Question

If $-1+\sqrt{-3}=r{e}^{i\theta }$, then $\theta$ is equal to :

• A. $\frac{2\pi }{3}$
• B. $-\frac{2\pi }{3}$
• C. $\frac{\pi }{3}$
• D. $-\frac{\pi }{3}$

Answer 1

The correct option is A $\frac{2\pi }{3}$

Given:$-1+\sqrt{-3}=r{e}^{i\theta }$

Let $z=-1+\sqrt{-3}$

$\Rightarrow z=-1+i\sqrt{3}$ ..........$\left(1\right)$

Let polar form be

$z=r(\cos \theta +i\sin \theta )$ ..........$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$-1+i\sqrt{3}=r\cos \theta +ir\sin \theta$

Comparing real part and imaginary parts,we get

$r\cos \theta =-1,r\sin \theta =\sqrt{3}$

Squaring,we get

$\Rightarrow r^2{\cos }^2\theta =1,r^2{\sin }^2\theta =3$

$\Rightarrow r^2{\cos }^2\theta =1$ .........$\left(3\right)$

and $r^2{\sin }^2\theta =3$ .........$\left(4\right)$

Adding $\left(3\right)$ and $\left(4\right)$ we get

$r^2{\cos }^2\theta +r^2{\sin }^2\theta =1+3$

$\Rightarrow r^2({\cos }^2\theta +{\sin }^2\theta )=4$

$\Rightarrow r^2=4$ since $({\cos }^2\theta +{\sin }^2\theta =1)$

$\Rightarrow r=2$

Again,$r\cos \theta =-1,r\sin \theta =\sqrt{3}$

$\Rightarrow 2\cos \theta =-1,2\sin \theta =\sqrt{3}$ since $r=2$

$\Rightarrow \cos \theta =\frac{-1}{2},\sin \theta =\frac{\sqrt{3}}{2}$

Solutions of $\cos \theta =\frac{-1}{2}$ are $\{\pi -\frac{\pi }{3},\pi +\frac{\pi }{3}\}$

Solutions of $\sin \theta =\frac{\sqrt{3}}{2}$ are $\{\pi -\frac{\pi }{3},\frac{\pi }{3}\}$

$\Rightarrow \theta =\{\pi -\frac{\pi }{3},\pi +\frac{\pi }{3}\}\cap \{\pi -\frac{\pi }{3},\frac{\pi }{3}\}$

$\Rightarrow \theta =\{\frac{2\pi }{3},\frac{4\pi }{3}\}\cap \{\frac{2\pi }{3},\frac{\pi }{3}\}$

$\therefore \theta =\frac{2\pi }{3}$