复变函数 第2次作业
Chasse_neige
第二章习题 初等函数与多值函数
1.请证明下列等式
(b)
$$
\sin(z_1 + z_2) = \sin z_1 \cos z_2 + \cos z_1 \sin z_2
$$
$$
\cos(z_1 + z_2) = \cos z_1 \cos z_2 - \sin z_1 \sin z_2
$$
证明
\[
\begin{aligned}
\sin(z_{1} + z_{2}) &= \frac{e^{i (z_{1} + z_{2})} - e^{-i (z_{1} + z_{2})}}{2i} \\\\
&= \frac{(e^{i z_{1}} + e^{-i z_{1}}) (e^{i z_{2}} - e^{-i z_{2}}) + (e^{i z_{1}} - e^{-i z_{1}})(e^{i z_{2}} + e^{-i z_{2}})}{4i} \\\\
&= \frac{e^{i z_{1}} + e^{-i z_{1}}}{2} \frac{e^{i z_{2}} - e^{-i z_{2}}}{2i} + \frac{e^{i z_{1}} - e^{-i z_{1}}}{2i} \frac{e^{i z_{2}} + e^{-i z_{2}}}{2} \\\\
&= \sin z_1 \cos z_2 + \cos z_1 \sin z_2
\end{aligned}
\]
\[
\begin{aligned}
\cos (z_{1} + z_{2}) &= \frac{e^{i (z_{1} + z_{2})} + e^{-i (z_{1} + z_{2})}}{2} \\\\
&= \frac{(e^{i z_{1}} + e^{-i z_{1}}) (e^{i z_{2}} + e^{-i z_{2}}) + (e^{i z_{1}} - e^{-i z_{1}})(e^{i z_{2}} - e^{-i z_{2}})}{4} \\\\
&=
\frac{e^{i z_{1}} + e^{-i z_{1}}}{2} \frac{e^{i z_{2}} + e^{-i z_{2}}}{2} - \frac{e^{i z_{1}} - e^{-i z_{1}}}{2i} \frac{e^{i z_{2}} - e^{-i z_{2}}}{2i} \\\\
&= \cos z_1 \cos z_2 - \sin z_1 \sin z_2
\end{aligned}
\]
(d)
\[
\cosh^2 z - \sinh^2 z = 1, \quad 1 - \tanh^2 z = \text{sech}^2 z
\]
证明
$$
\begin{aligned}
\cosh^{2} z - \sinh^{2} z &= \left(\frac{e^{z} + e^{-z}}{2}\right)^{2} - \left(\frac{e^{z} - e^{-z}}{2}\right)^{2} \\
&= \frac{e^{2z} + e^{-2z} + 2}{4} - \frac{e^{2z} + e^{-2z} - 2}{4} = 1
\end{aligned}
$$
\[
1 - \tanh^{2} z = 1 - \frac{\sinh^{2} z}{\cosh^{2} z} = \frac{\cosh^{2} z - \sinh^{2} z}{\cosh^{2} z} = \frac{1}{\cosh^{2} z} = \text{sech}^{2} z
\]
(f)
$$
\frac{d}{dz} \sinh z = \cosh z, \quad \frac{d}{dz} \cosh z = \sinh z
$$
证明
\[
\frac{d}{dz} \sinh z = \frac{d}{dz} \frac{e^{z} - e^{-z}}{2} = \frac{e^{z} + e^{-z}}{2} = \cosh z
\]
\[
\frac{d}{dz} \cosh z = \frac{d}{dz} \frac{e^{z} + e^{-z}}{2} = \frac{e^{z} - e^{-z}}{2} = \sinh z
\]
2.请证明不等式 (\(x\) 和 \(y\) 均为实数)
$$
|\sinh y| \leq |\sin(x + iy)| \leq |\cosh y|
$$
\[
|\sinh y| \leq |\cos(x + iy)| \leq |\cosh y|
\]
证明
$$
\sin (x + iy) = \sin x \cosh y + i \cos x \sinh y
$$
\[
