Question

If $|z|=|w|$, $\omega \ne 0$ and $arg\left(z\right)+arg\left(\omega \right)=\pi$, then $z=$ ..............

• A. $-\omega$
• B. $\omega$
• C. $\overline{\omega }$
• D. $-\overline{\omega }$

Answer 1

The correct option is D $-\overline{\omega }$

$z=|z|e^{arg\left(z\right)}$
$=|z|\left(e^{\pi -arg\left(w\right)}\right)$
$=|z|(e^{\pi }.e^{-arg\left(w\right)})$
$=|z|\left(\right(-1\left)e^{-arg\left(w\right)}\right)$
$=-|z|.e^{-arg\left(w\right)}$
Now
$|z|=|w|$
Hence
$z=-|w|e^{-arg\left(w\right)}$
$=-\overline{w}$