What is the residue of e^(2/z)/(1+e^z) at z = pi i?

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I need to find the residue of
<br /> \frac{e^{2/z}}{1+e^z}<br />

at z = \pi i

I've been scribbling over numerous papers trying to figure this out. So far I've tried to expand the denominator

<br /> \frac{e^{2/z}}{(1-e^z + \frac{e^{2z}}{2!} - ...)}<br />

I think maybe I've expanded that incorrectly, but I was thinking about pulling an e^z out of the denom and multiplying the entire function f(z) by the expanded 'leftovers', but I think that's incorrect too..
<br /> \frac{e^{2/z}}{e^z(\frac{1}{e^z}-1+1-...)}<br />

I feel like my steps are misguided because I can't seem to see where to go next.
 
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mateomy said:
I need to find the residue of
<br /> \frac{e^{2/z}}{1+e^z}<br />

at z = \pi i

I've been scribbling over numerous papers trying to figure this out. So far I've tried to expand the denominator

<br /> \frac{e^{2/z}}{(1-e^z + \frac{e^{2z}}{2!} - ...)}<br />

I think maybe I've expanded that incorrectly, but I was thinking about pulling an e^z out of the denom and multiplying the entire function f(z) by the expanded 'leftovers', but I think that's incorrect too..
<br /> \frac{e^{2/z}}{e^z(\frac{1}{e^z}-1+1-...)}<br />

I feel like my steps are misguided because I can't seem to see where to go next.

You can find residues by solving limit problems. Do you know that way? And yes, you are expanding incorrectly. But that's not really the way to do it anyway.
 
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Would it need to be a L'Hopital statetgy? (I'm going to try it). Also, I know this is the worst possible thing to say, according to a professor of mine, but I've gotten lazy and not practiced Taylor/Macluarin expansions in a while. I was under the impression that I could just expand 1/(1+x) and replace all the x's with e^x. What did I do wrong in there?
 


mateomy said:
Would it need to be a L'Hopital statetgy? (I'm going to try it). Also, I know this is the worst possible thing to say, according to a professor of mine, but I've gotten lazy and not practiced Taylor/Macluarin expansions in a while. I was under the impression that I could just expand 1/(1+x) and replace all the x's with e^x. What did I do wrong in there?

You can do that. You didn't do it right. What's the expansion of 1/(1+x)? But the expansion won't help you find the residue. Yes, write a limit and use l'Hopital.
 


Dick said:
...the expansion won't help you find the residue.

I know. Just looking to fill in memory lapses with my Calculus.

Thanks for the guidance.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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