# Remind me how to do this complex integral

1. Sep 7, 2014

integrate e^z/(1-cosz) dz over circle of radius, say 2

i can't seem to recall how it is done.

singularity at z=0

2*pi*i * res (at z=0) would be the solution

any shortcut to find this residue?

Last edited: Sep 7, 2014
2. Sep 7, 2014

btw, is there any kind of theorem saying integral of 1/(1-sinz) dz over circle of radius 2 would be zero? this one has 2 singularities at equal distance apart. i feel like there's a theorem somewhere. anymore care to refresh my memory?

3. Sep 8, 2014

4. Sep 8, 2014

### WWGD

The only one I know is the one where you use the sum of the residues at each of the singularities, so it would integrate to 0 only if the residue at one singularity is the negative
of the other one.

5. Sep 8, 2014

nvm what i said before. the only singularity inside our contour is z = 0. rest are outside and don't even matter.

and yes

integral = 2*pi*i * [sum of residues at each singularities]

6. Sep 8, 2014

i've tried several methods (that i can think of) for this problem and none of them seem to work.

also tried expanding as series and didn't get anywhere.

7. Sep 8, 2014

### WWGD

If you have done the expansion, the residue is the coefficient of 1/z , since all other terms integrate to 0.

8. Sep 8, 2014

why are you integrating?

this is what i was doing

e^z = 1+z+z^2/2 + O(z^3)
1-cosz = z^2/2-z^4/4! + O(z^6)

i couldn't really single out 1/z part

9. Sep 8, 2014

### WWGD

No I'm not integrating, just stating that if you did term-by-term integration of the series, the only term that would be nonzero is the term containing 1/z. Let me think it thru a bit more.