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No of complex numbers satisfying $\overline{z}=iz^2$ is

$\begin{array}{1 1}(A)\;1&(B)\;2\\(C)\;3&(D)\;4\end{array} $

1 Answer

$\overline{z}=iz^2$
$\Rightarrow |\overline{z}|=|iz^2|$
$\Rightarrow |z|=|i||z^2|\Rightarrow |z|=0$ or $| z|=1$
$|z|=0\Rightarrow z=0$
$|z|=1\Rightarrow z\overline{z}=1$
Thus,$\overline{z}=iz^2$
$\Rightarrow \large\frac{1}{z}$$=iz^2\Rightarrow z^3=-i$ or $i^3$
$\Rightarrow (\large\frac{z}{i})^3$$=1\Rightarrow z=i,i\omega,i\omega^2$
Hence (D) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

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