How do I find the polar form of #3sqrt2 - 3sqrt2i#?

1 Answer
Aug 7, 2018

Polar form is #6(cos(-pi/4)+isin(-pi/4))#

Explanation:

First just find the absolute value of the complex number. We have it as #3sqrt2-3sqrt2i# and its absolute value is

#sqrt((3sqrt2)^2+(-3sqrt2)^2)=sqrt(18+18)=sqrt36=6#

Hence number can be written as

#6((3sqrt2)/6-(3sqrt2)/6i)#

or #6(1/sqrt2-1/sqrt2i)#

As polar number is of the form #r(costheta+isintheta)#

we have #r=6# and #costheta=1/sqrt2# and #sintheta=-1/sqrt2#

As cosine is ratio is positive and sine ratio is negative,

#theta=pi/4#

Hence, polar form is #6(cos(-pi/4)+isin(-pi/4))#