复变函数 第1次作业
Chasse_neige
5 若 \(|z| = 1\)，请证明对任意复数 \(a\) 和 \(b\)，均有
$$
\left| \frac{a z + b}{\overline{b} z + \overline{a}} \right| = 1
$$
证明：由于 \(|z| = 1\)，所以 \(z \overline{z} = 1\)
$$
\left| \frac{a z + b}{\overline{b} z + \overline{a}} \right| = \frac{|a z + b|}{|\overline{b} z + \overline{a}|} = \frac{|a z + b|}{\left|\overline{\overline{b} z + \overline{a}}\right|} = \frac{|a z + b|}{|b \overline{z} + a|} = \frac{|a z + b|}{|z \overline{z}(b \overline{z} + a)|} = \frac{|a z + b|}{|\overline{z} (b + a z)|} = \frac{|a z + b|}{|\overline{z}| |a z + b|} = 1
$$
6 请利用单倍角的三角函数表示 \(\cos(n\theta)\) 和 \(\sin(n\theta)\)，并给出 \(\cos(3\theta)\) 和 \(\sin(3\theta)\) 的具体表达式。
\[
\begin{aligned}
\cos (n \theta) &= \frac{e^{i n \theta} + e^{- i n \theta}}{2} = \frac{1}{2} ((\cos \theta + i \sin \theta)^{n} + (\cos \theta - i \sin \theta)^{n}) \\\\
&= \frac{1}{2} (\sum_{k = 0}^{n} i^{k} C_{n}^{k} \cos^{n - k} \theta \sin^{k} \theta + \sum_{k = 0}^{n} (-i)^{k} C_{n}^{k} \cos^{n-k} \theta \sin^{k} \theta) \\\\
&= \frac{1}{2} (2 \sum_{k = 0}^{[\frac{n}{2}]} i^{2k} C_{n}^{2k} \cos^{n-2k} \sin^{2k} \theta) \\\\
&= \sum_{k = 0}^{[\frac{n}{2}]} i^{2k} C_{n}^{2k} \cos^{n-2k} \theta \sin^{2k} \theta
\end{aligned}
\]
\[
\begin{aligned}
\sin (n \theta) & = \frac{e^{i n \theta} - e^{- i n \theta}}{2 i} = \frac{1}{2i} ((\cos \theta + i \sin \theta)^{n} - (\cos \theta - i \sin \theta)^{n}) \\\\
&= \frac{1}{2i} (\sum_{k = 0}^{n} i^{k} C_{n}^{k} \cos^{n - k} \theta \sin^{k} \theta - \sum_{n = 0}^{k} (-i)^{k} \cos^{n - k} \theta \sin^{k} \theta) \\\\
&= \frac{1}{2i} (2i \sum_{n = 0}^{[\frac{n}{2}]} i^{2k} C_{n}^{2k + 1} \cos^{n - 2k - 1} \theta \sin^{2k + 1} \theta) \\\\
&=\sum_{n = 0}^{[\frac{n}{2}]} i^{2k} C_{n}^{2k + 1} \cos^{n - 2k - 1} \theta \sin^{2k + 1} \theta
\end{aligned}
\]
特别地，对于 \(n = 3\)
\[
\begin{aligned}
\cos (3 \theta) &= \sum_{k = 0}^{1} i^{2k} C_{3}^{2k} \cos^{3-2k} \theta \sin^{2k} \theta \\\\
&= \cos^{3} \theta - 3 \cos \theta \sin^{2} \theta
\end{aligned}
\]
\[
\begin{aligned}
\sin (3 \theta) &= \sum_{n = 0}^{1} i^{2k} C_{3}^{2k + 1} \cos^{3 - 2k - 1} \theta \sin^{2k + 1} \theta \\\\
&= 3 \cos^{2} \theta \sin \theta - \sin^{3} \theta
\end{aligned}
\]
7 请利用多倍角三角函数的线性关系表达 \(\cos^n\theta\) 和 \(\sin^n\theta\)，并给出 \(\cos^3\theta\) 和 \(\sin^3\theta\) 的具体表达式。
\[
\begin{aligned}
\cos^{n} \theta &= \left( \frac{e^{i \theta} + e^{- i \theta}}{2} \right)^{n} = \sum_{k = 0}^{[\frac{n + 1}{2}]} C_{n}^{k} \frac{e^{i (n - 2k) \theta} + e^{- i (n - 2k) \theta}}{2^{n}} \\\\
&= \frac{1}{2^{n - 1}} \sum_{k = 0}^{[\frac{n + 1}{2}]} C_{n}^{k} \cos ((n - 2k) \theta)
\end{aligned}
\]
\[
\begin{aligned}
\sin^{n} \theta &= \left( \frac{e^{i \theta} - e^{- i \theta}}{2 i} \right)^{n} = \sum_{k = 0}^{[\frac{n + 1}{2}]} C_{n}^{k} \frac{(-1)^{k} e^{i (n - 2k) \theta} + (-1)^{n - k} e^{- i (n - 2k) \theta}}{(2i)^{n}} \\\\
&=
\begin{cases}
