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

  • Thread starter Aranc
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In summary, the conversation is about a student seeking help with a calculus exam question involving the argument of a complex number, and another person guiding them through the solution. They determine that the complex number z must be on the negative imaginary axis and eventually arrive at the solution z = (i-ri)/(-ri-1), with r being a positive natural number. The student expresses gratitude and hopes to do well on the exam.
  • #1

Homework Statement

Hey all. I am trying to get ready for my upcomming calculus exam, but I am 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:
then multiply it with its conjugated complex
then I get:
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
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  • #2
Welcome to PF, Aranc! :smile:

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

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

Can you solve z from this?
  • #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?:)
  • #6
Yep! :smile:
  • #7
thanks a lot! now I hope I can manage the questions on the exam about complex numbers:) don't have to think to difficult..

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