ディガンマ凾数の相反公式
ディガンマ凾数の相反公式
ディガンマ凾数の定義
$$\begin{eqnarray} \psi\left(z\right)&=&\frac{\mathrm{d}}{\mathrm{d}z} \log{\left(\Gamma\left(z\right)\right)} \\&=&\frac{\Gamma^\prime\left(z\right)}{\Gamma\left(z\right)} \end{eqnarray}$$\(1-z\)のディガンマ凾数
$$\begin{eqnarray} \psi\left(1-z\right) &=&\frac{\mathrm{d}}{\mathrm{d}z} \log{\left(\Gamma\left(1-z\right)\right)} \\&=&\frac{\mathrm{d}}{\mathrm{d}u} \log{\left(\Gamma\left(u\right)\right)}\frac{\mathrm{d}u}{\mathrm{d}z} \;\ldots\;u=1-z,\;\frac{\mathrm{d}u}{\mathrm{d}z}=-1 \\&=&-\frac{\mathrm{d}}{\mathrm{d}z}\log\left(\Gamma\left(1-z\right)\right) \\&=&-\frac{\mathrm{d}}{\mathrm{d}z}\log\left(\frac{\Gamma\left(z\right)}{\Gamma\left(z\right)}\Gamma\left(1-z\right)\right) \\&=&-\frac{\mathrm{d}}{\mathrm{d}z}\log\left(\frac{1}{\Gamma\left(z\right)}\frac{\pi}{\sin{\left(\pi z\right)}}\right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/blog-post_26.html}{\Gamma\left(z\right)\Gamma\left(1-z\right)=\frac{\pi}{\sin{\left(\pi z\right)}}} \\&=&-\left\{ \frac{\mathrm{d}}{\mathrm{d}z} \log \left(\pi\right) -\frac{\mathrm{d}}{\mathrm{d}z} \log \left(\sin{\left(\pi z\right)}\right) -\frac{\mathrm{d}}{\mathrm{d}z} \log \left(\Gamma\left(z\right)\right) \right\} \\&=&-\left\{0-\pi\cot{\left(\pi z\right)}-\psi\left(z\right)\right\} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/logsinz.html}{\frac{\mathrm{d}}{\mathrm{d}z} \log \sin{\left(\pi z\right)}=\pi\cot{\left(\pi z\right)}} \\&&\;\ldots\;\frac{\mathrm{d}}{\mathrm{d}z} \log \Gamma\left(z\right)=\psi\left(z\right),\;(ディガンマ凾数の定義) \\&=&\pi\cot{\left(\pi z\right)}+\psi\left(z\right) \end{eqnarray}$$ディガンマ凾数の相反公式
$$\begin{eqnarray} \psi\left(1-z\right) &=&\pi\cot{\left(\pi z\right)}+\psi\left(z\right) \\\psi\left(1-z\right)-\psi\left(z\right)&=&\pi\cot{\left(\pi z\right)} \end{eqnarray}$$log(sin(πz))の微分
\(\log{\left(\sin{\left(\pi z\right)}\right)}\)の微分
$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\log{\left(\sin{\left(\pi z\right)}\right)} &=&\frac{\mathrm{d}}{\mathrm{d}f}\log{\left(\sin{\left(f\right)}\right)}\frac{\mathrm{d}f}{\mathrm{d}z} \;\ldots\;f=\pi z,\;f\in\mathbb{C} \\&=&\frac{\mathrm{d}}{\mathrm{d}g}\log{\left(g\right)}\frac{\mathrm{d}g}{\mathrm{d}f}\frac{\mathrm{d}f}{\mathrm{d}z} \;\ldots\;g=\sin{\left(f\right)},\;g\in\mathbb{C} \\&=&\frac{1}{g}\;\cos{\left(f\right)}\;\pi \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/logz.html}{\frac{\mathrm{d}}{\mathrm{d}g}\log{\left(g\right)}=\frac{1}{g}} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/sinz.html}{\frac{\mathrm{d}}{\mathrm{d}f}\sin{\left(f\right)}=\cos{\left(f\right)}} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/az.html}{\frac{\mathrm{d}}{\mathrm{d}z}\pi z=\pi} \\&=&\frac{1}{\sin\left(f\right)}\;\cos{\left(f\right)}\;\pi \\&=&\frac{1}{\sin\left(\pi z\right)}\;\cos{\left(\pi z\right)}\;\pi \\&=&\pi\frac{\cos{\left(\pi z\right)}}{\sin\left(\pi z\right)} \\&=&\pi\frac{1}{\tan\left(\pi z\right)} \;\ldots\;\tan\left(\pi z\right)=\frac{\sin\left(\pi z\right)}{\cos{\left(\pi z\right)}} \\&=&\pi\cot{\left(\pi z\right)} \end{eqnarray}$$sin(z)の微分
\(\sin{\left(z\right)}\)の微分
\(u+iv\)で表す
