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Help me with the residue.

  1. Jun 20, 2013 #1
    Hello guys,
    I just wanna confirm about this problem ..Find the residue of this function: f(z)=e1/z/(1-z)

    Thx in advance.
     
  2. jcsd
  3. Jun 20, 2013 #2

    HallsofIvy

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    "Confirm"? You mean you have already found an answer? Great! Tell us what you got and we will try to confirm it for you.
     
  4. Jun 20, 2013 #3
    Right :p
    the answer should be "exp"?
     
  5. Jun 20, 2013 #4

    Office_Shredder

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    The answer should be a number, not a function.

    Also, there are two different points where you might be interested in the residue - at z=0 and z=1. Which one are you supposed to find?
     
  6. Jun 20, 2013 #5
    at z°=1

    here is how I've done it ..

    Res=limit [f(z)*(z°-z)] as z goes to z°
    =limit [e1/z/[STRIKE](1-z)[/STRIKE]*[STRIKE](z°-z)[/STRIKE]] as z goes to z°
    =limit [e1/z] as z goes to z° ''which equals to 1''
    and finally I plugged in z=z°=1 and I got: Res=e1/1=e≈2.72

    am I wrong somewhere?
     
  7. Jun 20, 2013 #6

    Office_Shredder

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    That looks correct except that it should be f(z)*(z-z0) which means you're off by a minus sign.

    Usually "exp" refers to the function ex, not the number e itself, which is why I wrote the first part of my previous post.
     
  8. Jun 20, 2013 #7
    Thx guys for these replies..I just want to know one last thing. using the residue theorem, what is the integral of f(z) ...at |z|=1/2

    Thx again :)
     
  9. Jun 20, 2013 #8

    Office_Shredder

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    Your question isn't very well posed... Are you integrating around the circle defined by |z|=1/2?
     
  10. Jun 21, 2013 #9
    Im sry about that, yes integration around the circle .
     
  11. Jun 23, 2013 #10
    anyone can help me with the integration plz? f(z)=e1/z/(1-z)
    ∫f(z)dz around the circle |z|=1/2 ?
     
  12. Jun 23, 2013 #11
    You need to identify the poles within the circle |z| = 1/2. You might consider expanding the exponential in a series.
     
  13. Jun 23, 2013 #12
    All I need is the result, so can u provide me with that? thx
     
  14. Jun 23, 2013 #13

    Office_Shredder

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    No, we can't. We can help guide your work but we will not tell you what the answer is.

    There is a single pole at z=0 inside of the circle (it should be fairly obvious from looking at the function). Can you find the residue at that pole?
     
  15. Jun 23, 2013 #14
    But I thought the pole is z=1, no?
     
  16. Jun 24, 2013 #15
    This function has more than one pole. Look at the 1/z in the exponential.
     
  17. Jun 24, 2013 #16
    Right, ty person ;)
     
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