Browse Questions

# Find the modulus and argument of the complex number $\;\large\frac{1+2i}{1-3i}$

$(a)\;\large\frac{1}{\sqrt{2}}\;and\;\large\frac{3 \pi}{4}\qquad(b)\;\large\frac{1}{\sqrt{2}}\;and\;\large\frac{5\pi}{4}\qquad(c)\;\large\frac{1}{\sqrt{2}}\;and\;\large\frac{7 \pi}{4}\qquad(d)\;0\;and\;\large\frac{3 \pi}{4}$

Answer : $\;\large\frac{1}{\sqrt{2}}\;and\;\large\frac{3 \pi}{4}$
Explanation :
Let $\;z=\large\frac{1+2i}{1-3i}\;$, then
$\;z=\large\frac{1+2i}{1-3i} \times \large\frac{1+3i}{1+3i}$
$= \large\frac{1+3i+2i+6i^{2}}{1^{2}-(3i)^{2}}$
$= \large\frac{1+5i-6}{1-9(-1)} = \large\frac{-5+5i}{10}$
$= -\large\frac{1}{2} + \large\frac{1}{2} i$
Let $\; z= r cos \theta + i \;r sin \theta$
i.e $\;cos \theta = - \large\frac{1}{2} \;and\; sin \theta = \large\frac{1}{2}$
On squaring and adding , we obtain
$r^{2}\;(cos^{2} \theta + sin^{2} \theta )^{2} = (-\large\frac{1}{2})^{2}+(\large\frac{1}{2})^{2}$
$r^{2} = \large\frac{1}{2}$
$r = \large\frac{1}{\sqrt{2}} \qquad$ [conventionally , r> 0]
$\large\frac{1}{\sqrt{2}} cos \theta = -\large\frac{1}{2}\;$ and $\; \large\frac{1}{\sqrt{2}} sin \theta = \large\frac{1}{2}$
$cos \theta = -\large\frac{1}{\sqrt{2}}\;$ and $\; sin \theta = \large\frac{1}{\sqrt{2}}$
$\theta = \pi - \large\frac{\pi}{4} = \large\frac{3 \pi}{4} \qquad$ [As $\;theta\;$ lies in the $\;2^{nd}\;$ quadrant]
Therefore , the modulus and argument of the complex number are $\;\large\frac{1}{\sqrt{2}}\;$ and $\;\large\frac{3 \pi}{4}\;$ respectively .