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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
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docs/kakomonn/kyushu_university/ISEE/kyotsu_2019_complex_function_theory.md
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comments: false | ||
title: 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 | ||
tags: | ||
- Kyushu-University | ||
--- | ||
# 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 | ||
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## **Author** | ||
Zero | ||
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## **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. | ||
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$$ | ||
\frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial v}{\partial \theta},\frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta} | ||
$$ | ||
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## **Kai** | ||
コーシー・リーマンの方程式は以下の式表せる。 | ||
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$$ | ||
\left \{ | ||
\begin{align} | ||
&\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \tag{①} \\ | ||
&\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \tag{②} \\ | ||
\end{align} | ||
\right. | ||
$$ | ||
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$$ | ||
\begin{align} | ||
&\frac{\partial x}{\partial r} = \cos\theta ,\frac{\partial y}{\partial r} = \sin\theta \notag \\ | ||
&\frac{\partial x}{\partial \theta} = -r\sin\theta, \notag \\ | ||
&\frac{\partial y}{\partial \theta} = r\cos\theta \Leftrightarrow \cos\theta = \frac{1}{r}\frac{\partial y}{\partial \theta} \tag{③} | ||
\end{align} | ||
$$ | ||
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① の両辺に $\frac{\partial x}{\partial r} = \cos\theta$ をかける | ||
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$$ | ||
\begin{aligned} | ||
&\frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial r} = \frac{\partial v}{\partial y} \cdot \cos\theta \\ | ||
&\frac{\partial u}{\partial r} = \frac{\partial v}{\partial y} \cdot \cos\theta \\ | ||
&\frac{\partial u}{\partial r} = \frac{\partial v}{\partial y} \cdot \frac{1}{r}\frac{\partial v}{\partial \theta} \\ | ||
\therefore &\frac{\partial u}{\partial r} = \frac{1}{r} \cdot \frac{\partial v}{\partial \theta} | ||
\end{aligned} | ||
$$ | ||
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② の両辺に $-\frac{\partial x}{\partial r} = -\cos\theta$ をかける | ||
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$$ | ||
\begin{aligned} | ||
&\frac{\partial u}{\partial y} \cdot (-\cos\theta) = \frac{\partial v}{\partial x} \cdot \frac{\partial x}{\partial r} \\ | ||
&-\frac{\partial u}{\partial y} \cdot \cos\theta = \frac{\partial v}{\partial r} \\ | ||
&\frac{\partial v}{\partial r} = -\frac{\partial u}{\partial y} \cdot \cos\theta \\ | ||
&\frac{\partial v}{\partial r} = -\frac{\partial u}{\partial y} \cdot \frac{1}{r} \frac{\partial u}{\partial \theta} \\ | ||
\therefore &\frac{\partial v}{\partial r} = -\frac{1}{r} \cdot \frac{\partial u}{\partial \theta} | ||
\end{aligned} | ||
$$ |
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