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Need help

  1. Feb 25, 2007 #1
    Complex Analysis:
    Find all solutions of:
    (a) e^z = 1
    (b) e^z = 1 + i
     
  2. jcsd
  3. Feb 25, 2007 #2
    Have you at least made attempts to solve them?
     
  4. Feb 25, 2007 #3

    arildno

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    Dearly Missed

    How fascinating!
    We are all desirous to see what you have done so far! :smile:
     
  5. Feb 27, 2007 #4
    So far I got:
    e^z = 1
    I know e^z = e^x (cos y + i Sin y)
    Then,
    e^x cos y = 1
    e^x sin y = 0
    I know that for sin y = 0, y = 2 pi * K, where k is an integer.
    After this step I'm stuck!

    The second problem: e^z = 1 + i
    e^x cos y = 1
    e^x sin y = 1
    I order for sin y = 1, y = pi/2. Now what...
     
  6. Feb 28, 2007 #5
    can you express the RHS in polar form?
     
  7. Mar 1, 2007 #6

    HallsofIvy

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    Actually y= [itex]\pi K[/itex], since [itex]sin(\pi)= 0[/itex] also. Now, what values can cos y have? And then what is ex?


    No, exsin y= 1 does NOT tell you that sin y= 1! You might try this: square each equation and add. That will get rid of y so you can find ex (you don't really need to find x itself).
     
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