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

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Answer 1 · 2017-04-18T08:25:37

#-2sqrt3-2i=4(cos((4pi)/3)+isin((4pi)/3))=4e^(i(4pi)/3)#

Modulus or absolute value of #-2sqrt3-2i# is

#sqrt((-2sqrt3)^2+(-2)^2)#

= #sqrt(4×3+4)=sqrt16=4#

Hence #-2sqrt3-2i#

= #4(-(2sqrt3)/4-2/4i)#

= #4(-sqrt3/2-1/2i)#

= #4(cos((4pi)/3)+isin((4pi)/3))=4e^(i(4pi)/3)#