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

$(a)\;-\large\frac{2 \pi}{3}\qquad(b)\;-\large\frac{4 \pi}{3}\qquad(c)\;-\large\frac{ \pi}{3}\qquad(d)\;-\large\frac{ \pi}{2}$

Can you answer this question?

Answer : $\;-\large\frac{2 \pi}{3}$
Explanation :
Let$\;z=-\normalsize 1- \normalsize i\sqrt{3}$
Let $\;r cos \theta = -1\;$ and $\;r sin \theta = -\sqrt{3}$
on squaring and adding , we obtain
$(r cos \theta)^{2}+(r sin \theta)^{2} = (-1)^{2}+(-\sqrt{3})^{2}$
$r^{2}(cos^{2} \theta + sin^{2} \theta) = 1+ 3$
$r^{2} = 4 \qquad [cos^{2} \theta + sin^{2} \theta=1]$
$r =\sqrt{4} = 2 \qquad [conventionally , r > 0 ]$
Therefore , modulus = 2
Therefore , $\; 2 cos \theta = -1 \;$ and $\; 2 sin \theta = - \sqrt{3}$
$cos \theta = \large\frac{-1}{2}\;$ and $\;sin \theta = \large\frac{-\sqrt{3}}{2}$
Since both the values of $\;sin \theta\;$ and $\;cos \theta\;$ are negative and $\;sin \theta\;$ and $\;cos \theta\;$ are negative in $\;3^{rd}\;$ quadrant
Thus , the modulus and the argument of the complex number $\;z=-\normalsize 1- \normalsize i\sqrt{3}\;$ are $\;2\;$ and$-\large\frac{2 \pi}{3}$
answered Apr 10, 2014 by