Question

Find the modulus and argumrent of the following complex numbers and hence express each of them in the polar form:
$\frac{1+2i}{1-3i}$

Answer 1

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

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

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

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

$\frac{-5+5i}{10}$

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

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

$arg\left(z\right)=\sqrt{a^2+b^2}=\sqrt{{\left(\frac{-1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}$

$\therefore arg\left(z\right)=\frac{1}{\sqrt{2}}$

$r^2=a^2+b^2\Rightarrow r^2={\left(\frac{-1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2$

$\therefore r=\frac{1}{\sqrt{2}}$

$\theta ={\tan }^{-1}\left(\frac{b}{a}\right)+{180}^{\circ }\because a<0$

$\therefore \theta ={\tan }^{-1}\left(\frac{\frac{-1}{2}}{\frac{1}{2}}\right)+{180}^{\circ }$

$\theta =-{45}^{\circ }+{180}^{\circ }={135}^{\circ }=\frac{3\pi }{4}$

$\therefore z=r(\cos \theta +\sin \theta )$

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