Question

If $z_1$ and $z_2$ be complex numbers such that $z_1+i\left(\overline{z_2}\right)=0$ and $arg\left(\overline{z_1}{z}_2\right)=\frac{\pi }{3}$. Then, $arg\left(\overline{z_1}\right)$ is equal to

• A. $\frac{\pi }{3}$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{5\pi }{12}$
• E. $\frac{5\pi }{6}$

Answer 1

Answer: (D)

$\frac{5\pi }{12}$

Given, $z_1+i\overline{\left(z_2\right)}=0$
$\Rightarrow z_1=(-i)\overline{z_2}$
Taking argument on both sides, we get
$arg\left(z_1\right)=arg\left\{\right(-i)\cdot \overline{z_2}\}$
$\Rightarrow arg\left(z_1\right)=arg(-i)+arg\left(\overline{z_2}\right)$ (by property)
$\Rightarrow arg\left(z_1\right)-arg\left(\overline{z_2}\right)={\tan }^{-1}\left(\frac{-1}{0}\right)=-\frac{\pi }{2}$
$\Rightarrow arg\left(z_1\right)+arg\left(z_2\right)=\frac{-\pi }{2}.....\left(i\right)$
$[\because arg(\overline{z})=-arg(z\left)\right]$
and $arg\left(\overline{z_1}{z}_2\right)=\frac{\pi }{3}$
$\Rightarrow arg\left(\overline{z_1}\right)+arg\left(z_2\right)=\frac{\pi }{3}$
$\Rightarrow -arg\left(z_1\right)+arg\left(z_2\right)=\frac{\pi }{3}...\left(ii\right)$
On adding Eqs. (i) and (ii), we get
$2arg\left(z_2\right)=\frac{-\pi }{6}$
$\Rightarrow arg\left(z_2\right)=\frac{-\pi }{12}$
Therefore, from Eq. (i),
$arg\left(z_1\right)=\frac{-\pi }{2}+\frac{\pi }{12}=\frac{-5\pi }{12}$
$\Rightarrow -arg\left(z_1\right)=\frac{5\pi }{12}$
$\Rightarrow arg\left(z_1\right)=\frac{5\pi }{12}$.