Question

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

Answer 1

Given data:
let $z=\frac{1+2i}{1-3i}$
Do rationalisation
on multiplying numerator and denominator by
$(1+3i)$

$z=\frac{1+2i}{1-3i}\times \frac{1+3i}{1+3i}$

$=\frac{1+3i+2i+6i^2}{(1{)}^2-(3i{)}^2}$

$=\frac{1+5i-6}{1+9}$

$=\frac{-5+5i}{10}$ $\{\because i^2=-1\}$

$=-\frac{1}{2}+\frac{1}{2}i$

$z=\frac{-1}{2}+\frac{1}{2}i$

$\text{Apply}x=r\cos \theta \text{and}y=r\sin \theta$

Using the diagram

$r\cos \theta =-\frac{1}{2}.....\left(i\right)$

And $r\sin \theta =\frac{1}{2}.....\left(ii\right)$

Modulus of complex number

Squaring and adding equation (1) and equation (2)

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

$\Rightarrow r^2({\cos }^2\theta +{\sin }^2\theta )=\frac{1}{4}+\frac{1}{4}\{\because {\sin }^2\theta +{\cos }^2\theta =1\}$

$z=\frac{-1}{2}+\frac{1}{2}i$

r cos $\theta =-\frac{1}{2}$.....(i)

And $rsin\theta =\frac{1}{2}$.....(ii)


$\Rightarrow r^2=\frac{1}{2}[\because {\cos }^2\theta +{\sin }^2\theta =1]$

$\Rightarrow r=\frac{1}{\sqrt{2}}\{\because \text{r is always positive}\}$

Argument of the complex number

Divide equation (2) by equation (1)

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

$\Rightarrow \tan \theta =-1$
$\Rightarrow \theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$ { As $\theta$ lies in the second quadrant}

$\therefore$ the modulus and argument of the given complex number are $\frac{1}{\sqrt{2}}$ and $\frac{3\pi }{4}$ respectively