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

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Answer 1 · 2018-07-26T10:50:19

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

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

Where,

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

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

$\therefore 9+12i$

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

How do you convert 9 + 12i9 + 12i to polar form?