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.