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

Question

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

Answer 1

$\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.