Question

If arg$\left(\frac{z_1-\frac{z}{\left|z\right|}}{\frac{z}{\left|z\right|}}\right)=\frac{\pi }{2}$ and $\left|\frac{z}{\left|z\right|}-z_1\right|=3$ then $\left|z_1\right|$ equals to

• A. $\sqrt{26}$
• B. $\sqrt{10}$
• C. $\sqrt{3}$
• D. $2\sqrt{2}$

Answer 1

Answer: (B)

$\sqrt{10}$

$arg\left(\frac{z_1-\frac{z}{|z|}}{\frac{z}{|z|}}\right)=\frac{\pi }{2}$ ...(1)
$|\frac{z}{\left|z\right|}-z_1|=3$ ...(2)
Let $z=r_{1}cis\left(A\right)\&z_1=r_{2}cis\left(B\right)$
Substitute in eq. (1) we get,
$arg\left(\frac{r_{2}cis\left(B\right)-cis\left(A\right)}{cis\left(A\right)}\right)=\frac{\pi }{2}$
$arg(r_{2}cis(B-A)-1)=\frac{\pi }{2}$
$⟹\arctan \left(\frac{r_2\sin (B-A)}{r_2\cos (B-A)-1}\right)=\frac{\pi }{2}$
$⟹\cos (B-A)=\frac{1}{r_2}$ ...(3)
Now substitute $z\&z_1$ in eq. (2)
$|cis\left(A\right)-r_{2}cis(B\left)\right|=3$
$⟹{(\cos A-r_2\cos B)}^2+{(\sin A-r_2\sin B)}^2=9$
$⟹1+{r_2}^2-2r_2(\cos A\cos B+\sin B\sin A)=9$
$⟹1+{r_2}^2-2r_2\left(\cos (A-B)\right)=9$
$⟹1+{r_2}^2-2=9$ ...{from (3)}
$⟹1+{r_2}^2-2=9⟹r_2=\sqrt{10}$
Hence, option 'B' is correct.