Question

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

• A. $-i$
• B. $1$
• C. $-1$
• D. $i$

Answer 1

The correct option is A $-i$

$\left|zw\right|=1$

$\therefore \left|z\right|.\left|w\right|=1$

as argument difference between 'z' and $w=\frac{\pi }{2}$

we can write,

$\frac{z}{\left|z\right|}=\frac{w}{\left|w\right|}.e^{i\pi /2}.......\left(e^{i\pi /2}rotatesthecomplexnowby\pi /2\right)$

$\therefore z=\left|z\right|.\frac{w}{\left|w\right|}.e^{i\pi /2}$

$\therefore z=\frac{1}{\left|w\right|}.\frac{w}{\left|w\right|}.e^{i\pi /2}\to \frac{w}{{\left|w\right|}^2}{e}^{i\pi /2}$

thus we have
$\therefore \overline{z}=\overline{w}.\frac{1}{{\left|w\right|}^2}{e}^{-i\pi /2}..................(z=e^{i\theta }then,\overline{z}=e^{-i\theta })$

$\overline{z}\times w=\overline{w}.\frac{1}{{\left|w\right|}^2}{e}^{-i\pi /2}\times w$

$=w.\overline{w}.\frac{1}{{\left|w\right|}^2}{e}^{-i\pi /2}$

$={\left|w\right|}^2.\frac{1}{{\left|w\right|}^2}{e}^{-i\pi /2}..................z.\overline{z}={\left|z\right|}^2$

$=e^{-i\pi /2}$

$=-i$