$$\begin{eqnarray}
z&=&x+iy\;\ldots\;z\in\mathbb{Z},\;x,y\in\mathbb{R}
\\&=&re^{i\theta}\;\ldots\;r,\theta\in\mathbb{R},\;r\geq0,\;-\pi\lt\theta\leq\pi
\\r&=&|z|=\sqrt{x^2+y^2}
\\\arg{\left(z\right)}&=&\theta+2n\pi\;\left(n\in\mathbb{Z}\right)
\\&=&\mathrm{Arg}{\left(z\right)}+2n\pi\;\left(n\in\mathbb{Z}\right)
\end{eqnarray}$$
$$\begin{eqnarray}
\mathrm{Arg}{\left(z\right)}&=&\mathrm{Arg}{\left(x+iy\right)}
\\&=&\begin{cases}
\tan^{-1}{\left(\frac{y}{x}\right)} & (x\gt0)
\\\tan^{-1}{\left(\frac{y}{x}\right)+\pi} & (x\lt0\;かつ\;y\geq0)
\\\tan^{-1}{\left(\frac{y}{x}\right)-\pi} & (x\lt0\;かつ\;y\lt0)
\\\frac{\pi}{2} & (x=0\;かつ\;y\gt0)
\\-\frac{\pi}{2} & (x=0\;かつ\;y\lt0)
\\不定 & (x=0\;かつ\;y=0)
\end{cases}
\end{eqnarray}$$
xでの偏微分
$$\begin{eqnarray}
\frac{\partial}{\partial x}\arg{\left(z\right)}&=&\frac{\partial}{\partial x}\left\{\theta+2n\pi\right\}
\\&=&\frac{\partial}{\partial x}\left\{\mathrm{Arg}{\left(z\right)}+2n\pi\right\}
\\&=&\frac{\partial}{\partial x}\mathrm{Arg}{\left(z\right)}+\frac{\partial}{\partial x}2n\pi
\\&=&\frac{\partial}{\partial x}\mathrm{Arg}{\left(z\right)}\;\ldots\;\frac{\partial}{\partial x} C=0\;(Cは定数．xの凾数でない)
\end{eqnarray}$$
$$\begin{eqnarray}
\frac{\partial}{\partial x}\mathrm{Arg}{\left(z\right)}&=&\frac{\partial}{\partial x}\mathrm{Arg}{\left(x+iy\right)}
\\&=&\begin{cases}
\frac{\partial}{\partial x}\tan^{-1}{\left(\frac{y}{x}\right)} &=&\frac{-y}{x^2+y^2}& (x\gt0)
\\\frac{\partial}{\partial x}\tan^{-1}{\left(\frac{y}{x}\right)+\pi} &=&\frac{-y}{x^2+y^2}& (x\lt0\;かつ\;y\geq0)
\\\frac{\partial}{\partial x}\tan^{-1}{\left(\frac{y}{x}\right)-\pi} &=&\frac{-y}{x^2+y^2}& (x\lt0\;かつ\;y\lt0)
\\\frac{\partial}{\partial x}\frac{\pi}{2} &=&0& (x=0\;かつ\;y\gt0)
\\\frac{\partial}{\partial x}-\frac{\pi}{2} &=&0& (x=0\;かつ\;y\lt0)
\\不定 &&& (x=0\;かつ\;y=0)
\end{cases}
\ldots\href{https://shikitenkai.blogspot.com/2021/07/tan^{-1}yx.html}{\frac{\partial}{\partial x}\tan^{-1}{\left(\frac{y}{x}\right)}=\frac{-y}{x^2+y^2}}
\end{eqnarray}$$
yでの偏微分
$$\begin{eqnarray}
\frac{\partial}{\partial y}\arg{\left(z\right)}&=&\frac{\partial}{\partial y}\left\{\theta+2n\pi\right\}
\\&=&\frac{\partial}{\partial y}\left\{\mathrm{Arg}{\left(z\right)}+2n\pi\right\}
\\&=&\frac{\partial}{\partial y}\mathrm{Arg}{\left(z\right)}+\frac{\partial}{\partial y}2n\pi
\\&=&\frac{\partial}{\partial y}\mathrm{Arg}{\left(z\right)}\;\ldots\;\frac{\partial}{\partial y} C=0\;(Cは定数．xの凾数でない)
\end{eqnarray}$$
$$\begin{eqnarray}
\frac{\partial}{\partial y}\mathrm{Arg}{\left(z\right)}&=&\frac{\partial}{\partial y}\mathrm{Arg}{\left(x+iy\right)}
\\&=&\begin{cases}
\frac{\partial}{\partial y}\tan^{-1}{\left(\frac{y}{x}\right)} &=&\frac{x}{x^2+y^2}& (x\gt0)
\\\frac{\partial}{\partial y}\tan^{-1}{\left(\frac{y}{x}\right)+\pi} &=&\frac{x}{x^2+y^2}& (x\lt0\;かつ\;y\geq0)
\\\frac{\partial}{\partial y}\tan^{-1}{\left(\frac{y}{x}\right)-\pi} &=&\frac{x}{x^2+y^2}& (x\lt0\;かつ\;y\lt0)
\\\frac{\partial}{\partial y}\frac{\pi}{2} &=&0& (x=0\;かつ\;y\gt0)
\\\frac{\partial}{\partial y}-\frac{\pi}{2} &=&0& (x=0\;かつ\;y\lt0)
\\不定 &&& (x=0\;かつ\;y=0)
\end{cases}
\ldots\href{https://shikitenkai.blogspot.com/2021/07/tan^{-1}yx.html}{\frac{\partial}{\partial y}\tan^{-1}{\left(\frac{y}{x}\right)}=\frac{x}{x^2+y^2}}
\end{eqnarray}$$
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