复变函数第2次作业

Chasse_neige

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{matrix} \sin (z_{1} + z_{2}) = \frac{e^{i (z_{1} + z_{2})} - e^{- i (z_{1} + z_{2})}}{2 i} = \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 i} \\ = \frac{e^{i z_{1}} + e^{- i z_{1}}}{2} \frac{e^{i z_{2}} - e^{- i z_{2}}}{2 i} + \frac{e^{i z_{1}} - e^{- i z_{1}}}{2 i} \frac{e^{i z_{2}} + e^{- i z_{2}}}{2} = \sin z_{1} \cos z_{2} + \cos z_{1} \sin z_{2} \end{matrix}

\begin{matrix} \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}}}{2 i} \frac{e^{i z_{2}} - e^{- i z_{2}}}{2 i} = \cos z_{1} \cos z_{2} - \sin z_{1} \sin z_{2} \end{matrix}

(d)

\cosh^{2} z - \sinh^{2} z = 1, 1 - \tanh^{2} z = {sech}^{2} z

证明

\begin{matrix} \cosh^{2} z - \sinh^{2} z = {(\frac{e^{z} + e^{- z}}{2})}^{2} - {(\frac{e^{z} - e^{- z}}{2})}^{2} \\ = \frac{e^{2 z} + e^{- 2 z} + 2}{4} - \frac{e^{2 z} + e^{- 2 z} - 2}{4} = 1 \end{matrix}

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} = \sech^{2} z

(f)

\frac{d}{d z} \sinh z = \cosh z, \frac{d}{d z} \cosh z = \sinh z

证明

\frac{d}{d z} \sinh z = \frac{d}{d z} \frac{e^{z} - e^{- z}}{2} = \frac{e^{z} + e^{- z}}{2} = \cosh z

\frac{d}{d z} \cosh z = \frac{d}{d z} \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 + i y) | \leq | \cosh y |

| \sinh y | \leq | \cos (x + i y) | \leq | \cosh y |

证明

\sin (x + i y) = \sin x \cosh y + i \cos x \sinh y

| \sin (x + i y) | = \sqrt{\sin^{2} x \cosh^{2} y + \cos^{2} x \sinh^{2} y}

\begin{matrix} \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{matrix}

\begin{matrix} \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{matrix}

∴ | \sinh y | \leq | \sin (x + i y) | \leq | \cosh y |

\cos (x + i y) = \cos x \cosh y - i \sin x \sinh y

| \cos (x + i y) | = \sqrt{\cos^{2} x \cosh^{2} y + \sin^{2} x \sinh^{2} y}

\begin{matrix} \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{matrix}

\begin{matrix} \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{matrix}

∴ | \sinh y | \leq | \cos (x + i y) | \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

(2) $\sqrt[3]{z^{2} - 1}$

是多值函数

可能支点： $z = 1, z = - 1, z = \infty$

判断

(3) $\sqrt{\cos z}$

是多值函数

可能支点： $\cos z = 0$ ，即 $z = \frac{π}{2} + k π$

判断

$z = \frac{π}{2} + k π (k \in Z)$ ：此时 $z$ 绕行一圈后 $w$ 的辐角变化 $π$ ，是支点

复变函数 第2次作业 ​