Argument of - 1 - isqrt 3 is | vedclass

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Question

(d) Let $z = - 1 - i\sqrt 3 $
then $\alpha = {	an ^{ - 1}}\left| {\,\frac{b}{a}\,} ight| = {	an ^{ - 1}}\left| {\, - \frac{{\sqrt 3 }}{1}\,} ig

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) Let $z = - 1 - i\sqrt 3 $
then $\alpha = {	an ^{ - 1}}\left| {\,\frac{b}{a}\,} ight| = {	an ^{ - 1}}\left| {\, - \frac{{\sqrt 3 }}{1}\,} ight| = \frac{\pi }{3}$
Clearly, $z$ is in $III$ quadrant.
Therefore argument $	heta = - (\pi - \alpha ) = - (\pi - \pi /3) = \frac{{ - 2\pi }}{3}$.

Argument of $ - 1 - i\sqrt 3 $ is

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