Question

Z = - 16
1) o módulo de Z e o argumento de Z (ângulo 0)
2) A forma trigonométrica de Z
3) As raízes quartas de z, isto é

Answer 1

Resposta:

Explicação passo-a-passo:

1) Temos que z = -16, assim | z | = √(-16)² = 16

cos Ф = -16/16 = -1

sen Ф = 0/16 = 0

Ф = π

2) z = 16(cos π + isen π)

3) $\sqrt[4]{z}=\sqrt[4]{16}(cos\frac{\pi+2k\pi}{4}+sen\frac{\pi+2k\pi}{4})$

$\sqrt[4]{z}=2(cos\frac{\pi+2k\pi}{4}+sen\frac{\pi+2k\pi}{4})$

com k = 0, 1, 2 e 3

k = 0 => $z_{0}$$2(cos\frac{\pi+2.0.\pi}{4}+sen\frac{\pi+2.0.\pi}{4})=2(cos\frac{\pi}{4}+isen\frac{\pi}{4})=cos\frac{\pi}{2}+isen\frac{\pi}{2}=0+i=i$

k = 1 =>$z_{1}$$2(cos\frac{\pi+2.1.\pi}{4}+sen\frac{\pi+2.1.\pi}{4})=2(cos\frac{3\pi}{4}+isen\frac{3\pi}{4})=cos\frac{3\pi}{2}+isen\frac{3\pi}{2}=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$

k = 2 =>$z_{2}=2(cos\frac{\pi+2.2.\pi}{4}+sen\frac{\pi+2.2.\pi}{4})=2(cos\frac{5\pi}{4}+sen\frac{5\pi}{4})=cos\frac{5\pi}{2}+sen\frac{5\pi}{2}=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$

k = 3 =>$z_{3}=2(cos\frac{\pi+2.3.\pi}{4}+sen\frac{\pi+2.3.\pi}{4})=2(cos\frac{\pi+6\pi}{4}+sen\frac{\pi+6\pi}{4})=2(cos\frac{7\pi}{4}+sen\frac{7\pi}{4})=cos\frac{7\pi}{2}+sen\frac{7\pi}{2}=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$