Question

Find the modulus and the argument of the complex number $z=-1-i\sqrt{3}$

Answer 1

$z=-1-i\sqrt{3}$
Let $r\cos \theta =-1andr\sin \theta =-\sqrt{3}$
On squaring and adding, we obtain
${\left(r\cos \theta \right)}^2+{\left(r\sin \theta \right)}^2={(-1)}^2+{(-\sqrt{3})}^2$
$\Rightarrow r^2({\cos }^2\theta +{\sin }^2\theta )=1+3$
$\Rightarrow r^2=4[{\cos }^2\theta +{\sin }^2\theta =1]$
$\Rightarrow r=\sqrt{4}=2$ (Conventionally,r>0 )
$\therefore$ Modulus of $z$ i.e $|z|=2$
$\therefore 2\cos \theta =-1and2\sin \theta =-\sqrt{3}$
Since both the values of $\sin \theta$ and $\cos \theta$ are negative and $\sin \theta$ and $\cos \theta$ are negative in III quadrant,
Argument $=-(\pi -\frac{\pi }{3})=\frac{-2\pi }{3}$
Thus, the modules and argument of the complex number $-1-\sqrt{3i}$ are 2 and $\frac{-2\pi }{3}$ respectively.