\frac{1}{(2i)^{n - 1}} \sum_{k = 0}^{[\frac{n}{2}]} (-1)^{k} C_{n}^{k} \sin ((n - 2k) \theta) & (\text{n为奇数}) \\\\
\frac{1}{(2i)^{n}} \sum_{k = 0}^{[\frac{n}{2}]} 2 (-1)^{k} C_{n}^{k} \cos ((n - 2k) \theta) & (\text{n为偶数})
\end{cases}
\end{aligned}
\]
特别地，对于 \(n = 3\)
\[
\cos^{3} \theta = \frac{1}{2^{2}} \sum_{k = 0}^{1} C_{3}^{k} \cos ((3 - 2k) \theta) = \frac{1}{4} (\cos (3 \theta) + 3 \cos \theta)
\]
\[
\sin^{3} \theta = \frac{1}{(2i)^{2}} \sum_{k = 0}^{1} (-1)^{k} C_{3}^{k} \sin ((3 - 2k) \theta) = - \frac{1}{4} (\sin (3 \theta) - 3 \sin \theta)
\]
11 证明下列 Cauchy-Riemann 条件，其中 \(w = u + iv = \rho e^{i\varphi}\)，\(z = x + iy = r e^{i\theta}\)：
(a)
$$
\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}, \quad \frac{\partial v}{\partial r} = -\frac{1}{r} \frac{\partial u}{\partial \theta}
$$
考虑沿着径向的逼近方式：
$$
\lim_{\Delta z \to 0} \frac{\Delta w}{\Delta z} = \lim_{\Delta r \to 0} \frac{\Delta u + i \Delta v}{e^{i \theta} \Delta r} =\frac{\partial u}{\partial r} \cos \theta + \frac{\partial v}{\partial r} \sin \theta + i (\frac{\partial v}{\partial r} \cos \theta - \frac{\partial v}{\partial r} \sin \theta)
$$
再考虑沿着角向的逼近方式：
$$
\lim_{\Delta z \to 0} \frac{\Delta w}{\Delta z} = \lim_{\Delta \theta \to 0} \frac{\Delta u + i \Delta v}{r (i \cos \theta - \sin \theta) \Delta \theta} = \frac{1}{r} ((\frac{\partial v}{\partial \theta} \cos \theta - \frac{\partial u}{\partial \theta} \sin \theta) - i (\frac{\partial u }{\partial \theta} \cos \theta + \frac{\partial v}{\partial \theta} \sin \theta))
$$
对比上述二式，得到
\[
\begin{aligned}
\frac{\partial u}{\partial r} \cos \theta + \frac{\partial v}{\partial r} \sin \theta &= \frac{1}{r} (\frac{\partial v}{\partial \theta} \cos \theta - \frac{\partial u}{\partial \theta} \sin \theta) \\\\
\frac{\partial v}{\partial r} \cos \theta - \frac{\partial v}{\partial r} \sin \theta &= - \frac{1}{r} (\frac{\partial u }{\partial \theta} \cos \theta + \frac{\partial v}{\partial \theta} \sin \theta)
\end{aligned}
\]
即
$$
\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}, \quad \frac{\partial v}{\partial r} = -\frac{1}{r} \frac{\partial u}{\partial \theta}
$$
14 已知函数 \(f(z) = u + iv\) 在区域 \(G\) 中解析，则对于常实数 \(a\) 和 \(b\)，\(au + bv\) 在区域中也解析的条件为什么？并据此证明若在 \(G\) 中 \(\text{Im}\, f(z) = 0\)，则 \(f(z)\) 为常函数。
也解析的条件：由于 \(f (z)\) 解析，所以 \(f(z)\) 的各个偏导数在区域 \(G\) 中连续，即 \(au + bv\) 的各个偏导数在区域 \(G\) 中连续，所以 \(au + bv\) 在区域中也解析的充要条件即为 Cauchy-Riemann 条件：
$$
a \frac{\partial u}{\partial x} + b \frac{\partial v}{\partial x} = \frac{\partial u}{\partial y} + b \frac{\partial v}{\partial y} = 0
$$
若在 \(G\) 中 \(\text{Im}\, f(z) = 0\)，则 \(v = 0\)，即有
$$
\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = 0
$$
由于由于 \(f (z)\) 解析，所以在区域 \(G\) 中\(u \in C^{1}\)，所以在区域 \(G\) 中 \(\mathrm{d} u = 0\)
所以对于任意路径的积分有
\[
u (\mathbf{x}) - u(\mathbf{x}\_{0}) = \int_{\mathbf{x}_{0}}^{\mathbf{x}} \mathbf{d} u =0
\]
即\(f(z)\) 为常函数。