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))#