log(z)の微分

\(\log{\left(z\right)}\)の微分

\(u+iv\)で表す

$$\begin{eqnarray} \log{\left(z\right)} &=&\log{\left(x+iy\right)} \;\ldots\;z=x+iy,\;z\in\mathbb{C},\;x,y\in\mathbb{R} \\&=&\log{\left( |z| e^{i\arg{\left(z\right)}} \right)} \\&&\;\ldots\;z=|z| e^{i\arg{\left(z\right)}},\;|z|=|x+iy|=\sqrt{x^2+y^2}は実数,\:\arg{\left(z\right)}は実数で多価(集合) \\&&\;\ldots\;\arg{\left(z\right)}=\mathrm{Arg}{\left(z\right)}+2n\pi,\;n\in\mathbb{Z} \\&&\;\ldots\;-\pi\lt\mathrm{Arg}{\left(z\right)}\leq\pi,\;\mathrm{Arg}{\left(z\right)}は実数で一価 \\&=&\log{\left(|z|\right)}+\log{\left(e^{i\arg{\left(z\right)}}\right)} \\&=&\log{\left(|z|\right)}+\log{\left(e^{i\left(\mathrm{Arg}{\left(z\right)}+2n\pi\right)}\right)} \\&=&\log{\left(|z|\right)}+i\left(\mathrm{Arg}{\left(z\right)}+2n\pi\right) \\&=&u(x,y)+iv(x,y)\;\ldots\;v(x,y)は実数で多価(集合) \end{eqnarray}$$ $$\left\{\begin{eqnarray} u(x,y)&=&\log{\left(\sqrt{x^2+y^2}\right)} \\v(x,y)&=&\mathrm{Arg}{\left(z\right)}+2n\pi=\begin{cases} \arctan{\left(\frac{y}{x}\right)}&+2n\pi & (x\gt0) \\\arctan{\left(\frac{y}{x}\right)}+\pi&+2n\pi & (x\lt0\;かつ\;y\geq0) \\\arctan{\left(\frac{y}{x}\right)}-\pi&+2n\pi & (x\lt0\;かつ\;y\lt0) \\\frac{\pi}{2}&+2n\pi & (x=0\;かつ\;y\gt0) \\-\frac{\pi}{2}&+2n\pi & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate} && (x=0\;かつ\;y=0) \end{cases} \end{eqnarray}\;\ldots\;x,y\in\mathbb{R},\;n\in\mathbb{Z}\right.$$

