How do you convert #((2sqrt3) 2i)/(1-sqrt3i)# to polar form?

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Answer 1 · 2016-06-30T12:31:43

#((2sqrt3)2i)/(1-sqrt3i)=2sqrt3(cos(-pi/6)+isin(-pi/6))#

#((2sqrt3)2i)/(1-sqrt3i)#

Let us first rationalize it by multiplying numerator and denominator by complex conjugate of denominator. Then above becomes

#(((2sqrt3)2i)(1+sqrt3i))/((1-sqrt3i)(1+sqrt3i))#

= #(4sqrt3i+12i^2)/(1^2-(sqrt3i)^2)#

= #(4sqrt3i+12i^2)/(1^2-(3i^2))=(4sqrt3i+12i^2)/(1+3)#

= #(-12+4sqrt3i)/4=-3+sqrt3i#

Now when a complex number #a+bi# is written in polar form as #r(costheta+isintheta)#, #r=sqrt(a^2+b^2)# and #theta=arctan(b/a)#

Hence, here #r=sqrt((-3)^2+(sqrt3)^2)=sqrt(9+3)=sqrt12=2sqrt3#

and #theta=arctan(sqrt3/(-3))=arctan(-1/sqrt3)=-pi/6#

Hence, #((2sqrt3)2i)/(1-sqrt3i)=2sqrt3(cos(-pi/6)+isin(-pi/6))#