Question

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

Answer 1

Let, $z=\frac{1+2i}{1-3i}$
$z=\frac{1+2i}{1-3i}\times \frac{1+3i}{1+3i}$
$=\frac{1+3i+2i+6i^2}{1^2+3^2}$
$=\frac{1+5i+6(-1)}{1+9}$
$=\frac{-5+5i}{10}$
$=\frac{-1}{2}+\frac{1}{2}i$
Let $z=r\cos \theta +ir\sin \theta$

$i.e.,r\cos \theta =\frac{-1}{2}$ and $r\sin \theta =\frac{1}{2}$

On squaring and adding, we obtain

$r^2({\cos }^2\theta +{\sin }^2\theta )={\left(\frac{-1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2$

$\Rightarrow r^2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

As $r>0$

$\Rightarrow r=\frac{1}{\sqrt{2}}$
$\therefore \frac{1}{\sqrt{2}}\cos \theta =\frac{-1}{2}$ and $\frac{1}{\sqrt{2}}\sin \theta =\frac{1}{2}$

$\Rightarrow \cos \theta =\frac{-1}{\sqrt{2}}$

and $\sin \theta =\frac{1}{\sqrt{2}}$

$\therefore \theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

[As $\theta$ lies in the II quadrant ]

Therefore the modulus and argument of the given complex number are

$\frac{1}{\sqrt{2}}$ and $\frac{3\pi }{4}$ respectively.