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

1 Answer
Oct 24, 2016

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

Explanation:

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))