Question

if |z| = 1 and w = (z-1)/(z+1) z not equal -1 then Re(w) is?

Answer 1

\[z=x+iy\]\[x^2+y^2=1\]\[w=\frac{x-1+iy}{x+1+iy}\]rationalize the denum see what happens

Answer 2

lol...i did nt see any thing.

Answer 3

lol...me too

Answer 4

answer will not be a number ha?

Answer 5

|z|=1 so we can write\[z=e^{i\theta}\]right?

Answer 6

The answer shuld be 0

Answer 7

thats it

Answer 8

\[w=\frac{x-1+iy}{x+1+iy}=\frac{x-1+iy}{x+1+iy}\frac{x+1-iy}{x+1-iy}=\frac{x^2+y^2-1+2xyi}{(x+1)^2+y^2}\]but x^2+y^2=1 so\[w=\frac{2xyi}{(x+1)^2+y^2}\]ok

Answer 9

Yup..

Answer 10

well we are done

Answer 11

thxx

Answer 12

second is better try to work it out\[w=\frac{e^{i\theta}-1}{e^{i\theta}+1}=\frac{e^{i\theta}-1}{e^{i\theta}+1}\times\frac{e^{-i\theta}+1}{e^{-i\theta}+1}\]