Integrate complex function over unit circle

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  • #1
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Homework Statement


Calculate ##\int _Kz^2exp(\frac{2}{z})dz## where ##K## is unit circle.


Homework Equations





The Attempt at a Solution



Hmmm, I am having some troubles here. Here is how I tried:

In general ##\int _\gamma f(z)dz=2\pi i\sum_{k=1}^{n}I(\gamma,a_k)Res(f,a_k)## where in my case ##I(\gamma , a_k)=1##.

Now ##Res(f,a_k)##:

##f(z)=z^2exp(\frac{2}{z})=z^2\sum_{n=0}^{\infty}\frac{1}{n!}(\frac{2}{z})^n=\sum_{n=0}^{\infty}\frac{1}{n!}\frac{2^n}{z^{n-2}}##

To get ##\frac{1}{z}## clearly ##n-2=1## so ##n=3##.

Which gives me ##Res(f , 0)=\frac{4}{3}##.

Therefore ##\int _Kz^2exp(\frac{2}{z})dz=2\pi i \frac{4}{3}##.


Or is that completely wrong? Thanks in advance!
 

Answers and Replies

  • #2
This is absolutely correct!

EDIT: Remember you can always verify your Laurent series (the hardest part in my opinion) using WolframAlpha :D
 
  • #3
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Thank you! =)
 
  • #4
HallsofIvy
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You know, I presume that [itex]e^x= 1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot[/itex].

So [itex]e^{2/z}= 1+ 2/z+ 2/z^2+ 4/(3z^3)+ \cdot\cdot\cdot[/itex] and then
[itex]z^2e^{2/z}= z^2+ 2z+ 2+ 4/(3z)+ \cdot\cdot\cdot[/itex]

The integral around the closed circle of any power of z (positive or negative), except -1, is 0 and the integral of [itex]z^{-1}[/itex] is [itex]2\pi i[/itex] (which is where the idea of "residue" comes from) so the integral of [itex]z^2e^{2/z}[/itex] around the unit circle is [itex](4/3)(2\pi i)= (8/3)\pi i[/itex].
 
  • #5
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hehe, nicely said Hallsoflvy. Everything is a bit more obvious now. Thank you!
 

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