(a) 证明当 $z^2\ne z^{-2}$ 时，

(b) 令 $z=\cos\theta+\mathrm{i}\sin\theta$，证明当 $\sin2\theta\ne0$ 时，

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。

将 $\sum z^{4r}$ 看成首项和公比都为 $z^4$ 的等比和。分子和分母再同乘 $z^{-2}$。

The condition $z^2\ne z^{-2}$ implies $z\ne0$ and $z^4\ne1$. The series has first term $z^4$, common ratio $z^4$, and last term $z^{4n}$. Thus

用 de Moivre 定理写出分子分母。分母为纯虚数，乘上 $-\mathrm{i}$ 化成实数后再取虚部。

For $z=\cos\theta+\mathrm{i}\sin\theta$, de Moivre’s theorem gives $z^2-z^{-2}=2\mathrm{i}\sin2\theta$. Write $a=\cos((4n+2)\theta)-\cos2\theta$ and $b=\sin((4n+2)\theta)-\sin2\theta$. By (a),

The imaginary part of the left-hand side is $\sum_{r=1}^n\sin(4r\theta)$. Equating imaginary parts gives