Question

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

Answer 1

Let $z=\frac{1+2i}{1-3i}$, then

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

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

Let $z=r\cos \theta +ir\sin \theta$

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

On squaring and adding we obtain

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

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

$\Rightarrow r=\frac{1}{\sqrt{2}}$ [Conventionally $r>0$]

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