The argument of z = -1 - isqrt{3} is: | vedclass

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(D) Let $z = -1 - i\sqrt{3}$.Here,the real part $a = -1$ and the imaginary part $b = -\sqrt{3}$.First,we find the reference angle $\alpha = 	an^{-1}\

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $z = -1 - i\sqrt{3}$.Here,the real part $a = -1$ and the imaginary part $b = -\sqrt{3}$.First,we find the reference angle $\alpha = 	an^{-1}\left|\frac{b}{a}ight| = 	an^{-1}\left|\frac{-\sqrt{3}}{-1}ight| = 	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.Since both $a The argument $	heta$ for a complex number in the $III$ quadrant is given by $	heta = -(\pi - \alpha)$.Therefore,$	heta = -(\pi - \frac{\pi}{3}) = -(\frac{2\pi}{3}) = -\frac{2\pi}{3}$.

The argument of $z = -1 - i\sqrt{3}$ is:

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