How do you convert #-9 - 9sqrt2 i # to polar form?

1 Answer
Shwetank Mauria
Aug 12, 2017

#-9-9sqrt2i=9sqrt3(costheta+isi ntheta)#, where #theta=tan^(-1)sqrt2# and #theta# is in #Q3#.

Explanation:

If a complex number #a+bi=rcostheta+irsintheta#, #a=rcostheta# and #b=rsintheta# and squaring and adding #a^2+b^2=r^2# or #r=sqrt(a^2+b^2)#.

Hence, as we have #-9-9sqrt2i#, #r=sqrt((-9)^2+(-9sqrt2)^2)=sqrt(81+162)=sqrt243=9sqrt3#.

Hence #costheta=-1/sqrt3# and #sintheta=-sqrt(2/3)# i.e. #theta# is in third quadrant and #tantheta=sqrt2#

and #-9-9sqrt2i=9sqrt3(costheta+isi ntheta)#, where #theta=tan^(-1)sqrt2# and #theta# is in #Q3#.

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