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

  • Thread starter Aranc
  • Start date
  • #1
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Homework Statement


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



Homework Equations


ehm none


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
 

Answers and Replies

  • #2
I like Serena
Homework Helper
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Welcome to PF, Aranc! :smile:

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