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

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Answer 1 · 2016-05-22T12:26:42

Polar form of $-6i$ is $(6,-pi/2)$

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

$rcostheta+irsintheta$

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

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

$r=sqrt()^2+(-6)^2)=sqrt36=6$ and hence

$costheta=0/6=0$ and $sintheta=-6/6=-1$

Hence, $theta=-pi/2$ and

Polar form of $-6i$ is $(6,-pi/2)$

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