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 $-2sqrt3-2i$

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

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

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

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