Question

Find the modulus and the argument of the complex number $z=-\sqrt{3}+i$

Answer 1

Given, $z=-\sqrt{3}+i$
Let $r\cos \theta =-\sqrt{3}$ and $r\sin \theta =1$
On squaring and adding, we obtain
$r^2{\cos }^2\theta +r^2{\sin }^2\theta ={(-\sqrt{3})}^2+1^2$
$\Rightarrow r^2=3+1=4$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$\Rightarrow r=\sqrt{4}=2$ (Conventionally r>0)
$\therefore$ Modulus of $z$ i.e. $|z|=2$
$\therefore 2\cos \theta =-\sqrt{3}$ and $2\sin \theta =1$
$\Rightarrow \cos \theta =\frac{-\sqrt{3}}{2}and\sin \theta =\frac{1}{2}$
$\therefore \theta =\pi -\frac{\pi }{6}=\frac{5\pi }{6}$ [As $\theta$ lies in the II quadrant]
Thus, the modulus and argument of the complex number $-\sqrt{3}+i$ are 2 and $\frac{5\pi }{6}$