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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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Myyura committed Dec 25, 2024
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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


## **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

$$
\begin{aligned}
&\frac{\partial u}{\partial y} \cdot (-\cos\theta) = \frac{\partial v}{\partial x} \cdot \frac{\partial x}{\partial r} \\
&-\frac{\partial u}{\partial y} \cdot \cos\theta = \frac{\partial v}{\partial r} \\
&\frac{\partial v}{\partial r} = -\frac{\partial u}{\partial y} \cdot \cos\theta \\
&\frac{\partial v}{\partial r} = -\frac{\partial u}{\partial y} \cdot \frac{1}{r} \frac{\partial u}{\partial \theta} \\
\therefore &\frac{\partial v}{\partial r} = -\frac{1}{r} \cdot \frac{\partial u}{\partial \theta}
\end{aligned}
$$
1 change: 1 addition & 0 deletions docs/kakomonn/kyushu_university/index.md
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- [解析学・微積分](ISEE/kyotsu_2019_analysis_calculus.md)
- [ベクトル解析](ISEE/kyotsu_2019_vector_analysis.md)
- [確率・統計](ISEE/kyotsu_2019_prob_stat.md)
- [複素関数論](ISEE/kyotsu_2019_complex_function_theory.md)
- 2018年度:
- [複素関数論](ISEE/kyotsu_2018_complex_function_theory.md)
- 2016年度:
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1 change: 1 addition & 0 deletions mkdocs.yml
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- 解析学・微積分: kakomonn/kyushu_university/ISEE/kyotsu_2019_analysis_calculus.md
- ベクトル解析: kakomonn/kyushu_university/ISEE/kyotsu_2019_vector_analysis.md
- 確率・統計: kakomonn/kyushu_university/ISEE/kyotsu_2019_prob_stat.md
- 複素関数論: kakomonn/kyushu_university/ISEE/kyotsu_2019_complex_function_theory.md
- 2018年度:
- 複素関数論: kakomonn/kyushu_university/ISEE/kyotsu_2018_complex_function_theory.md
- 2016年度:
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