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I am stuck calculating the integral $$\int_{\gamma_1} \frac{dz}{z(z-i)}$$ over $\gamma_1 = Re^{it}, R>1$.

If I had to integrate over $\gamma_2 = re^{it}, r < 1$, I could just expand the integrand into a power series (using the geometric series) around $z=0$, but with $R>1$ this approach won't work. I quite frankly don't have any other idea on how to approach this.

Any hints?

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The residue theorem is unavailable to you I take it? – anon May 13 '12 at 17:53
Use partial fractions and then use the geometric series expansion on both of them, if you're really that averse to using the residue theorem. – Potato May 13 '12 at 18:26
He didn't say "homework" so why would the Residue Theorem be off-limits? – GEdgar May 13 '12 at 18:49
The appropriate geometric series in this case is in powers of $1/R$. – Robert Israel May 13 '12 at 18:52
Yes, residue theorem is off-limits. – gnirx May 14 '12 at 18:26
up vote 1 down vote accepted

The easy answer here is to use the residue theorem. I assume because you are asking this question, you have not yet learned about it.

Instead, you could note $\frac{1}{z(z-i)}=i(\frac{1}{z}-\frac{1}{z-i})$. The $\frac{1}{z}$ term you can integrate directly. For the other term, note that on this contour we have $|z|>|i|$, so $1>|i/z|$. Then we can write the second term as $\frac{1}{z}\frac{1}{1-i/z}=\frac{1}{z}(1+i/z+(i/z)^2+...)$. You can then multiply out and integrate term by term.

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Thanks a lot. I actually tried using partial fraction decomposition, but was too dumb to note $|\frac{i}{z}|<1$. :) – gnirx May 14 '12 at 18:23

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