What is the Argument of (z+i)/(z-1) when it equals 3π/2?

[Aranc]

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:
(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

 

[I like Serena]

Welcome to PF, Aranc!

Suppose you had to solve arg(z)=3π/2.
What would z be?

 

[Aranc]

thanks for the welcome:)
z would be any number smaller then 0 on the I am axis?

 

[I like Serena]

Yes!

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

Can you solve z from this?

 

[Aranc]

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

 

[I like Serena]

Yep!

 

[Aranc]

thanks a lot! now I hope I can manage the questions on the exam about complex numbers:) don't have to think to difficult..