Question

Convert the complex number $\frac{(1+7i)}{(3-4i)}$ into polar form.

Answer 1

Given,

$\frac{1+7i}{3-4i}$

$\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}$

$=\frac{-25+25i}{3^2-(4i{)}^2}$

$=\frac{-25+25i}{25}$

$\therefore z=-1+i$

$r=\sqrt{a^2+b^2}=\sqrt{(-1{)}^2+(1{)}^2}=\sqrt{2}$

$\theta ={\tan }^{-1}\left(\frac{b}{a}\right)={\tan }^{-1}\left(\frac{-1}{1}\right)=-\frac{\pi }{4}$

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

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

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