Question

Z= -3/2 + √3/2, colocar na forma trigonometrica dos numeros complexos.

Answer 1

Z = -3/2 + √3/2 i
$|Z|= \sqrt{ \frac{(-3}{2})^2+ (\frac{ \sqrt{3} }{2})}=\ \textgreater \ |Z|= \sqrt{ \frac{9}{4}+ \frac{3}{4} }=\ \textgreater \ |Z|= \sqrt{ \frac{12}{4} } = \frac{2 \sqrt{3} }{2} = \sqrt{3} \\ \\ cos \alpha = \frac{a}{|Z|} = \frac{ \frac{-3}{2} }{ \sqrt{3} }= \frac{-3}{2 \sqrt{3} } = \frac{-3 \sqrt{3} }{6} =- \frac{ \sqrt{3} }{2} \\ \\ sen \alpha = \frac{b}{|Z|}= \frac{ \frac{ \sqrt{3} }{2} }{ \sqrt{3} } = \frac{ \sqrt{3} }{2 \sqrt{3} } = \frac{1}{2}$

Como cosα < 0 e senα > 0 => α ∈ 2° quadrante

α = π - π/6 = (6π - π)/6 = 5π/6

Z = |Z|(cosα + i.senα)

Z = √3(cos5π/6 + i.sen5π/6)