$$\begin{eqnarray} \sin{\left(z\right)} &=&\sin{\left(x\right)}\cos{\left(iy\right)}+\cos{\left(x\right)}\sin{\left(iy\right)} \\&=&\sin{\left(x\right)}\cosh{\left(y\right)}+\cos{\left(x\right)}i\sinh{\left(y\right)} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cosi-x-sini-x-cos-sin.html}{\cos\left(iy\right)=i\cosh\left(y\right),\;\sin\left(iy\right)=i\sinh\left(y\right)} \\&=&\sin{\left(x\right)}\cosh{\left(y\right)}+i\cos{\left(x\right)}\sinh{\left(y\right)} \\&=&u(x,y)+iv(x,y) \end{eqnarray}$$ $$\left\{\begin{eqnarray} u(x,y)&=&\sin{\left(x\right)}\cosh{\left(y\right)} \\v(x,y)&=&\cos{\left(x\right)}\sinh{\left(y\right)} \end{eqnarray}\;\ldots\;x,y\in\mathbb{R}\right.$$\(u,v\)を\(x,y\)で偏微分する
$$\begin{eqnarray} \frac{\partial u(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\sin{\left(x\right)}\cosh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\cosh{\left(y\right)}\frac{\partial}{\partial x}\sin{\left(x\right)} \\&=&\cosh{\left(y\right)}\cos{\left(x\right)} \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\sin{\left(\theta\right)}=\cos{\left(\theta\right)} \\&=&\cos{\left(x\right)}\cosh{\left(y\right)} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial u(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\sin{\left(x\right)}\cosh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\sin{\left(x\right)}\frac{\partial}{\partial y}\cosh{\left(y\right)} \\&=&\sin{\left(x\right)}\sinh{\left(y\right)} \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\cosh{\left(\theta\right)}=\sinh{\left(\theta\right)} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial x} &=&\frac{\partial}{\partial x}\cos{\left(x\right)}\sinh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\sinh{\left(y\right)}\frac{\partial}{\partial x}\cos{\left(x\right)} \\&=&\sinh{\left(y\right)}\left(-\sin{\left(x\right)}\right) \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\cos{\left(\theta\right)}=-\sin{\left(\theta\right)} \\&=&-\sin{\left(x\right)}\sinh{\left(y\right)} \end{eqnarray}$$ $$\begin{eqnarray} \frac{\partial v(x,y)}{\partial y} &=&\frac{\partial}{\partial y}\cos{\left(x\right)}\sinh{\left(y\right)} \;\ldots\;x,y\in\mathbb{R} \\&=&\cos{\left(x\right)}\frac{\partial}{\partial y}\sinh{\left(y\right)} \\&=&\cos{\left(x\right)}\cosh{\left(y\right)} \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}\theta}\sinh{\left(\theta\right)}=\cosh{\left(\theta\right)} \end{eqnarray}$$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
$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\sin{\left(z\right)} &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}} \\&=&\cos{\left(x\right)}\cosh{\left(y\right)}+i\left(-\sin{\left(x\right)}\sinh{\left(y\right)}\right) \\&=&\cos{\left(x\right)}\cosh{\left(y\right)}-i\sin{\left(x\right)}\sinh{\left(y\right)} \\&=&\cos{\left(x\right)}\cos{\left(iy\right)}-\sin{\left(x\right)}\sin{\left(iy\right)} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cosi-x-sini-x-cos-sin.html}{\cos\left(iy\right)=\cosh\left(y\right),\;\sin\left(iy\right)=i\sinh\left(y\right)} \\&=&\cos{\left(x+iy\right)} \\&=&\cos{\left(z\right)} \end{eqnarray}$$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
