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

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

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Answer 1 · 2016-04-13T10:45:08

$8 i = 8 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))$ $8i=8(cos(pi/2)+isin(pi/2))$

Explanation:

$a + b i$ $a+bi$ in polar form is $r cos θ + i r sin θ$ $rcostheta+irsintheta$ , where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ and $θ = arctan (\frac{b}{a})$ $theta=arctan(b/a)$

$8 i = 0 + 8 i$ $8i=0+8i$ hence $r = \sqrt{0^{2} + 8^{2}} = 8$ $r=sqrt(0^2+8^2)=8$ and $θ = arctan (\frac{8}{0}) = arctan \infty$ $theta=arctan(8/0)=arctanoo$ and one can say $θ = \frac{π}{2}$ $theta=pi/2$

$8 i = 8 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))$ $8i=8(cos(pi/2)+isin(pi/2))$

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

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