Question

Find the modulus and argument of mentioned below complex number and hence express in polar form:

$\frac{1+2i}{1-3i}$

Answer 1

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

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

$\Rightarrow z=\frac{1+3i+2i+6i^2}{1^2-9i^2}$

$\Rightarrow z=\frac{-5+5i}{10}$

$\Rightarrow z=-\frac{1}{2}+\frac{1}{2}i\dots \left(1\right)$

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

$\Rightarrow |z|=\sqrt{{(-\frac{1}{2})}^2+{\left(\frac{1}{2}\right)}^2}$

$\Rightarrow |z|=\sqrt{\frac{1}{4}+\frac{1}{4}}$

$\Rightarrow |z|=\frac{1}{\sqrt{2}}\dots \left(2\right)$

Let $\alpha$ be acute angle such that-

$\tan \alpha =\frac{|\text{Im}\left(z\right)|}{|\text{Re}\left(z\right)|}$

$\Rightarrow \tan \alpha =\frac{|\frac{1}{2}|}{|-\frac{1}{2}|}$

$\Rightarrow \tan \alpha =1$

$\Rightarrow \alpha =\frac{\pi }{4}$

Since, $z$ lies in the $2^{nd}$ quadrant,

$\therefore arg\left(z\right)=\pi -\alpha$

$\Rightarrow arg\left(z\right)=\pi -\frac{\pi }{4}$

$\Rightarrow arg\left(z\right)=\pi -\frac{\pi }{4}$

$\Rightarrow arg\left(z\right)=\frac{3\pi }{4}$

Hence, Polar form of $z$

$z=|z|\left[\cos arg\right(z)+i\sin arg(z\left)\right]$

$\Rightarrow z=\frac{1}{\sqrt{2}}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$

Therefore, Polar form of $\left(\frac{1+2i}{1-3i}\right)\text{is}\frac{1}{\sqrt{2}}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$