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}