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

sorry I made a mistake in typing the question, its ARG((z+i)/(z-1))=3pi/2

Somewhat confusing as I first looked at 3pi/2 as (pi) being the real constant 3.14159..... Once I lost that confusion and looked at 3/2pi as a*i with a=3p/2 with p (and a) as a real number it became a method of how to express z as a complex number, which is algebra.

(z+i)/(z-1)=ai
z+i=ai(z+1)
z+i=azi+ai
z-azi=ai-i
z(1-ai)=(a-1)i
z=(a-1)i/(1-ai)
now make the denominator real by multiply both sides by 1= (1+ai)/(1+ai).

z=(a-1)(1+ai)i/(1+a^2) with a=3p/2

BTW further simplification is possible.

Rap

Ive been thinking a bit more, and it seems to be a lot more simple then I actually thought:
arg((z+i)/(z-1))=3pi/2
substitution:
arg(w)=3pi/2
then w = -ri with r being any positive number..
so
w=(z+i)/(z-1)
(z+i)/(z-1)=-ri
-ri(z-1)=z+i
-riz-z=i-ri
z(-ri-1)=i-ri
z=(i-ri)/(-ri-1)
it quite simple actualle:)

Oh rap, didnt see your post until I finished this one.. with pi I did mean 3.14159 a.s.o. but thanks a lot, you gave me a whole new insight into some other problems:)