If -1 + sqrt{-3} = re^{itheta},then theta is | vedclass

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(C) Given the complex number $z = -1 + i\sqrt{3}$.We represent this in polar form as $z = r(\cos 	heta + i\sin 	heta) = re^{i	heta}$.Comparing the

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$\frac{2\pi}{3}$

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(C) Given the complex number $z = -1 + i\sqrt{3}$.We represent this in polar form as $z = r(\cos 	heta + i\sin 	heta) = re^{i	heta}$.Comparing the real and imaginary parts:$r\cos 	heta = -1$ and $r\sin 	heta = \sqrt{3}$.Since the real part is negative and the imaginary part is positive,the complex number lies in the second quadrant.$	an 	heta = \frac{r\sin 	heta}{r\cos 	heta} = \frac{\sqrt{3}}{-1} = -\sqrt{3}$.In the second quadrant,$	heta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.Thus,$	heta = \frac{2\pi}{3}$.

List-$I$ (Complex number)	List-$II$ (Polar form)
$(i) \sqrt{3}-i$	$(a) 2 \operatorname{cis} \frac{\pi}{6}$
$(ii) \sqrt{3}+i$	$(b) 2 \operatorname{cis} \frac{5 \pi}{6}$
$(iii) -\sqrt{3}+i$	$(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$
$(iv) -\sqrt{3}-i$	$(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$

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

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