cos(i x), sin(i x) (純虚数に対するcos, sin)
\(\cos{(i x)}, \sin{(i x)}\) (純虚数に対する\(\cos, \sin\))
\(\cos{\left(i x\right)}\)
$$\begin{eqnarray} \cos{\left(i x\right)} &=&\frac{e^{i\left(ix\right)}+e^{-i\left(ix\right)}}{2 }\;\ldots\;x\in\mathbb{R} \\&=&\frac{e^{-x}+e^{x}}{2 } \\&=&\cosh{\left(x\right)} \end{eqnarray}$$\(\sin{\left(i x\right)}\)
$$\begin{eqnarray} \sin{\left(ix\right)} &=&\frac{e^{i\left(ix\right)}-e^{-i\left(ix\right)}}{2i}\;\ldots\;x\in\mathbb{R} \\&=&\frac{e^{-x}-e^{x}}{2i} \\&=&\frac{1}{i}\frac{-\left(e^{x}-e^{-x}\right)}{2} \\&=&\frac{i}{i}\frac{-1}{i}\sinh{\left(x\right)} \\&=&i\sinh{\left(x\right)} \end{eqnarray}$$\(cosh{(i x)}, sinh{(i x)}\) (純虚数に対する\(\cosh, \sinh\))
\(cosh{(i x)}, sinh{(i x)}\) (純虚数に対する\(\cosh, \sinh\))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(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
$$\left\{ \begin{eqnarray} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y} \\\frac{\partial v}{\partial x}&=&-\frac{\partial u}{\partial y} \end{eqnarray} \right.$$cosh(i x), sinh(i x) (純虚数に対するcosh, sinh)
\(\cosh{\left(ix\right)}\)
$$\begin{eqnarray} \cosh{\left(x\right)}&=&\frac{e^x+e^{-x}}{2} \\\cosh{\left(ix\right)}&=&\frac{e^{ix}+e^{-ix}}{2} \\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}+\left\{\cos{\left(-x\right)}+i\sin{\left(-x\right)}\right\}}{2} \\&&\;\ldots\;e^{ix}=\cos{\left(x\right)}+i\sin{\left(x\right)} \\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}+\left\{\cos{\left(x\right)}-i\sin{\left(x\right)}\right\}}{2} \\&&\;\ldots\;\cos{\left(-x\right)}=\cos{\left(x\right)},\;\sin{\left(-x\right)}=-\sin{\left(-x\right)} \\&=&\frac{\cos{\left(x\right)}\color{red}{+i\sin{\left(x\right)}}\color{black}{+\cos{\left(x\right)}}\color{red}{-i\sin{\left(x\right)}}}{2} \\&=&\frac{\cos{\left(x\right)}+\cos{\left(x\right)}}{2} \\&=&\frac{2\cos{\left(x\right)}}{2} \\&=&\cos{\left(x\right)} \end{eqnarray}$$\(\sinh{\left(ix\right)}\)
$$\begin{eqnarray} \sinh{\left(x\right)}&=&\frac{e^x-e^{-x}}{2} \\\sinh{\left(ix\right)}&=&\frac{e^{ix}-e^{-ix}}{2} \\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}-\left\{\cos{\left(-x\right)}+i\sin{\left(-x\right)}\right\}}{2} \\&&\;\ldots\;e^{ix}=\cos{\left(x\right)}+i\sin{\left(x\right)} \\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}-\left\{\cos{\left(x\right)}-i\sin{\left(x\right)}\right\}}{2} \\&&\;\ldots\;\cos{\left(-x\right)}=\cos{\left(x\right)},\;\sin{\left(-x\right)}=-\sin{\left(-x\right)} \\&=&\frac{\color{red}{\cos{\left(x\right)}}\color{black}{+i\sin{\left(x\right)}}\color{red}{-\cos{\left(x\right)}}\color{black}{+i\sin{\left(x\right)}}}{2} \\&=&\frac{i\sin{\left(x\right)}+i\sin{\left(x\right)}}{2} \\&=&\frac{2i\sin{\left(x\right)}}{2} \\&=&i\sin{\left(x\right)} \end{eqnarray}$$\(cos{(i x)}, sin{(i x)}\) (純虚数に対する\(\cos, \sin\))
\(cos{(i x)}, sin{(i x)}\) (純虚数に対する\(\cos, \sin\))複素数の行列表現
複素数の行列表現
$$ \begin{eqnarray} a+bi &\leftrightarrow& \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \\ \\積 \\ (a+bi)\cdot(c+di) &=&ac+adi+bic+bidi \\&=&ac-bd+adi+bci \\&=&(ac-bd)+(ad+bc)i \\ \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \cdot \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} &=& \begin{bmatrix} ac-bd&-(ad+bc)\\ ad+bc&ac-bd\\ \end{bmatrix} \\ \\積(交換) \\ (c+di)\cdot(a+bi) &=&(ac-bd)+(ad+bc)i \\ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} \cdot \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} &=& \begin{bmatrix} ac-bd&-(ad+bc)\\ ad+bc&ac-bd\\ \end{bmatrix} \;\cdots\;行列の積は一般に非可換だが，この形の行列では可換である \\ \\和 \\ (a+ib)+(c+id) &=&(a+c)+(b+d)i \\ \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} + \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} &=& \begin{bmatrix} a+c&-(b+d)\\ b+d&a+c\\ \end{bmatrix} \\ \\差 \\ (a+bi)-(c+di) &=&(a-c)+(b-d)i \\ \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} - \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} &=& \begin{bmatrix} a-c&-(b-d)\\ b-d&a-c\\ \end{bmatrix} \\ \\逆数 \\ \frac{1}{c+di} &=&\frac{1}{c+di}\frac{c-di}{c-di} \\&=&\frac{c-di}{c^2+d^2} \\ \frac{1}{ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} } &=& \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix}^{-1} \;\cdots\;\frac{1}{A}=A^{-1} \\&=& \frac{1}{ \begin{vmatrix} c&d\\ -d&c\\ \end{vmatrix} } \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \;\cdots\; \begin{bmatrix} a&b\\ c&d\\ \end{bmatrix}^{-1} = \frac{1}{ \begin{vmatrix} a&b\\ c&d\\ \end{vmatrix} } \begin{bmatrix} d&-b\\ -c&a\\ \end{bmatrix} \\&=& \frac{1}{c^2+d^2} \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \\ \\商 \\ \frac{a+bi}{c+di} &=&\frac{(a+bi)(c-di)}{(c+di)(c-di)} \\&=&\frac{(ac+bd)+(bc-ad)i}{c^2+d^2} \\ \frac{ \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} }{ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} } &=& \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \cdot \frac{1}{ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} } \\&=& \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \cdot \left( \frac{1}{c^2+d^2} \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \right) \;\cdots\; \frac{1}{ \begin{bmatrix} c&-d\\ d&c\\ \end{bmatrix} } = \frac{1}{c^2+d^2} \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \\&=& \frac{1}{c^2+d^2} \begin{bmatrix} a&-b\\ b&a\\ \end{bmatrix} \cdot \begin{bmatrix} c&d\\ -d&c\\ \end{bmatrix} \\&=& \frac{1}{c^2+d^2} \begin{bmatrix} ac+bd&-(bc-ad)\\ bc-ad&ac+bd\\ \end{bmatrix} \end{eqnarray} $$ド・モアブルの定理(de Moivre's theorem)と三角凾数(trigonometric function)の3倍角の公式
ド・モアブルの定理
$$\displaystyle (\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$$3倍角の公式(Triple-angle formulae)
$$\begin{array}{rcl} \displaystyle (\cos\theta+i\sin\theta)^3 &=& \displaystyle \cos^3\theta+3\cos^2\theta(i\sin\theta)+3\cos\theta(i\sin\theta)^2+(i\sin\theta)^3\\ &=& \displaystyle \cos^3\theta+i3\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\\ &=& \displaystyle \left(\cos^3\theta-3\cos\theta\sin^2\theta\right)+i\left(3\cos^2\theta\sin\theta-\sin^3\theta\right)\\ &=& \displaystyle \cos(3\theta)+i\sin(3\theta)\,\dotso\,これがド・モアブルの定理の結果と等しい． \end{array}$$ $$\begin{array}{rcl} \displaystyle \cos(3\theta)&=& \cos^3\theta-3\cos\theta\sin^2\theta\\ &=& \cos^3\theta-3\cos\theta(1-\cos^2\theta)\\ &=& \cos^3\theta-3\cos\theta+3\cos^3\theta\\ &=& 4\cos^3\theta-3\cos\theta\\ \displaystyle \sin(3\theta)&=& 3\cos^2\theta\sin\theta-\sin^3\theta\\ &=& 3\sin\theta(1-\sin^2\theta)-\sin^3\theta\\ &=& 3\sin\theta-3\sin^3\theta-\sin^3\theta\\ &=& 3\sin\theta-4\sin^3\theta\\ \end{array}$$4倍角の場合
$$\begin{array}{rcl} \displaystyle (\cos\theta+i\sin\theta)^4 &=& \displaystyle \cos^4\theta+4\cos^3\theta(i\sin\theta)+6\cos^2\theta(i\sin\theta)^2+4\cos\theta(i\sin\theta)^3+(i\sin\theta)^4\\ &=& \displaystyle \cos^4\theta+i4\cos^3\theta\sin\theta-6\cos^2\theta\sin^2\theta-i4\cos\theta\sin^3\theta+\sin^4\theta\\ &=& \displaystyle \left(\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\right)+i\left(4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\right)\\ &=& \displaystyle \cos(4\theta)+i\sin(4\theta)\,\dotso\,これがド・モアブルの定理の結果と等しい． \end{array}$$ $$\begin{array}{rcl} \displaystyle \cos(4\theta)&=&\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\ &=&\cos^4\theta-6\cos^2\theta(1-\cos^2\theta)+(1-\cos^2\theta)^2\\ &=&\cos^4\theta-6\cos^2\theta+6\cos^4\theta+1-2\cos^2\theta+\cos^4\theta\\ &=&\cos^4\theta+6\cos^4\theta+\cos^4\theta-6\cos^2\theta-2\cos^2\theta+1\\ &=&8\cos^4\theta-8\cos^2\theta+1\\ \displaystyle \sin(4\theta)&=&4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\\ &=&\cos\theta(4\cos^2\theta\sin\theta-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta(1-\sin^2\theta)-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta-4\sin^3\theta-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta-8\sin^2\theta)\\ \end{array}$$
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