九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2019年度 複素関数論

Author

Zero

Description

解析関数 \(f(z) = u + iv\) を考える．ただし，\(z = x + iy\) は複素数，\(x\) と \(y\) は実数，\(u\) と \(v\) は実数値関数，\(i = \sqrt{-1}\) である．x と y が極形式 \(x = r\cos\theta\) と \(y = r\sin\theta\) で表されるとき，極形 式のコーシー・リーマンの方程式は以下の式で書けることを示せ．

\[ \frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial v}{\partial \theta},\frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta} \]

Kai

コーシー・リーマンの方程式は以下の式表せる。

\[ \left \{ \begin{align} &\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \tag{①} \\ &\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \tag{②} \\ \end{align} \right. \]

\[ \begin{align} &\frac{\partial x}{\partial r} = \cos\theta ,\frac{\partial y}{\partial r} = \sin\theta \notag \\ &\frac{\partial x}{\partial \theta} = -r\sin\theta, \notag \\ &\frac{\partial y}{\partial \theta} = r\cos\theta \Leftrightarrow \cos\theta = \frac{1}{r}\frac{\partial y}{\partial \theta} \tag{③} \end{align} \]

① の両辺に \(\frac{\partial x}{\partial r} = \cos\theta\) をかける

\[ \begin{aligned} &\frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial r} = \frac{\partial v}{\partial y} \cdot \cos\theta \\ &\frac{\partial u}{\partial r} = \frac{\partial v}{\partial y} \cdot \cos\theta \\ &\frac{\partial u}{\partial r} = \frac{\partial v}{\partial y} \cdot \frac{1}{r}\frac{\partial v}{\partial \theta} \\ \therefore &\frac{\partial u}{\partial r} = \frac{1}{r} \cdot \frac{\partial v}{\partial \theta} \end{aligned} \]

② の両辺に \(-\frac{\partial x}{\partial r} = -\cos\theta\) をかける

\[ \begin{aligned} &\frac{\partial u}{\partial y} \cdot (-\cos\theta) = \frac{\partial v}{\partial x} \cdot \frac{\partial x}{\partial r} \\ &-\frac{\partial u}{\partial y} \cdot \cos\theta = \frac{\partial v}{\partial r} \\ &\frac{\partial v}{\partial r} = -\frac{\partial u}{\partial y} \cdot \cos\theta \\ &\frac{\partial v}{\partial r} = -\frac{\partial u}{\partial y} \cdot \frac{1}{r} \frac{\partial u}{\partial \theta} \\ \therefore &\frac{\partial v}{\partial r} = -\frac{1}{r} \cdot \frac{\partial u}{\partial \theta} \end{aligned} \]