Question

How do you convert `9 + 12i` to polar form?

Answer 1

`15(cos(\tan^{-1}(4/3))+isin(\tan^{-1}(4/3)))`

Explanation:

`9+12i=r(\cos\theta+i\sin\theta)`

Where,

`r=\sqrt{9^2+12^2}=15`

`\theta =\tan^{-1}(12/9)=\tan^{-1}(4/3)`

`\therefore 9+12i`

`=15(cos(\tan^{-1}(4/3))+isin(\tan^{-1}(4/3)))`