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Argument of (z+i)/(z-1)=3pi/2

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Hey all. Im trying to get ready for my upcomming calculus exam, but Im kind of stuck here. I have to solve: arg((z+i)/(z-1))=3π/2



    2. Relevant equations
    ehm none


    3. The attempt at a solution
    Tried to fill in z=a+bi:
    (a+bi+i)/(a-1+bi)
    then multiply it with its conjugated complex
    (a-1-bi)/(a-1-bi)
    then I get:
    (a2-a-abi+abi-bi-b2+ai-i-b)/(a2-a-abi-a+1+bi+abi-bi+b2)
    simplify:
    =(a2-a-b2-b-bi+ai-i)/(a2-2a+1+b2)..
    sure, I got rid of the complex part in the denumerator, but I have no idea what to do next..any help would be appriciated
     
  2. jcsd
  3. Nov 4, 2011 #2

    I like Serena

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    Welcome to PF, Aranc! :smile:

    Suppose you had to solve arg(z)=3π/2.
    What would z be?
     
  4. Nov 4, 2011 #3
    thanks for the welcome:)
    z would be any number smaller then 0 on the Im axis?
     
    Last edited: Nov 4, 2011
  5. Nov 4, 2011 #4

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    Yes! :wink:

    So ((z+i)/(z-1)=-r i for some real number r > 0.

    Can you solve z from this?
     
  6. Nov 4, 2011 #5
    I think so,
    (z+i)/(z-1)= -r i with r = a positive natural number
    -r i (z-1) = z + i
    -r i z + r i = z + i
    -r i z - z = i - r i
    z (-r 1 -1) = i - r i
    so z = (i-ri)/(-ri-1) with r = a positive natural number

    this is it right?:)
     
  7. Nov 4, 2011 #6

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  8. Nov 4, 2011 #7
    thanks a lot! now I hope I can manage the questions on the exam about complex numbers:) dont have to think to difficult..
     
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