Question

If $arg(z+a)=\frac{\pi }{6}$ and $arg(z-a)=\frac{2\pi }{3}\left(aϵR^{+}\right)$, then

Answer 1

For line $CAarg(z+a)=\frac{\pi }{6}\Rightarrow \angle CAB={30}^{\circ }$

For line $BCarg(z-a)=\frac{2\pi }{3}\Rightarrow \angle ABC={60}^{\circ }$

$OA=OB=a$

$\angle ACB={90}^{\circ }$

$\therefore \Delta ACB$ is a right-angle triangle

$AC:y-0=\frac{1}{\sqrt{3}}(x+a)$

$\Rightarrow \sqrt{3}y=x+a$

$\Rightarrow x-\sqrt{3}y=-a$ …$\left(1\right)$

$BC:y-0=-\sqrt{3}(x-a)$

$\sqrt{3}x+y=a\sqrt{3}$ …$\left(2\right)$

From equation $\left(1\right)$ & $\left(2\right)$

$C(\frac{a}{2},\frac{\sqrt{3}a}{2})$

$\therefore$ Distance $OC=a$

$\therefore \angle COB={60}^{\circ }$

From line $OC:|z|=a$

And
$arg\left(z\right)=\angle COB=\frac{\pi }{3}$

Hence, option $\left(A\right)$ & $\left(D\right)$ are correct.