Question

If z and $\omega$ are two non-zero complex numbers such that $|z\omega |=1$, and $Arg\left(z\right)-Arg\left(\omega \right)=\pi /2$, then $\overline{z}\omega$ is equal to

• A. 1
• B. -1
• C. i
• D. -i

Answer 1

The correct option is D -i

Let $z=r_1(\cos {\theta }_1+i\sin {\theta }_1)$ and

$w=r_2(cos{\theta }_2+i\sin {\theta }_2)$

$\Rightarrow$ $|z|=r_1,|w=r_2,arg\left(z\right)={\theta }_1$ and $arg\left(w\right)={\theta }_2$

$|zw|=r_1{r}_2=1$ and ${\theta }_1-{\theta }_2=\frac{\pi }{2}$ [ Given ]

$\overline{z}=r(\cos {\theta }_1-i\sin {\theta }_1)$

$\overline{z}=r_1\left(\cos \right(-{\theta }_1)+i\sin (-{\theta }_1\left)\right)$

$arg\left(\overline{z}\right)=-{\theta }_1$

$\Rightarrow$ $\overline{z}w=r_1{r}_2\left(\cos \right(-{\theta }_1+{\theta }_2)+i\sin (-{\theta }_1+{\theta }_2\left)\right)$

$\Rightarrow$ $\overline{z}w=1(\cos \left(\frac{-\pi }{2}\right)+i\sin (-\frac{\pi }{2}))$

$\Rightarrow$ $\overline{z}w=-i$