what is z= -2-2square root of 3i in trigonometric form.

Answer 1

-2 - 2√3 i corresponds to the point (-2,-2√3)

so, tanθ = y/x = -2√3/-2 = √3
that puts θ in QIII, so θ = 4pi/3

r^2 = 4+12 = 16, so r = 4

so, -2 - 2√3 i = (4,4pi/3)