Residue Theorem: Finding the Integral of z^3e^(-1/z^2) over |z|=5

[cragar]

Homework Statement

use the residue theorem to find the value of the integral,
integral of $z^3e^{\frac{-1}{z^2}}$ over the contour |z|=5

The Attempt at a Solution

When I first look at this I see we have a pole at z=0 , because we can't 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 can't seem to get it using the residue theorem.

 

[FactChecker]

Homework Statement

use the residue theorem to find the value of the integral,
integral of $z^3e^{\frac{-1}{z^2}}$ over the contour |z|=5

The Attempt at a Solution

When I first look at this I see we have a pole at z=0 , because we can't 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,

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

 

[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, I am trying to find the value of that integral over the circle |z|=5, and I am trying to use the residue theorem ,

 

[FactChecker]

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, I am trying to find the value of that integral over the circle |z|=5, and I am trying to use the residue theorem ,

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) = lim_z->0(z⋅f(z))? Or are you trying some other way to find the coefficient of the 1/z term of the Laurent series?

 

[cragar]

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

 

[FactChecker]

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

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

 

[cragar]

thanks for helping me clear it up.