# $\frac{\partial }{\partial z}$ and $\frac{\partial }{\partial \bar z}$: Wirtinger derivative?

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.

• 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.. – Fabian Apr 6 '19 at 8:19
• @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. – 04170706 Apr 6 '19 at 8:25