|\sin (x + iy)| = \sqrt{\sin^{2} x \cosh^{2} y + \cos^{2} x \sinh^{2} y}
\]
\[
\begin{aligned}
&\sqrt{\sin^{2} x \cosh^{2} y + \cos^{2} x \sinh^{2} y}
\\\\
=& \sqrt{\sin^{2} x \cosh^{2} y + \cos^{2} x (\cosh^{2} y - 1)} \\\\
=& \sqrt{\cosh^{2} y - \cos^{2} x} \leq |\cosh y|
\end{aligned}
\]
\[
\begin{aligned}
&\sqrt{\sin^{2} x \cosh^{2} y + \cos^{2} x \sinh^{2} y} \\\\
&= \sqrt{\sin^{2} x (1 + \sinh^{2} y) + \cos^{2} x \sinh^{2} y} \\\\
&= \sqrt{\sin^{2} x + \sinh^{2} y} \geq |\sinh y|
\end{aligned}
\]
\[
\therefore \,\, |\sinh y| \leq |\sin(x + iy)| \leq |\cosh y|
\]
\[
\cos (x + iy) = \cos x \cosh y - i \sin x \sinh y
\]
\[
|\cos (x + iy)| = \sqrt{\cos^{2} x \cosh^{2} y + \sin^{2} x \sinh^{2} y}
\]
\[
\begin{aligned}
&\sqrt{\cos^{2} x \cosh^{2} y + \sin^{2} x \sinh^{2} y} \\\\
&= \sqrt{\cos^{2} x \cosh^{2} y + \sin^{2} x (\cosh^{2} y - 1)} \\\\
&=
\sqrt{\cosh^{2} y - \sin^{2} x} \leq |\cosh y|
\end{aligned}
\]
\[
\begin{aligned}
&\sqrt{\cos^{2} x \cosh^{2} y + \sin^{2} x \sinh^{2} y} \\\\
&= \sqrt{\cos^{2} x (1 + \sinh^{2} y) + \sin^{2} x \sinh^{2} y} \\\\
&=\sqrt{\cos^{2} x + \sinh^{2} y} \geq |\sinh y|
\end{aligned}
\]
\[
\therefore \,\, |\sinh y| \leq |\cos(x + iy)| \leq |\cosh y|
\]
4.判断下列函数是否为多值函数, 若为多值函数, 请分析其支点的情况
(1) \(\sqrt{1 - z^3}\)
是多值函数
可能支点：\(z = 1, z = - \frac{1}{2} + \frac{\sqrt{3}}{2} i, - \frac{1}{2} - \frac{\sqrt{3}}{2} i, z = \infty\)
判断
- \(z=1\)：此时\(z\)绕行一圈后\(w\) 的辐角变化\(\pi\)，是支点
- \(z = - \frac{1}{2} + \frac{\sqrt{3}}{2} i\)：此时\(z\)绕行一圈后\(w\) 的辐角变化\(\pi\)，是支点
- \(z = - \frac{1}{2} - \frac{\sqrt{3}}{2} i\)：此时\(z\)绕行一圈后\(w\) 的辐角变化\(\pi\)，是支点
- 无穷远点：此时\(z\)绕行一圈后\(w\) 的辐角变化\(3 \pi\)，是支点
(2) \(\sqrt[3]{z^2 - 1}\)
是多值函数
可能支点：\(z = 1, z = -1, z = \infty\)
判断
- \(z = 1\)：此时\(z\)绕行一圈后\(w\) 的辐角变化\(\frac{2}{3} \pi\)，是支点
- \(z = - 1\)：此时\(z\)绕行一圈后\(w\) 的辐角变化\(\frac{2}{3} \pi\)，是支点
- 无穷远点：此时\(z\)绕行一圈后\(w\) 的辐角变化\(\frac{4}{3} \pi\)，是支点
(3) \(\sqrt{\cos z}\)
是多值函数
可能支点：\(\cos z = 0\)，即 \(z = \frac{\pi}{2} + k \pi\)
判断
\(z = \frac{\pi}{2} + k \pi \quad (k \in \mathbb{Z})\)：此时\(z\)绕行一圈后\(w\) 的辐角变化\(\pi\)，是支点