求z^4=sqrt(3)+i的根.

将复数 $\sqrt{3} + i$ 转化为极坐标形式，得 $r = 2$，$\theta = \frac{\pi}{6}$。 方程 $z^4 = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$ 的解为： \[ z_k = \sqrt[4]{2} \left( \cos\left(\frac{\pi}{24} + \frac{k\pi}{2}\right) + i\sin\left(\frac{\pi}{24} + \frac{k\pi}{2}\right) \right), \quad k = 0, 1, 2, 3 \] 具体解为： \[ \boxed{ \begin{array}{ll} z_0 = \sqrt[4]{2} \left( \cos\frac{\pi}{24} + i\sin\frac{\pi}{24} \right), \\ z_1 = \sqrt[4]{2} \left( \cos\frac{13\pi}{24} + i\sin\frac{13\pi}{24} \right), \\ z_2 = \sqrt[4]{2} \left( \cos\frac{25\pi}{24} + i\sin\frac{25\pi}{24} \right), \\ z_3 = \sqrt[4]{2} \left( \cos\frac{37\pi}{24} + i\sin\frac{37\pi}{24} \right). \end{array} } \] 或等价表示为： \[ \boxed{ \sqrt[4]{2} e^{i\frac{\pi}{24}}, \sqrt[4]{2} e^{i\frac{13\pi}{24}}, \sqrt[4]{2} e^{i\frac{25\pi}{24}}, \sqrt[4]{2} e^{i\frac{37\pi}{24}} } \]