How do you convert $- 6 i$ $-6i$ to polar form?

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Answer 1 · 2016-05-22T12:26:42

Polar form of $- 6 i$ $-6i$ is $(6, - \frac{π}{2})$ $(6,-pi/2)$

A complex number $a + i b$ $a+ib$ in polar form is written as

$r cos θ + i r sin θ$ $rcostheta+irsintheta$

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

As $- 6 i$ $-6i$ can be written as $0 - 6 i$ $0-6i$

$r = {\sqrt{}}^{2} + {(- 6)}^{2}) = \sqrt{36} = 6$ $r=sqrt()^2+(-6)^2)=sqrt36=6$ and hence

$cos θ = \frac{0}{6} = 0$ $costheta=0/6=0$ and $sin θ = - \frac{6}{6} = - 1$ $sintheta=-6/6=-1$

Hence, $θ = - \frac{π}{2}$ $theta=-pi/2$ and

Polar form of $- 6 i$ $-6i$ is $(6, - \frac{π}{2})$ $(6,-pi/2)$

How do you convert −6i -6i to polar form?