I am studying the following notations: $$ \frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right),\qquad \frac{\partial}{\partial\bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right). \tag{1} $$

It might be obtained if we would use the chain rule (I don't know if it is right).

Note that, for $z=x+iy,$ $x=\frac{z+ \bar z}{2}~\hbox{and}~y=\frac{z-\bar z}{2i}=-\frac{i}{2}(z-\bar z).$ Then $$ \frac{\partial x}{\partial z}=\frac{1}{2},~\frac{\partial x}{\partial \bar z}=\frac{1}{2},~\frac{\partial y}{\partial z}=-\frac{i}{2},~\frac{\partial y}{\partial \bar z}=\frac{i}{2}. $$

Let $f(x,y)=u(x,y)+iv(x,y)$. Then $\hat f(z,\bar z)=u(x,y)+iv(x,y),$ and \begin{eqnarray*} \frac{\partial \hat f}{\partial z}=\frac{\partial f}{\partial z} &=&\frac{\partial u}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial z}+i\left[\frac{\partial v}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial z}\right] \\ &=& \frac{1}{2}[u_x -i u_y +i(v_x-iv_y)] \\ &=& \frac{1}{2}\left[ u_x+iv_x-i(u_y+iv_y)\right] =\frac{1}{2} (f_x-i f_y), \end{eqnarray*} where $f_x=u_x+iv_x$ and $f_y=u_y+iv_y.$ Similarly, \begin{eqnarray*} \frac{\partial f}{\partial \bar z} &=&\frac{1}{2}(f_x+i f_y) =\frac{1}{2}\left[ u_x-v_y +i(u_y+v_x)\right]. \end{eqnarray*}

First Question. Are the above assertions true or just give the intuition for $(1)$?

Using $(1)$, it is easy to check that $$ \frac{\partial z}{\partial z} = \frac{\partial \bar{z}}{\partial \bar{z}} = 1,\qquad \frac{\partial \bar{z}}{\partial z} = \frac{\partial z}{\partial \bar{z}} = 0. $$

Second question. I don't know if the following assertions are true: $$ \frac{\partial}{\partial z} f(x,y) = D_{1}\hat f(z, \bar{z}),\qquad \frac{\partial}{\partial \bar{z}} f(x,y) = D_{2}\hat f(z, \bar{z}), $$ If they are true, how to prove them? Is it easy or do I know something more complicated? It seems that this question is related to the first one. May I use the chain rule used above? If so, I can understand all of it.

I would be grateful if you give any comments for my questions. Thanks in advance.

  • $\begingroup$ You are right, that this is all just the chain rule. You have a mapping $\mathbb{R}^2\to \mathbb{R}^2$ given by $(x,y) \mapsto (z,\bar z)$. The only special thing is that the partial derivatives are not real numbers but complex numbers.. $\endgroup$ – Fabian Apr 6 '19 at 8:19
  • 1
    $\begingroup$ @Fabian Thanks for your comment. Is it right that the mapping you defined is from $\mathbb{R}^2$ to $\mathbb{R}^2$ or to $\mathbb{C}^2$? I think I miss something trivial. $\endgroup$ – 04170706 Apr 6 '19 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.