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# The complex number $z$ is such that $|z|=1,\:z\neq1\:and\:w=\large|\frac{z-1}{z+1}|.$ Then the real part of $w$ is ?

$\begin{array}{1 1}(A) \;\large\frac{1}{|z+1|}^2 \\(B)\; \large\frac{-1}{|z+1|}^2 \\(C)\;\large\frac{\sqrt 2}{|z+1|}^2\\(D)\;0\end{array}$

Can you answer this question?

Let $z=x+iy$
Given $|z|=1\Rightarrow\:x^2+y^2=1$
$w=\large\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$
Rationalising the denominator we get
$w=\large\frac{x-1+iy}{x+1+iy}\times\frac{x+1-iy}{x+1-iy}$
$=\large\frac{(x^2-1+y^2)+i(xy+y-xy+y)}{x^2+1+2x+y^2}$
Substituting $x^2+y^2=1$ we get
$w=\large\frac{0+2yi}{2(1+x)}=0+\frac{y}{1+x}i$
$\Rightarrow$ Real part of $w =0$
answered Jul 5, 2013