Question

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

Answer 1

Let $z=\frac{1+2i}{1-3i}\times \frac{1+3i}{1+3i}=\frac{1+3i+2i+6i^2}{1-9i^2}=\frac{-5+5i}{10}=\frac{-1+i}{2}$

Now $z=\frac{-1}{2}+\frac{i}{1}=r(cos\theta +isin\theta )$

$\therefore rcos\theta =\frac{-1}{2}andrsin\theta =\frac{1}{2}...\left(i\right)$

Squaring both sides of (i) and adding

$r^2(co{s}^2\theta +si{n}^2\theta )=\frac{1}{4}+\frac{1}{4}$

$\Rightarrow r^2=\frac{1}{2}\Rightarrow r=\frac{1}{\sqrt{2}}$

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

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

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

Since $sin\theta$ is positive and $cos\theta$ is negative

$\therefore \theta$ lies in second quadrant

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

$\therefore$ Modulus of $z=\frac{1}{\sqrt{2}}$ and argument of $z=\frac{3\pi }{4}$