How do you convert $3 sqrt 2 - 3 sqrt2i$ to polar form?

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Answer 1 · 2016-10-24T07:43:19

In polar form $3sqrt2-3sqrt2i$ is $6(cos(-pi/4)+isin(-pi/4))$

A complex number $a+bi$ can be written in polar form as $r(costheta+isintheta)$ i.e. $a=rcostheta$ and $b=rsintheta$ .

Squaring and adding the last two, we get $a^2+b^2=r^2cos^2theta+r^2sin^2theta$

= $r^2(cos^2theta+sin^2theta)=r^2xx1=r^2$

and $b/a=(rsintheta)/(rcostheta)=tantheta$

Here in the complex number $3sqrt2-3sqrt2i$ , we have $a=3sqrt2$ and $b=-3sqrt2$ and hence $r=sqrt(a^2+b^2)=sqrt((3sqrt2)^2+(-3sqrt2)^2)$

= $sqrt(18+18)=sqrt36=6$ and

$costheta=(3sqrt2)/6=1/sqrt2$ and $costheta=(-3sqrt2)/6=-1/sqrt2$

i.e. $theta=(-pi/4)$

Hence in polar form $3sqrt2-3sqrt2i$ is $6(cos(-pi/4)+isin(-pi/4))$

How do you convert 3 sqrt 2 - 3 sqrt2i3 sqrt 2 - 3 sqrt2i to polar form?