How do I represent #z=-3+3sqrt3 i# in polar form?

Question

A) #6(sin((2pi) /3) +icos((2pi) /3))#

B) #6(-cos((2pi) /3) +isin((2pi) /3))#

C) #6(cos((2pi) /3) +isin((2pi) /3))#

D) #6(cos((2pi) /3) - isin((2pi) /3))#

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Answer 1 · 2017-07-28T05:02:43

The answer is option #(a)#, i.e, #=6(cos(2/3pi)+isin(2/3pi))#

The polar form of a complex number is

#z=r(costheta+isintheta)....................#(1)#

Our complex number is

#z=-3+3sqrt3i#

#r=|z|=sqrt((-3)^2+(3sqrt3)^2)=sqrt(9+27)=sqrt36=6#

Therefore,

#z=6(-3/6+3/6sqrt3i)#

#=6(-1/2+sqrt3/2i)#

Comparing this equation to equation #(1)#

#costheta=-1/2# and #sintheta=sqrt3/2#

So,

we are in the quadrant #II#

#theta=2/3pi#

The polar form is

#z=6(cos(2/3pi)+isin(2/3pi))#