(a) 写出当 $z\ne 1$ 时级数 $z+z^2+z^3+\dots+z^n$ 的和。

(b) 求方程 $1+z+z^2+\dots+z^6=0$ 的所有根，写成 $\mathrm{e}^{\mathrm{i}q\pi}$ 的形式，其中 $q$ 为有理数。

(c) 现改为已知 $z=\mathrm{e}^{-1+\mathrm{i}\theta}$，用棣莫弗定理证明

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。

有限等比和的首项和公比都是 $z$，项数为 $n$。

For $z\ne1$, the sum is $\boxed{z(1-z^n)/(1-z)}$.

先排除 $z=1$，因为直接代入时原式为 $7$。其余根可由乘上 $1-z$ 后得到的七次单位根方程求出。

For $z\ne1$, $1+z+\cdots+z^6=(1-z^7)/(1-z)$. Thus the equation holds precisely at the seventh roots of unity other than $1$. Their modulus is $1$ and argument spacing is $2\pi/7$; $z=1$ gives $7$ in the original expression and is excluded. The roots are

因为 $|z|=e^{-1}<1$，可用无穷等比和。由 de Moivre 定理取实部得到所求级数，再将分母乘共轭化成实数。

Here $z=e^{-1}(\cos\theta+\mathrm{i}\sin\theta)$ has modulus $e^{-1}<1$. Hence $\sum_{m=1}^{\infty}z^m=z/(1-z)$. By de Moivre’s theorem, $\operatorname{Re}(z^m)=e^{-m}\cos(m\theta)$, so the desired sum is the real part of $z/(1-z)$. Multiplying by the conjugate of the denominator gives

Indeed, $z(1-\bar z)=z-|z|^2$ and $(1-z)(1-\bar z)=|1-z|^2=1-2\operatorname{Re}z+|z|^2$. The denominator is positive because $|z|<1$, so $z\ne1$. Taking real parts and multiplying numerator and denominator by $e^2$ gives