Residue Calculation for Integral of e^(1/z^2) over Circle |z|=2

[player1_1_1]

Homework Statement

\int\limits_Ce^{\frac{1}{z^2}}\mbox{d}z where C is circle |z|=2

The Attempt at a Solution

\sum\limits^{\infty}_{j=0}\frac{z^{-2j}}{j!}=1+\frac{1}{z^2}+\frac{1}{2!z^4}+\ldots
I=2\pi ic_{-1}=2\pi i
correct?

 

[jackmell]

c_(-1)=0

This is what you may want to consider: just solve it numerically to check your answer. It's easy in Mathematica:

NIntegrate[Exp[1/z^2] 2 I Exp[I t]/.z->2 Exp[I t],{t,0,2 Pi}]

 

[HallsofIvy]

No, that is not correct. Since that series contains an infinite number of terms with negative exponent, z= 0 is an "essential singularity", not a pole. You cannot use residues to find the integral.

 

[player1_1_1]

thanks for answer, why c_{-1}=0 then? maybe i wrote this series wrong?

 

[player1_1_1]

okay, and what if i had \int\limits_Ce^{\frac1z}\mbox{d}z, can i use residues here? is there pole in z=0?

 

[jackmell]

No, that is not correct. Since that series contains an infinite number of terms with negative exponent, z= 0 is an "essential singularity", not a pole. You cannot use residues to find the integral.

I believe you are in error sir. It is indeed an essential singularity but that does not change in the slightest the fact that the Residue Theorem still applies.

 

[jackmell]

okay, and what if i had \int\limits_Ce^{\frac1z}\mbox{d}z, can i use residues here? is there pole in z=0?

Yes you can but don't want to pick a fight with Hall. Now what do you think the answer is based on it's series representation? How can you now change the Mathematica code above and confirm your answer?

 

[jackmell]

thanks for answer, why c_{-1}=0 then? maybe i wrote this series wrong?

Well that one too: c_(-1) is the 1/z term right as in -1. So c_(-2) is the 1/z^2 term right? Same dif with the rest of them.