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$z=x+iy$ and $\omega=\large\frac{1-iz}{z-i}$ then $|\omega|=1$ implies that in complex plane

$\begin{array}{1 1}(A)\;\text{z lies on real axis}\\(B)\;\text{z lies on imaginary axis}\\(C)\;\text{z lies on a unit circle}\\(D)\;\text{None of these}\end{array} $

1 Answer

$|\omega|=1\Rightarrow \big|\large\frac{1-iz}{z-i}\big|$$=1$
$\Rightarrow |-i^2-iz\mid=\mid z-i\mid$
$\Rightarrow \mid (-i)(z+i)\mid=\mid z-i\mid$
$\Rightarrow \mid z+i\mid=\mid z-i\mid$
$\Rightarrow z$ lies on perpendicular bisector of line segment joining i & -i
$\Rightarrow z$ lies on real axis
Hence (A) is the correct answer.
answered Apr 10, 2014 by sreemathi.v

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