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}$$