Question

Match the equation in $z$ in $Column-I$ with the corresponding value of $arg\left(z\right)$ in $column-II$.

$Column-I$ (equation in z)	$Column-II$ (principal value of arg(z))
$z^2-z+1=0$	$-2\pi /3$
$z^2+z+1=0$	$-\pi /3$
${2z}^2+1+i\sqrt{3}=0$	$\pi /3$
${2z}^2+1-i\sqrt{3}=0$	$2\pi /3$

Answer 1

(1) $z^2-z+1=0$

$\frac{+1\pm \sqrt{1-4}\left(1\right)}{2\left(1\right)}=\frac{1\pm \sqrt{3}i}{2}\Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(\frac{\pi }{\sqrt{3}}\right)$

$={60}^o=\frac{\pi }{3}$

(2) $z^2+z+1=0$

$\frac{-1\pm \sqrt{3}i}{2}\Rightarrow {\tan }^{-1}\left(\frac{-\sqrt{3}}{1}\right)=\frac{2\pi }{3}$

(3) $2z^2+1+i\sqrt{3}=0$

$z^2=\sqrt{\frac{-1-i\sqrt{3}}{2}}\Rightarrow {\tan }^{-1}\left(\sqrt{3}\right)=\frac{\pi }{3}$

(4) $2z^2+1-i\sqrt{3}=0$

$z=\sqrt{\frac{1-i\sqrt{3}}{2}}\Rightarrow {\tan }^{-1}\left(\sqrt{3}\right)=\frac{-2\pi }{3}$