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))#
