[SOLVED] Application of Cauchy's Integral Formula

**davismj** · February 16th 2010, 01:40 PM

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?

**mabruka** · February 16th 2010, 03:55 PM

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.

**davismj** · February 16th 2010, 04:13 PM

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.

**mabruka** · February 16th 2010, 05:33 PM

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