log(z)の微分

$\log \left(z\right)$の微分

$u+iv$で表す

$$\begin{array}{rcl}\log \left(z\right)& =& \log (x+iy)\dots z=x+iy,z\in \mathbb{C},x,y\in \mathbb{R}\\ & =& \log (|z|e^{i\arg \left(z\right)})\\ & & \dots z=|z|e^{i\arg \left(z\right)},|z|=|x+iy|=\sqrt{x^2+y^2}\mathrm{は}\mathrm{実}\mathrm{数},\arg \left(z\right)\mathrm{は}\mathrm{実}\mathrm{数}\mathrm{で}\mathrm{多}\mathrm{価}(\mathrm{集}\mathrm{合})\\ & & \dots \arg \left(z\right)=\mathrm{Arg}\left(z\right)+2n\pi ,n\in \mathbb{Z}\\ & & \dots -\pi <\mathrm{Arg}\left(z\right)\le \pi ,\mathrm{Arg}\left(z\right)\mathrm{は}\mathrm{実}\mathrm{数}\mathrm{で}\mathrm{一}\mathrm{価}\\ & =& \log (|z|)+\log \left(e^{i\arg \left(z\right)}\right)\\ & =& \log (|z|)+\log \left(e^{i(\mathrm{Arg}\left(z\right)+2n\pi )}\right)\\ & =& \log (|z|)+i(\mathrm{Arg}\left(z\right)+2n\pi )\\ & =& u(x,y)+iv(x,y)\dots v(x,y)\mathrm{は}\mathrm{実}\mathrm{数}\mathrm{で}\mathrm{多}\mathrm{価}(\mathrm{集}\mathrm{合})\end{array}$$ $$\{\begin{array}{rcl}u(x,y)& =& \log \left(\sqrt{x^2+y^2}\right)\\ v(x,y)& =& \mathrm{Arg}\left(z\right)+2n\pi =\{\begin{array}{lll}\arctan \left(\frac{y}{x}\right)& +2n\pi & (x>0)\\ \arctan \left(\frac{y}{x}\right)+\pi & +2n\pi & (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \arctan \left(\frac{y}{x}\right)-\pi & +2n\pi & (x<0\mathrm{か}\mathrm{つ}y<0)\\ \frac{\pi }{2}& +2n\pi & (x=0\mathrm{か}\mathrm{つ}y>0)\\ -\frac{\pi }{2}& +2n\pi & (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& & (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\end{array}\dots x,y\in \mathbb{R},n\in \mathbb{Z}$$