$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\log{\left(z\right)} &=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}} \\&=&\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2} \\&=&\frac{x-iy}{x^2+y^2} \\&=&\frac{\cancel{x-iy}}{(x+iy)\cancel{(x-iy)}} \\&&\;\ldots\;(x+iy)(x-iy)=x^2\cancel{+ixy}\cancel{-ixy}-i^2y^2=x^2+y^2 \\&=&\frac{1}{x+iy} \\&=&\frac{1}{z}\;\ldots\;z=x+iy \end{eqnarray}$$cot(az)の微分
\(cot\left(az\right)\)の微分
\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}z}\cot{\left(az\right)} &=&\frac{\mathrm{d}}{\mathrm{d}w}\cot{\left(w\right)}\frac{\mathrm{d}w}{\mathrm{d}z} \\&&\;\ldots\;a\in\mathbb{R},\;z\in\mathbb{C} \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/wz.html}{w=az,\;\frac{\mathrm{d}w}{\mathrm{d}z}=a} \\&=&\frac{-1}{\sin^2{\left(az\right)}}\cdot a \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/cotz.html}{\frac{\mathrm{d}}{\mathrm{d}w}\cot{\left(w\right)}=\frac{-1}{\sin^2{\left(w\right)}}} \\&=&-\frac{a}{\sin^2{\left(az\right)}} \end{eqnarray}cot(z)の微分
\(u(x,y)\)を\(x\)で偏微分する
$$\begin{eqnarray} \frac{\partial u}{\partial x} &=&\frac{\partial}{\partial x}\href{https://shikitenkai.blogspot.com/2021/07/cotzuxyivxy.html}{\frac{\cos{\left(x\right)}\sin{\left(x\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}} \\&=&\frac{\sinh^2\left(y\right)(\cos^2\left(x\right)-\sin^2\left(x\right))-\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cos^2\left(x\right)\sinh^2\left(y\right)-\sinh^2\left(y\right)\sin^2\left(x\right)-\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cos^2\left(x\right)\sinh^2\left(y\right)-\sin^2\left(x\right)(1+\sinh^2\left(y\right))}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cos^2\left(x\right)\sinh^2\left(y\right)-\sin^2\left(x\right)\cosh^2\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \end{eqnarray}$$\(u(x,y)\)を\(y\)で偏微分する
$$\begin{eqnarray} \frac{\partial u}{\partial y} &=&\frac{\partial}{\partial y}\href{https://shikitenkai.blogspot.com/2021/07/cotzuxyivxy.html}{\frac{\cos{\left(x\right)}\sin{\left(x\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}} \\&=&-\frac{2\sin\left(x\right)\cos\left(x\right)\sinh\left(y\right)\cosh\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \end{eqnarray}$$\(v(x,y)\)を\(x\)で偏微分する
$$\begin{eqnarray} \frac{\partial v}{\partial x} &=&\frac{\partial}{\partial x}\href{https://shikitenkai.blogspot.com/2021/07/cotzuxyivxy.html}{\frac{-\cosh{\left(y\right)}\sinh{\left(y\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}} \\&=&\frac{2\sin\left(x\right)\cos\left(x\right)\sinh\left(y\right)\cosh\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \end{eqnarray}$$\(v(x,y)\)を\(y\)で偏微分する
