Browse Questions

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

$(a)\;\large\frac{5 \pi}{6}\qquad(b)\;\large\frac{ \pi}{6}\qquad(c)\;\large\frac{6 \pi}{5}\qquad(d)\;\pi$

Answer : $\;\large\frac{5 \pi}{6}$
Explanation :
Let $\;z=-\normalsize \sqrt{3} + \normalsize i$
$r cos \theta = -\sqrt{3}\;$ and $\; r sin \theta = 1$
on squaring and adding , we obtain
$(r cos \theta)^{2}+(r sin \theta)^{2} = (-\sqrt{3})^{2}+(1)^{2}$
$r^{2}(cos^{2} \theta + sin^{2} \theta) = 3+1$
$r^{2} = 4 \qquad [cos^{2} \theta + sin^{2} \theta=1]$
$r =\sqrt{4} = 2 \qquad [conventionally , r > 0 ]$
Therefore , modulus = 2
Therefore , $\; 2 cos \theta = -\sqrt{3} \;$ and $\; 2 sin \theta = 1$
$cos \theta = \large\frac{-\sqrt{3}}{2}\;$ and $\;sin \theta = \large\frac{1}{2}$
$\theta = \pi - \large\frac{\pi}{6} = \large\frac{5 \pi}{6} \qquad [As \; \theta \;lies\;in\;the\;2^{nd}\;quadrant]$
Thus , the modulus and the argument of the complex number $\;z=-\normalsize \sqrt{3} + \normalsize i$ are 2 and $\;\large\frac{5 \pi}{6}\;$ respactively