Circle difference in contour integral

Question

Let's say I am integrating a function over $|z| = 1$ and $|z-1| = 1$, is there any difference? I think the answer for both cases will be same, as in both cases,

$$ z = \exp^{i\Theta} $$ and $$ dz = i \exp^{i\Theta} d\Theta $$

Please correct me if I am wrong.

Thanks @Dr.MV, I thought the center won't be included in the equation of z. This solves my problem. — simple, Mar 30 '15 at 03:43

score 0 · Accepted Answer · answered Mar 30 '15 at 03:45

0

$$ \int\frac{dz}{z+\frac 1 2} $$

Integrated over $|z-1|=1$, this is $0$, since it is holomorphic everywhere on and inside that circle. Integrated over $|z|=1$, running counterclockwise, this is $2\pi i$, because of its pole inside that circle.

answered Mar 30 '15 at 03:45

Michael Hardy

1

Why will it be $2\pi i$? – simple Mar 30 '15 at 03:52
Ok got it now, Cauchy integral formula. Thanks. – simple Mar 30 '15 at 04:04

Circle difference in contour integral

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