$$\begin{eqnarray} \frac{\partial v}{\partial y} &=&\frac{\partial}{\partial y}\href{https://shikitenkai.blogspot.com/2021/07/cotzuxyivxy.html}{\frac{-\cosh{\left(y\right)}\sinh{\left(y\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}} \\&=&\frac{\cosh^2\left(y\right)(\sinh^2\left(y\right)-\sin^2\left(x\right))-\sinh^2\left(y\right)(\sinh^2\left(y\right)+\sin^2\left(x\right))}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cosh^2\left(y\right)\sinh^2\left(y\right)-\cosh^2\left(y\right)\sin^2\left(x\right)-\sinh^2\left(y\right)\sinh^2\left(y\right)-\sinh^2\left(y\right)\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\sinh^2\left(y\right)(\cosh^2\left(y\right)-\sinh^2\left(y\right))-\cosh^2\left(y\right)\sin^2\left(x\right)-\sinh^2\left(y\right)\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\sinh^2\left(y\right)-\cosh^2\left(y\right)\sin^2\left(x\right)-\sinh^2\left(y\right)\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\sinh^2\left(y\right)(1-\sin^2\left(x\right))-\cosh^2\left(y\right)\sin^2\left(x\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\cos^2\left(x\right)\sinh^2\left(y\right)-\sin^2\left(x\right)\cosh^2\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \end{eqnarray}$$コーシー・リーマンの関係式を満たす
$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{ \left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial u}{\partial y}&=&-\frac{\partial v}{\partial x} \end{eqnarray} \right. }$$実軸(x)方向の微分
$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}\cot{}&=&\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{\cos^2\left(x\right)\sinh^2\left(y\right)-\sin^2\left(x\right)\cosh^2\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} +i\frac{2\sin\left(x\right)\cos\left(x\right)\sinh\left(y\right)\cosh\left(y\right)}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\left(\cos\left(x\right)\sinh\left(y\right)+i\sin\left(x\right)\cosh\left(y\right)\right)^2}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\left(\cos\left(x\right)\sinh\left(y\right)+i\sin\left(x\right)\cosh\left(y\right)\right)^2}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\left(\cos\left(x\right)\sinh\left(y\right)+i\sin\left(x\right)\cosh\left(y\right)\right)^2}{\left(\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}\right)^2} \\&=&\frac{\left(\cos\left(x\right)\frac{1}{i}\sin\left(iy\right)+i\sin\left(x\right)\cos\left(iy\right)\right)^2} {\left(\sin^2{\left(x\right)}+\left(\frac{1}{i}\sin\left(iy\right)\right)^2\right)^2} \;\ldots\;\cos\left(iy\right)=i\cosh\left(y\right),\;\sin\left(iy\right)=i\sinh\left(y\right),\;\sinh\left(y\right)=\frac{1}{i}\sin\left(iy\right) \\&=&\frac{i^2}{i^2}\frac{\left(\cos\left(x\right)\frac{1}{i}\sin\left(iy\right)+i\sin\left(x\right)\cos\left(iy\right)\right)^2} {\left(\sin^2{\left(x\right)}+\left(\frac{1}{i}\right)^2\left(\sin\left(iy\right)\right)^2\right)^2} \\&=&\frac{1}{i^2}\frac{\left(i\left(\cos\left(x\right)\frac{1}{i}\sin\left(iy\right)+i\sin\left(x\right)\cos\left(iy\right)\right)\right)^2} {\left(\sin^2{\left(x\right)}-\sin^2\left(iy\right)\right)^2} \\&=&\frac{1}{-1}\frac{\left(\cos\left(x\right)\sin\left(iy\right)-\sin\left(x\right)\cos\left(iy\right)\right)^2} {\left(\sin^2{\left(x\right)}-\sin^2\left(iy\right)\right)^2} \\&=&-\frac{\left(-\left(-\cos\left(x\right)\sin\left(iy\right)+\sin\left(x\right)\cos\left(iy\right)\right)\right)^2} {\left(\sin^2{\left(x\right)}-\sin^2\left(iy\right)\right)^2} \\&=&-\frac{\left(-\sin(x-iy)\right)^2} {\