) = 2\pi i$

Cool, thanks.
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