$u,v$を$x,y$で偏微分する

$$\begin{array}{rcl}\frac{\partial u(x,y)}{\partial x}& =& \frac{\partial }{\partial x}\log \left(\sqrt{x^2+y^2}\right)\\ & =& \frac{\partial }{\partial f}\log \left(f\right)\frac{\partial f}{\partial x}\dots f=\sqrt{x^2+y^2}={(x^2+y^2)}^{\frac{1}{2}},f\in \mathbb{R}\\ & =& \frac{\partial }{\partial f}\log \left(f\right)\frac{\partial f}{\partial g}\frac{\partial g}{\partial x}\dots g=x^2+y^2,g\in \mathbb{R}\\ & =& \frac{1}{f}\frac{1}{2}{\left(g\right)}^{-\frac{1}{2}}2x\\ & =& \frac{1}{\sqrt{x^2+y^2}}\frac{1}{2\sqrt{x^2+y^2}}2x\\ & =& \frac{x}{x^2+y^2}\end{array}$$ $$\begin{array}{rcl}\frac{\partial u(x,y)}{\partial y}& =& \frac{\partial }{\partial y}\log \left(\sqrt{x^2+y^2}\right)\\ & =& \frac{\partial }{\partial f}\log \left(f\right)\frac{\partial f}{\partial y}\dots f=\sqrt{x^2+y^2}={(x^2+y^2)}^{\frac{1}{2}},f\in \mathbb{R}\\ & =& \frac{\partial }{\partial f}\log \left(f\right)\frac{\partial f}{\partial g}\frac{\partial g}{\partial y}\dots g=x^2+y^2,g\in \mathbb{R}\\ & =& \frac{1}{f}\frac{1}{2}{\left(g\right)}^{-\frac{1}{2}}2y\\ & =& \frac{1}{\sqrt{x^2+y^2}}\frac{1}{2\sqrt{x^2+y^2}}2y\\ & =& \frac{y}{x^2+y^2}\end{array}$$ $$\begin{array}{rcl}\frac{\partial v(x,y)}{\partial x}& =& \frac{\partial }{\partial x}\mathrm{Arg}(x+iy)+2n\pi \\ & =& \{\begin{array}{ll}\frac{\partial }{\partial x}(\arctan \left(\frac{y}{x}\right)+2n\pi )& (x>0)\\ \frac{\partial }{\partial x}(\arctan \left(\frac{y}{x}\right)+\pi +2n\pi )& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{\partial }{\partial x}(\arctan \left(\frac{y}{x}\right)-\pi +2n\pi )& (x<0\mathrm{か}\mathrm{つ}y<0)\\ \frac{\partial }{\partial x}(\frac{\pi }{2}+2n\pi )& (x=0\mathrm{か}\mathrm{つ}y>0)\\ \frac{\partial }{\partial x}(-\frac{\pi }{2}+2n\pi )& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\\ & =& \{\begin{array}{ll}\frac{-y}{x^2+y^2}& (x>0)\\ \frac{-y}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{-y}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y<0)\\ 0& (x=0\mathrm{か}\mathrm{つ}y>0)\\ 0& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\end{array}$$ $$\begin{array}{rcl}\frac{\partial v(x,y)}{\partial y}& =& \frac{\partial }{\partial y}\mathrm{Arg}(x+iy)+2n\pi \\ & =& \{\begin{array}{ll}\frac{\partial }{\partial y}(\arctan \left(\frac{y}{x}\right)+2n\pi )& (x>0)\\ \frac{\partial }{\partial y}(\arctan \left(\frac{y}{x}\right)+\pi +2n\pi )& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{\partial }{\partial y}(\arctan \left(\frac{y}{x}\right)-\pi +2n\pi )& (x<0\mathrm{か}\mathrm{つ}y<0)\\ \frac{\partial }{\partial y}(\frac{\pi }{2}+2n\pi )& (x=0\mathrm{か}\mathrm{つ}y>0)\\ \frac{\partial }{\partial y}(-\frac{\pi }{2}+2n\pi )& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\\ & =& \{\begin{array}{ll}\frac{x}{x^2+y^2}& (x>0)\\ \frac{x}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y\ge 0)\\ \frac{x}{x^2+y^2}& (x<0\mathrm{か}\mathrm{つ}y<0)\\ 0& (x=0\mathrm{か}\mathrm{つ}y>0)\\ 0& (x=0\mathrm{か}\mathrm{つ}y<0)\\ \mathrm{identerminate}& (x=0\mathrm{か}\mathrm{つ}y=0)\end{array}\end{array}$$

コーシー・リーマンの関係式を満たす

$$\{\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x}& =& -\frac{\partial u}{\partial y}\end{array}$$

実軸方向での微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}\log \left(z\right)& =& \frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}\\ & =& \frac{x}{x^2+y^2}+i\frac{-y}{x^2+y^2}\\ & =& \frac{x-iy}{x^2+y^2}\\ & =& \frac{\cancel{x-iy}}{(x+iy)\cancel{(x-iy)}}\\ & & \dots (x+iy)(x-iy)=x^2\cancel{+ixy}\cancel{-ixy}-i^2{y}^2=x^2+y^2\\ & =& \frac{1}{x+iy}\\ & =& \frac{1}{z}\dots z=x+iy\end{array}$$

虚軸方向での微分

$$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}z}\log \left(z\right)& =& \frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y}\\ & =& \frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}\\ & =& \frac{x-iy}{x^2+y^2}\\ & =& \frac{\cancel{x-iy}}{(x+iy)\cancel{(x-iy)}}\\ & & \dots (x+iy)(x-iy)=x^2\cancel{+ixy}\cancel{-ixy}-i^2{y}^2=x^2+y^2\\ & =& \frac{1}{x+iy}\\ & =& \frac{1}{z}\dots z=x+iy\end{array}$$