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Given $arg\bigg(z_1-\large\frac{\Large\frac{z}{|z|}}{\Large\frac{z}{|z|}}\bigg)=\frac{\pi}{2}$ and $\big|\large\frac{z}{|z|}-$$z_1\big|=3$,then $|z_1|$ equals

$\begin{array}{1 1}(A)\;\sqrt{26}&(B)\;\sqrt 3\\(C)\;2\sqrt 2&(D)\;\sqrt{10}\end{array} $

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$\large\frac{z}{|z|}=$$\cos\theta+i\sin \theta$
$arg(\large\frac{z}{|z|})=$$\theta$
$arg\big(z_1-\large\frac{z}{|z|}\big)$$-arg\big(\large\frac{z}{|z|}\big)=\frac{\pi}{2}$
$\Rightarrow arg\big(z_1-\large\frac{z}{|z|}\big)=\frac{\pi}{2}$$+\theta$
$\big|\large\frac{z}{|z|}$$-z_1\big|=3\Rightarrow \big|z_1-\large\frac{z}{|z|}\big|$$=3$
$\Rightarrow z_1-\large\frac{z}{|z|}$$=3\big[\cos(\large\frac{\pi}{2}$$+\theta)+i\sin(\large\frac{\pi}{2}$$+\theta)\big]$
$\Rightarrow z_1=\cos\theta+i\sin\theta+3[-\sin\theta+i\cos\theta]$
$\Rightarrow |z_1|=\sqrt{10}$
Hence (D) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

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