Question

Calcule as raízes quartas de z=-√3/2-1/2i

Answer 1

Vamos inicialmente reescrever $z$ na forma polar. Seu módulo é:

$\rho=\sqrt{(-\frac{\sqrt{3}}{2})^2+(-\frac{1}{2})^2}$

$\rho=\sqrt{\frac{3}{4}+\frac{1}{4}}=1$

Seu argumento é o ângulo $\theta\in[0,2\pi]$ para o qual:

$\sin\theta=\frac{-1/2}{\rho}\;|\;\cos\theta=\frac{-\sqrt{3}/2}{\rho}$

$\sin\theta=-\frac{1}{2}\;|\;\cos\theta=-\frac{\sqrt{3}}{2}$

Daí tiramos que $\theta=\frac{7\pi}{6}$. Temos então que $z=\cos(7\pi/6)+i\sin(7\pi/6)$. Aplicando a 2º lei de Moivre achamos que:

$\sqrt[4]{z}=\sqrt[4]{\rho}\left[\cos\left(\frac{\theta+2k\pi}{4}\right)+i\sin\left(\frac{\theta+2k\pi}{4}\right)\right],\;k\in \{0,1,2,3\}$

$\sqrt[4]{z}=\sqrt[4]{1}\left[\cos\left(\frac{7\pi/6+2k\pi}{4}\right)+i\sin\left(\frac{7\pi/6+2k\pi}{4}\right)\right],\;k\in \{0,1,2,3\}$

$\sqrt[4]{z}=\cos\left(\frac{7\pi+12k\pi}{24}\right)+i\sin\left(\frac{7\pi+12k\pi}{24}\right),\;k\in \{0,1,2,3\}$

Concluindo assim que as raízes quartas são:

$\sqrt[4]{z}_1=\cos\left(\frac{7\pi}{24}\right)+i\sin\left(\frac{7\pi}{24}\right)$

$\sqrt[4]{z}_2=\cos\left(\frac{19\pi}{24}\right)+i\sin\left(\frac{19\pi}{24}\right)$

$\sqrt[4]{z}_3=\cos\left(\frac{31\pi}{24}\right)+i\sin\left(\frac{31\pi}{24}\right)$

$\sqrt[4]{z}_4=\cos\left(\frac{43\pi}{24}\right)+i\sin\left(\frac{43\pi}{24}\right)$