Question 1134030
{{{-2 + 2sqrt(3)* i }}}

{{{z = a + i b }}}

{{{z=-2 + 2sqrt(3)* i}}}  => {{{a= -2}}}, {{{b=2sqrt(3)}}}


 trigonometric form:

{{{z= r ( cos(theta) + i sin(theta) )}}}


where:
{{{r = sqrt(a^2 + b^2)}}} and {{{tan(theta) = b / a}}}

{{{r = sqrt((-2)^2 + (2sqrt(3))^2)}}}

{{{r = sqrt(4 + 4(3))}}}

{{{r = sqrt(4 + 12)}}}

{{{r = sqrt(16)}}}

{{{r = 4}}}


{{{tan(theta) = 2sqrt(3) / -2}}}

{{{tan(theta) = -sqrt(3)}}}

{{{theta=pi+tan^-1(-sqrt(3))}}}

{{{theta= (2 pi)/3}}}

{{{theta=120}}}°

then,

{{{z= 4 ( cos(120) + i* sin(120) )}}}