Question

If |Z|=1 and
then $w= \frac{Z-1}{Z+1}$
where z≠-1
then Re(w)=?

Answer 1

put z=x+iy
$w= \frac{x-1+iy}{x+iy+1}= \frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}= \frac{2+iy}{2x+2}= \frac{1}{x+1}+ \frac{2yi}{2x+2}$
$Re(w)= \frac{1}{x+1}$
in terms of z
x=(z+z')/2  where z'=x-iy=1/z
therefore
$Re(w)= \frac{2}{z+z'+2}$

Answer 2

put z=cos α+i sinα
then solve for w
u can get
Re(z)=1/(cos α +1)