\(u,v\)を\(x,y\)で偏微分する

$$\begin{eqnarray} \frac{\partial u(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\log{\left(\sqrt{x^2+y^2}\right)} \\&=&\frac{\partial}{\partial f}\log{\left(f\right)}\frac{\partial f}{\partial x} \;\ldots\;f=\sqrt{x^2+y^2}=\left(x^2+y^2\right)^{\frac{1}{2}},\;f\in\mathbb{R} \\&=&\frac{\partial}{\partial f}\log{\left(f\right)} \frac{\partial f}{\partial g}\frac{\partial g}{\partial x} \;\ldots\;g=x^2+y^2,\;g\in\mathbb{R} \\&=&\frac{1}{f}\;\frac{1}{2}\left(g\right)^{-\frac{1}{2}}\;2x \\&=&\frac{1}{\sqrt{x^2+y^2}}\;\frac{1}{2\sqrt{x^2+y^2}}\;2x \\&=&\frac{x}{x^2+y^2} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial u(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\log{\left(\sqrt{x^2+y^2}\right)} \\&=&\frac{\partial}{\partial f}\log{\left(f\right)}\frac{\partial f}{\partial y} \;\ldots\;f=\sqrt{x^2+y^2}=\left(x^2+y^2\right)^{\frac{1}{2}},\;f\in\mathbb{R} \\&=&\frac{\partial}{\partial f}\log{\left(f\right)} \frac{\partial f}{\partial g}\frac{\partial g}{\partial y} \;\ldots\;g=x^2+y^2,\;g\in\mathbb{R} \\&=&\frac{1}{f}\;\frac{1}{2}\left(g\right)^{-\frac{1}{2}}\;2y \\&=&\frac{1}{\sqrt{x^2+y^2}}\;\frac{1}{2\sqrt{x^2+y^2}}\;2y \\&=&\frac{y}{x^2+y^2} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\mathrm{Arg}{\left(x+iy\right)}+2n\pi \\&=&\begin{cases} \frac{\partial}{\partial x}\left(\arctan{\left(\frac{y}{x}\right)}+2n\pi\right) & (x\gt0) \\\frac{\partial}{\partial x}\left(\arctan{\left(\frac{y}{x}\right)}+\pi+2n\pi\right) & (x\lt0\;かつ\;y\geq0) \\\frac{\partial}{\partial x}\left(\arctan{\left(\frac{y}{x}\right)}-\pi+2n\pi\right) & (x\lt0\;かつ\;y\lt0) \\\frac{\partial}{\partial x}\left(\frac{\pi}{2}+2n\pi\right) & (x=0\;かつ\;y\gt0) \\\frac{\partial}{\partial x}\left(-\frac{\pi}{2}+2n\pi\right) & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate} & (x=0\;かつ\;y=0) \end{cases} \\&=&\begin{cases} \href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{-y}{x^2+y^2}} & (x\gt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{-y}{x^2+y^2}} & (x\lt0\;かつ\;y\geq0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{-y}{x^2+y^2}} & (x\lt0\;かつ\;y\lt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{0} & (x=0\;かつ\;y\gt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{0} & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate}& (x=0\;かつ\;y=0) \end{cases} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\mathrm{Arg}{\left(x+iy\right)}+2n\pi \\&=&\begin{cases} \frac{\partial}{\partial y}\left(\arctan{\left(\frac{y}{x}\right)}+2n\pi\right) & (x\gt0) \\\frac{\partial}{\partial y}\left(\arctan{\left(\frac{y}{x}\right)}+\pi+2n\pi\right) & (x\lt0\;かつ\;y\geq0) \\\frac{\partial}{\partial y}\left(\arctan{\left(\frac{y}{x}\right)}-\pi+2n\pi\right) & (x\lt0\;かつ\;y\lt0) \\\frac{\partial}{\partial y}\left(\frac{\pi}{2}+2n\pi\right) & (x=0\;かつ\;y\gt0) \\\frac{\partial}{\partial y}\left(-\frac{\pi}{2}+2n\pi\right) & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate} & (x=0\;かつ\;y=0) \end{cases} \\&=&\begin{cases} \href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{x}{x^2+y^2}} & (x\gt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{x}{x^2+y^2}} & (x\lt0\;かつ\;y\geq0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{\frac{x}{x^2+y^2}} & (x\lt0\;かつ\;y\lt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{0} & (x=0\;かつ\;y\gt0) \\\href{https://shikitenkai.blogspot.com/2021/07/blog-post_9.html}{0} & (x=0\;かつ\;y\lt0) \\\mathrm{identerminate}& (x=0\;かつ\;y=0) \end{cases} \end{eqnarray}$$

コーシー・リーマンの関係式を満たす

$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial v}{\partial x}&=&-\frac{\partial u}{\partial y} \end{eqnarray} \right.}$$

実軸方向での微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\log{\left(z\right)} &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}} \\&=&\frac{x}{x^2+y^2}+i\frac{-y}{x^2+y^2} \\&=&\frac{x-iy}{x^2+y^2} \\&=&\frac{\cancel{x-iy}}{(x+iy)\cancel{(x-iy)}} \\&&\;\ldots\;(x+iy)(x-iy)=x^2\cancel{+ixy}\cancel{-ixy}-i^2y^2=x^2+y^2 \\&=&\frac{1}{x+iy} \\&=&\frac{1}{z}\;\ldots\;z=x+iy \end{eqnarray}$$

虚軸方向での微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\log{\left(z\right)} &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}} \\&=&\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2} \\&=&\frac{x-iy}{x^2+y^2} \\&=&\frac{\cancel{x-iy}}{(x+iy)\cancel{(x-iy)}} \\&&\;\ldots\;(x+iy)(x-iy)=x^2\cancel{+ixy}\cancel{-ixy}-i^2y^2=x^2+y^2 \\&=&\frac{1}{x+iy} \\&=&\frac{1}{z}\;\ldots\;z=x+iy \end{eqnarray}$$