Residue of an integral

1. Apr 24, 2017

cragar

1. The problem statement, all variables and given/known data
use the residue theorem to find the value of the integral,
integral of $z^3e^{\frac{-1}{z^2}}$ over the contour |z|=5
3. The attempt at a solution
When I first look at this I see we have a pole at z=0 , because we cant divide by zero in the exponential term.
and a pole of order 2, So I multiply the function by the function that causes the singularity , and take the first derivative of that and evaluate it at z=0,
this give me $2pi*i[3z^2]$ but this gives me 0,
my book says the answer should be i*pi , which I can find from the Laurent series , but I cant seem to get it using the residue theorem.

2. Apr 24, 2017

FactChecker

I'm not sure that I understand. Are you directly calculating Res(f,0) = limz->0(z⋅f(z))?

3. Apr 24, 2017

cragar

so the exponential function, does not have a pole or a removable singularity, since it does not have a pole , I think I just have to expand it in a Laurent series to find the residue, im trying to find the value of that integral over the circle |z|=5, and I am trying to use the residue theorem ,

4. Apr 24, 2017

FactChecker

I understand that you are trying to use the residue theorem, but what are the details of how you doing that? Are you directly calculating Res(f,0) = limz->0(z⋅f(z))? Or are you trying some other way to find the coefficient of the 1/z term of the Laurent series?

5. Apr 24, 2017

cragar

I expanded the exponential as a Laurent series , then I multiplied by z^3, and then found the 1/z term , other ways wont work because its not a pole ,

6. Apr 24, 2017

FactChecker

I'll buy that. It is an essential singularity.

7. Apr 24, 2017

cragar

thanks for helping me clear it up.