Calcule raiz cúbica de 1 + i

Para calcular a raiz cúbica de \(1 + i\), podemos usar a forma polar dos números complexos. 1. Encontrar o módulo: \[ r = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} \] 2. Encontrar o argumento: \[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \] 3. Escrever na forma polar: \[ 1 + i = \sqrt{2} \left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \] 4. Calcular a raiz cúbica: Para encontrar a raiz cúbica, usamos a fórmula: \[ z_k = r^{1/3} \left(\cos\left(\frac{\theta + 2k\pi}{3}\right) + i\sin\left(\frac{\theta + 2k\pi}{3}\right)\right) \] onde \(k = 0, 1, 2\). - Para \(k = 0\): \[ z_0 = (\sqrt{2})^{1/3} \left(\cos\left(\frac{\pi/4}{3}\right) + i\sin\left(\frac{\pi/4}{3}\right)\right) = 2^{1/6} \left(\cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right)\right) \] - Para \(k = 1\): \[ z_1 = (\sqrt{2})^{1/3} \left(\cos\left(\frac{\pi/4 + 2\pi}{3}\right) + i\sin\left(\frac{\pi/4 + 2\pi}{3}\right)\right) = 2^{1/6} \left(\cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right)\right) \] - Para \(k = 2\): \[ z_2 = (\sqrt{2})^{1/3} \left(\cos\left(\frac{\pi/4 + 4\pi}{3}\right) + i\sin\left(\frac{\pi/4 + 4\pi}{3}\right)\right) = 2^{1/6} \left(\cos\left(\frac{19\pi}{12}\right) + i\sin\left(\frac{19\pi}{12}\right)\right) \] Assim, as três raízes cúbicas de \(1 + i\) são: 1. \(z_0 = 2^{1/6} \left(\cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right)\right)\) 2. \(z_1 = 2^{1/6} \left(\cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right)\right)\) 3. \(z_2 = 2^{1/6} \left(\cos\left(\frac{19\pi}{12}\right) + i\sin\left(\frac{19\pi}{12}\right)\right)\) Essas são as raízes cúbicas de \(1 + i\).