# [SOLVED] Application of Cauchy's Integral Formula

• Feb 16th 2010, 01:40 PM
davismj
[SOLVED] Application of Cauchy's Integral Formula
http://i50.tinypic.com/2dhgx0x.jpg

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

The book says the answer is $\displaystyle 2\pi i$, not $\displaystyle -2\pi i$. Who is right?
• Feb 16th 2010, 03:55 PM
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.
• Feb 16th 2010, 04:13 PM
davismj
Quote:

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 $\displaystyle \int \! \frac{z^2 (\frac{1}{2 - z}) \, dz}{z - (-2)}$

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

Cool, thanks.
• Feb 16th 2010, 05:33 PM
mabruka
(Rofl)