left(\sin{\left(x+iy\right)}\sin{\left(x-iy\right)}\right)^2} \\&&\;\ldots\;\sin^2{\left(x\right)}-\sin^2\left(iy\right) \\&&\;\ldots\;=\sin^2{\left(x\right)}-\sin^2\left(iy\right)\color{red}{-\sin^2{\left(x\right)}\sin^2\left(iy\right)+\sin^2{\left(x\right)}\sin^2\left(iy\right)} \\&&\;\ldots\;=\sin^2{\left(x\right)}\left(1-\sin^2\left(iy\right)\right)-\sin^2{\left(iy\right)}\left(1-\sin^2\left(x\right)\right) \\&&\;\ldots\;=\sin^2{\left(x\right)}\cos^2\left(iy\right)-\sin^2{\left(iy\right)}\cos^2{\left(x\right)} \\&&\;\ldots\;=\left\{\sin{\left(x\right)}\cos\left(iy\right)+\sin{\left(iy\right)}\cos{\left(x\right)}\right\} \left\{\sin{\left(x\right)}\cos\left(iy\right)-\sin{\left(iy\right)}\cos{\left(x\right)}\right\} \\&&\;\ldots\;=\sin{\left(x+iy\right)}\sin{\left(x-iy\right)} \\&=&-\frac{\cancel{\sin^2(x-iy)}} {\sin^2{\left(x+iy\right)}\cancel{\sin^2{\left(x-iy\right)}}} \\&=&\frac{-1}{\sin^2{\left(x+iy\right)}} \\&=&\frac{-1}{\sin^2{\left(z\right)}} \end{eqnarray}$$cot(z)をu(x,y)+iv(x,y)で表す
$$\begin{eqnarray}
\cot{\left(z\right)}
&=&\frac{1}{\tan{\left(z\right)}}\;\ldots\;z\in\mathbb{C}
\\&=&\frac{\cos{\left(z\right)}}{\sin{\left(z\right)}}
\\&=&\frac{\cos{\left(x+iy\right)}}{\sin{\left(x+iy\right)}}\;\ldots\;x,y\in\mathbb{R}
\\&=&\frac{ \cos{\left(x\right)}\cos{\left(iy\right)} - \sin{\left(x\right)}\sin{\left(iy\right)}}
{\sin{\left(x\right)}\cos{\left(iy\right)}+\cos{\left(x\right)}\sin{\left(iy\right)}}
\\&&\;\ldots\;\cos{\left(\alpha+\beta\right)}=\cos{\left(\alpha\right)}\cos{\left(\beta\right)}-\sin{\left(\alpha\right)}\sin{\left(\beta\right)}
\\&&\;\ldots\;\sin{\left(\alpha+\beta\right)}=\cos{\left(\alpha\right)}\sin{\left(\beta\right)}+\sin{\left(\alpha\right)}\cos{\left(\beta\right)}
\\&=&\frac{ \cos{\left(x\right)}\cosh{\left(y\right)} - \sin{\left(x\right)}i\sinh{\left(y\right)}}
{\sin{\left(x\right)}\cosh{\left(y\right)}+\cos{\left(x\right)}i\sinh{\left(y\right)}}
\\&&\;\ldots\;\cos{\left(iy\right)}=\frac{e^{i\left(iy\right)}+e^{-i\left(iy\right)}}{2 }=\frac{e^{-y}+e^{y}}{2 }=\cosh{\left(y\right)}
\\&&\;\ldots\;\sin{\left(iy\right)}=\frac{e^{i\left(iy\right)}-e^{-i\left(iy\right)}}{2i}=\frac{e^{-y}-e^{y}}{2i}=\frac{1}{i}\frac{-\left(e^{y}-e^{-y}\right)}{2}=\frac{i}{i}\frac{-1}{i}\sinh{\left(y\right)}=i\sinh{\left(y\right)}
\\&=&\frac{ \cos{\left(x\right)}\cosh{\left(y\right)} - i\sin{\left(x\right)}\sinh{\left(y\right)}}
{\sin{\left(x\right)}\cosh{\left(y\right)}+i\cos{\left(x\right)}\sinh{\left(y\right)}}
\frac{\sin{\left(x\right)}\cosh{\left(y\right)}-i\cos{\left(x\right)}\sinh{\left(y\right)}}
{\sin{\left(x\right)}\cosh{\left(y\right)}-i\cos{\left(x\right)}\sinh{\left(y\right)}}
\\&=&\frac{
\cos{\left(x\right)}\cosh{\left(y\right)} \sin{\left(x\right)}\cosh{\left(y\right)}
+\cos{\left(x\right)}\cosh{\left(y\right)} (-i\cos{\left(x\right)}\sinh{\left(y\right)})
- i\sin{\left(x\right)}\sinh{\left(y\right)} \sin{\left(x\right)}\cosh{\left(y\right)}
- i\sin{\left(x\right)}\sinh{\left(y\right)} (-i\cos{\left(x\right)}\sinh{\left(y\right)})
}
{\sin^2{\left(x\right)}\cosh^2{\left(y\right)}+\cos^2{\left(x\right)}\sinh^2{\left(y\right)}}
\\&=&\frac{
