How do you convert $8i$ to polar form?

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Answer 1 · 2016-04-13T10:45:08

$8i=8(cos(pi/2)+isin(pi/2))$

$a+bi$ in polar form is $rcostheta+irsintheta$ , where $r=sqrt(a^2+b^2)$ and $theta=arctan(b/a)$

$8i=0+8i$ hence $r=sqrt(0^2+8^2)=8$ and $theta=arctan(8/0)=arctanoo$ and one can say $theta=pi/2$

$8i=8(cos(pi/2)+isin(pi/2))$

How do you convert 8i8i to polar form?