# What is the intuition behind the Wirtinger derivatives?

The Wirtinger differential operators are introduced in complex analysis to simplify differentiation in complex variables. Most textbooks introduce them as if it were a natural thing to do. However, I fail to see the intuition behind this. Most of the time, I even think they tend to make calculations harder.

Is there a simple interpretation of these operators? What mental picture do you have when you use them?

-

It is natural when dealing with a complex-valued function $f(x,y)$ to think of a change of variables from $(x,y)$ to $(z,\bar z)$, where $z = x+i y$. We need to know how the derivatives transform and we find $$\begin{eqnarray*} \frac{\partial}{\partial x} &=& \frac{\partial z}{\partial x} \frac{\partial}{\partial z} + \frac{\partial \bar z}{\partial x} \frac{\partial}{\partial \bar z} \\ &=& \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar z} \\ \frac{\partial}{\partial y} &=& \frac{\partial z}{\partial y} \frac{\partial}{\partial z} + \frac{\partial \bar z}{\partial y} \frac{\partial}{\partial \bar z} \\ &=& i\left(\frac{\partial}{\partial z} - \frac{\partial}{\partial \bar z}\right). \end{eqnarray*}$$ Solving for $(\partial/\partial z, \partial/\partial \bar z)$ we find the usual expressions for the Wirtinger derivatives. The derivatives act as they should, $$\frac{\partial z}{\partial z} = \frac{\partial \bar z}{\partial \bar z} = 1 \qquad \textrm{and}\qquad \frac{\partial \bar z}{\partial z} = \frac{\partial z}{\partial \bar z} = 0.$$
The derivative $\partial/\partial \bar z$ has the important property that if $$\frac{\partial}{\partial \bar z} f(z,\bar z) = 0$$ then $f$ satisfies the Cauchy-Riemann equations and so is holomorphic (and thus analytic). This is straightforward to verify, $$\begin{eqnarray*} \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) f(x,y) &=& \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) (u(x,y)+i v(x,y)) \\ &=& \frac{1}{2}((u_x - v_y) + i(u_y + v_x)), \end{eqnarray*}$$ where subscripts denote partial derivatives. Thus, if $\partial f(z,\bar z)/\partial \bar z = 0$ then the Cauchy-Riemann equations are satisfied, $u_x = v_y$ and $u_y = -v_x$. This gives us a very good intuition indeed. Roughly, if $f$ is not a function of $\bar z$, then $f$ is holomorphic.
@laovultai: As described above, if you solve $\partial_x=\partial_z+\partial_{\bar z}$ and $\partial_y = i(\partial_z-\partial_{\bar z})$ for $\partial_z$ and $\partial_{\bar{z}}$ you will find the usual expressions for the Wirtinger derivatives, $\partial_z = \frac{1}{2}(\partial_x-i\partial_y)$ and $\partial_{\bar z} = \frac{1}{2}(\partial_x+i\partial_y)$. – user26872 Nov 16 '13 at 19:04
@oen.. Do you agree that if $f$ is analytic, then $\frac{df}{dz}=\frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})$, but if eq. $f(x,y)=x^2-y^2+i(y^2-x^2)$ then $0=\frac{df}{dz} \neq \frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})$? So you would need assumption that $f$ is analytic in order that Wirtinger derivatives would hold. – alvoutila Nov 21 '13 at 11:57
@laovultai: If you calculate carefully you will find that $\partial f/\partial \bar z = (1-i)\bar z = (1-i)(x-i y)$. You will find this result whether you use the Wirtinger derivative or instead making the transformation $(x,y)\to(z,\bar z)$ in $f$. – user26872 Nov 21 '13 at 15:33
@oen.. So $f$ can be expressed as $f(z)$ in this case? What about in case if $f(x,y)=x^2+i(y^2-x^2)$ then $0=\frac{df}{dz} \neq \frac{1}{2}(\partial/\partial x + i \partial / \partial y)$ and you need analyticity in order WDO to hold. Right? – alvoutila Nov 22 '13 at 7:58