\cos{\left(x\right)}\sin{\left(x\right)}\cosh^2{\left(y\right)}
-i\cos^2{\left(x\right)}\cosh{\left(y\right)}\sinh{\left(y\right)}
- i\sin^2{\left(x\right)}\cosh{\left(y\right)}\sinh{\left(y\right)}
- \cos {\left(x\right)}\sin {\left(x\right)}\sinh{\left(y\right)}
}
{\sin^2{\left(x\right)}\cosh^2{\left(y\right)}+\left(1-\sin^2{\left(x\right)}\right)\sinh^2{\left(y\right)}}
\\&=&\frac{
\cos{\left(x\right)}\sin{\left(x\right)}\cosh^2{\left(y\right)}
-i\cos^2{\left(x\right)}\cosh{\left(y\right)}\sinh{\left(y\right)}
- i\sin^2{\left(x\right)}\cosh{\left(y\right)}\sinh{\left(y\right)}
- \cos {\left(x\right)}\sin {\left(x\right)}\sinh{\left(y\right)}
}
{\sin^2{\left(x\right)}\cosh^2{\left(y\right)}+\sinh^2{\left(y\right)}-\sin^2{\left(x\right)}\sinh^2{\left(y\right)}}
\\&=&\frac{
\cos{\left(x\right)}\sin{\left(x\right)}\left(\cosh^2{\left(y\right)}-\sinh{\left(y\right)}\right)
-i\left\{
\left(\cos^2{\left(x\right)}+\sin^2{\left(x\right)}\right)\cosh{\left(y\right)}\sinh{\left(y\right)}
\right\}
}
{\sin^2{\left(x\right)}\left(\cosh^2{\left(y\right)}-\sinh^2{\left(y\right)}\right)+\sinh^2{\left(y\right)}}
\\&=&\frac{ \cos{\left(x\right)}\sin{\left(x\right)}-i\cosh{\left(y\right)}\sinh{\left(y\right)}}
{\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}
\\&=&\frac{\cos{\left(x\right)}\sin{\left(x\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}
+i\frac{-\cosh{\left(y\right)}\sinh{\left(y\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}
\\&=&u(x,y)+iv(x,y)
\end{eqnarray}$$
$$\left\{\begin{eqnarray}
u(x,y)&=&\frac{\cos{\left(x\right)}\sin{\left(x\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}
\\v(x,y)&=&\frac{-\cosh{\left(y\right)}\sinh{\left(y\right)}} {\sin^2{\left(x\right)}+\sinh^2{\left(y\right)}}
\end{eqnarray}\right.$$
wzの微分
\(wz\)の微分
\(u+iv\)で表す
$$\begin{eqnarray} wz&=&(a+ib)(x+iy)\;\ldots\;a,b,x,y\in\mathbb{R},\;w,z\in\mathbb{C} \\&=&ax-by+i(ay+bx) \\&=&u(x,y)+iv(x,y) \end{eqnarray}$$\(u,v\)を\(x,y\)で偏微分する
$$\begin{eqnarray} \frac{\partial u(x,y)}{\partial x}&=&\frac{\partial }{\partial x}(ax-by) \\&=&a \\\frac{\partial u(x,y)}{\partial y}&=&\frac{\partial }{\partial y}(ax-by) \\&=&-b \\\frac{\partial v(x,y)}{\partial x}&=&\frac{\partial }{\partial x}(ay+bx) \\&=&b \\\frac{\partial v(x,y)}{\partial y}&=&\frac{\partial }{\partial y}(ay+bx) \\&=&a \end{eqnarray}$$コーシー・リーマンの関係式を満たす
$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{ \left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial u}{\partial y}&=&-\frac{\partial v}{\partial x} \end{eqnarray} \right. }$$実軸(x)方向の微分
$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}wz&=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}} \\&=&a+ib \\&=&w \end{eqnarray}$$虚軸(y)方向の微分
$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}wz&=&\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{\frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}} \\&=&a-i(-b) \\&=&a+ib \\&=&w \end{eqnarray}$$azの微分
\(az\)の微分
\(u+iv\)で表す
$$\begin{eqnarray} az&=&a(x+iy)\;\ldots\;a,x,y\in\mathbb{R},\;z\in\mathbb{C} \\&=&ax+iay \\&=&u(x,y)+iv(x,y) \end{eqnarray}$$ $$\left\{ \begin{eqnarray} u(x,y)&=&ax \\v(x,y)&=&ay \end{eqnarray} \right.$$\(u,v\)を\(x,y\)で偏微分する
