(a) 已知 $z_1=r_1\mathrm{e}^{\mathrm{i}\theta_1}$ 且 $z_2=r_2\mathrm{e}^{\mathrm{i}\theta_2}$。

证明 $(z_1z_2)^*=z_1^*z_2^*$。

(b) $z=3\mathrm{e}^{\frac14\pi\mathrm{i}}$ 是方程 $z^2+bz+c=0$ 的一个根，其中 $b$ 和 $c$ 为实数。

写出另一个根，并由此求 $b$ 和 $c$ 的值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

指数形式中共轭使辐角取相反数，分别计算乘积的共轭和共轭的乘积即可。

From the given exponential forms,

Its conjugate is

On the other hand,

The expressions are identical, so $\boxed{(z_1z_2)^*=z_1^*z_2^*}$.

实系数二次方程的非实根成共轭对，再用两根之和与积求系数。

Since the coefficients are real, the other root is

The sum of the roots is

and their product is $9$. Comparing with the sum $-b$ and product $c$ gives