Question

Let $z_1$ and $z_2$ are two complex number such that $(1-i)z_1=2z_2$ and $\arg \left(z_1{z}_2\right)=\frac{\pi }{2}$ then $\arg \left(z_2\right)$ is equal

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{8}$
• C. $\frac{5\pi }{8}$
• D. $\frac{-7\pi }{8}$

Answer 1

The correct option is A $\frac{\pi }{12}$

Let $arg\left(z_1\right)=a,arg\left(z_2\right)=b$

$z_1=r_1{e}^{ia},z_2=e^{ib}$

$z_1(1-i)=2z_2$

$\frac{z_2}{z_1}=\frac{1-i}{2}=\frac{1}{2}-\frac{i}{2}=\cos (-\frac{\pi }{3})+i\sin (-\frac{\pi }{3})$

$\frac{z_2}{z_1}=e^{-\pi /3i}$

$arg\left(\frac{z_2}{z_1}\right)=arg\left(e^{-\frac{\pi }{3}i}\right)=-\frac{\pi }{3}$

$arg\left(z_2\right)-arg\left(z_1\right)=-\pi /3$

$b-a=-\frac{\pi }{3}\to 1$

$arg\left(z_1{z}_2\right)=arg\left(z_1\right)+arg\left(z_2\right)=\frac{\pi }{2}$

$a+b=\frac{\pi }{2}\to 2$

Solving $1$ and $2$, we get

$b=\frac{\pi }{12}$