$$\begin{eqnarray} \frac{\partial u(x,y)}{\partial x}&=&\frac{\partial }{\partial x}ax \\&=&a \\\frac{\partial u(x,y)}{\partial y}&=&\frac{\partial }{\partial y}ax \\&=&0 \\\frac{\partial v(x,y)}{\partial x}&=&\frac{\partial }{\partial x}ay \\&=&0 \\\frac{\partial v(x,y)}{\partial y}&=&\frac{\partial }{\partial y}ay \\&=&a \end{eqnarray}$$コーシー・リーマンの関係式を満たす
$$\href{https://shikitenkai.blogspot.com/2021/07/blog-post_19.html}{ \left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial u}{\partial y}&=&-\frac{\partial v}{\partial x} \end{eqnarray} \right. }$$実軸方向での微分
\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}az&=&\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}} \\&=&a+i0 \\&=&a \end{eqnarray}虚軸方向での微分
\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} z}az&=&\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}} \\&=&a-i0 \\&=&a \end{eqnarray}コーシー・リーマンの関係式
コーシー・リーマンの関係式
複素平面の実軸方向の微分(偏微分)
$$\begin{eqnarray} \lim_{\Delta z\rightarrow0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z} &=& \lim_{\Delta x\rightarrow0}\frac{\left\{u\left(x_0+\Delta x, y_0\right)+iv\left(x_0+\Delta x, y_0\right)\right\} -\left\{u\left(x_0, y_0\right)+iv\left(x_0, y_0\right)\right\}}{\Delta x} \;\ldots\;z_0,\Delta zin\mathbb{C},\;x_0,y_0,\Delta x\in\mathbb{R} \\&=& \lim_{\Delta x\rightarrow0}\left\{ \frac{ u\left(x_0+\Delta x, y_0\right) -u\left(x_0, y_0\right) }{\Delta x} +i\frac{ v\left(x_0+\Delta x, y_0\right) -v\left(x_0, y_0\right) }{\Delta x} \right\} \\&=&\frac{\partial u\left(x_0, y_0\right)}{\partial x}+i\frac{\partial v\left(x_0, y_0\right)}{\partial x} \end{eqnarray}$$複素平面の虚軸方向の微分(偏微分)
$$\begin{eqnarray} \lim_{\Delta z\rightarrow0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z} &=& \lim_{\Delta y\rightarrow0}\frac{\left\{u\left(x_0, y_0+\Delta y\right)+iv\left(x_0, y_0+\Delta y\right)\right\} -\left\{u\left(x_0, y_0\right)+iv\left(x_0, y_0\right)\right\}}{i\Delta y} \;\ldots\;z_0,\Delta zin\mathbb{C},\;x_0,y_0,\Delta y\in\mathbb{R} \\&=& \lim_{\Delta y\rightarrow0}\left\{ \frac{ u\left(x_0, y_0+\Delta y\right) -u\left(x_0, y_0\right) }{i\Delta y} +i\frac{ v\left(x_0, y_0+\Delta y\right) -v\left(x_0, y_0\right) }{i\Delta y} \right\} \\&=&\frac{1}{i}\frac{\partial u\left(x_0, y_0\right)}{\partial y}+\frac{i}{i}\frac{\partial v\left(x_0, y_0\right)}{\partial y} \\&=&\frac{i}{i}\frac{1}{i}\frac{\partial u\left(x_0, y_0\right)}{\partial y}+\frac{\partial v\left(x_0, y_0\right)}{\partial y} \\&=&\frac{i}{-1}\frac{\partial u\left(x_0, y_0\right)}{\partial y}+\frac{\partial v\left(x_0, y_0\right)}{\partial y} \\&=&-i\frac{\partial u\left(x_0, y_0\right)}{\partial y}+\frac{\partial v\left(x_0, y_0\right)}{\partial y} \\&=&\frac{\partial v\left(x_0, y_0\right)}{\partial y}+i\left\{-\frac{\partial u\left(x_0, y_0\right)}{\partial y}\right\} \end{eqnarray}$$複素平面の各軸微分結果の実部同士，虚部同士が等しくなる場合という関係
$$\left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial v}{\partial x}&=&-\frac{\partial u}{\partial y} \end{eqnarray} \right.$$
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