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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 ?

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

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Let $z=x+iy$
Given $|z|=1\Rightarrow\:x^2+y^2=1$
Rationalising the denominator we get
Substituting $x^2+y^2=1 $ we get
$\Rightarrow$ Real part of $w =0$
answered Jul 5, 2013 by rvidyagovindarajan_1

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