# Is there a way to "proof" the Wirtinger derivates?

I'm a bit confused about the wirtinger derivates, as I understand they define:

$$df/dz := 1/2 (df/dx - idf/dy)$$

and

$$df/d\bar z := 1/2 (df/dx +idf/dy)$$

Is there actually a way to derive a holomorph function with $d/d\bar z$, as of the limes definition of the derivate? Or is it just that there are there some benefits if we introduce the operater $d/d\bar z$ defined like that?

If so what does this enable us to do?

I'd like to offer an answer different to the one already provided. I think the perspective will serve you better.

Let's say for the sake of simplicity that $f \in C^{\infty}(\Omega)$, that is, $f$ is a smooth (infinitely differentiable) function on an open set $\Omega \subset \mathbb{C}$. Then, just like you wrote, the first and second Wirtinger derivatives of $f$ are defined to be $$\frac{\partial f}{\partial z} := \frac{1}{2} \left(\frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y} \right).$$ $$\frac{\partial f}{\partial \bar z} := \frac{1}{2} \left(\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right).$$ Let's pause for a second to understand what is happening here. The Wirtinger derivatives make sense for any smooth function, since it has partial derivatives in the $x$ and $y$ directions. A priori, they do not need to have connections to complex analysis.

So why have we defined these derivatives this way? Well, simply looking at the definition and notation of the first formula, we would hope that $\frac{\partial f}{\partial z}$ coincides with the definition of the complex derivative of $f$, $f'(z)$, given by $$f'(z) = \lim_{h \rightarrow 0} \frac{f(z+h)-f(z)}{h}.$$ However, we need to be very careful. Not all smooth functions are holomorphic. So even though our definition makes sense for all smooth functions $f$, it does not follow that $\frac{\partial f}{\partial z}$ is the complex derivative.

However, magic happens when we assume further that $f = u + iv$ is holomorphic, i.e, its complex derivative defined above exists for all $z \in \Omega$. In this case, the real and imaginary parts of $f$ satisfy the Cauchy Riemann equations $$u_x = v_y \,\,\,\,\mbox{and} \,\,\, v_x = -u_y.$$ In this case, we compute

\begin{eqnarray*} \frac{\partial f}{\partial z} :&=& \frac{1}{2} \left(\frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y} \right) \\ &=& \frac{1}{2} \left(u_x + iv_x - iu_y + v_y \right) = \frac{1}{2} \left(u_x + v_y + i(v_x - u_y) \right) \\ &=&\frac{1}{2} \left(u_x + u_x + i(v_x +v_x) \right) = u_x + i v_x \\ &=& f'(z). \end{eqnarray*} In the above computation, I just applied the definition, then used to Cauchy Riemann equations. To conclude that this was equal to $f'(z)$, we note that since $f$ was holomorphic, the limit existed and was independent of how $h$ approached $0$, so approaching along the $x$-axis gives the complex derivative.

The moral of the story is that when we assumed further than $f$ is holomorphic, the Cauchy Riemann equations allowed us to conclude that the first Wirtinger derivative coincides with the complex derivative. We may do a similar computation using the Cauchy Riemann equations to show that $f$ is holomorphic if and only if $$\frac{\partial f}{\partial \bar z} = 0.$$ In other words, $f$ is holomorphic if and only if its second Wirtinger derivative is $0$. This is a great exercise and I recommend you try it.

To summarize, the Wirtinger derivatives are operators that makes sense for smooth functions, but have extra meaning when we apply them to holomorphic functions. In some sense, they are defined to encode the Cauchy Riemann equations in a "complex" way rather than using real partial derivatives. The benefits of introducing this point of view are tough to describe at this point (at least I don't have a quick answer, and this answer is already too long), but feel free to ask me in the comments.

• I did some more reading on my own, but that aswer was exactly what I was looking for! Mar 14, 2017 at 12:26
• I just used this again for exam preperation and understood it just fine and I don't think it's too long, actually I never found a text yet that was able to better explain me what Wirtinger derivatives do, thanks again!! Jun 26, 2017 at 16:27
• @Jack Burkart My prof actually derived this by writing $\frac{\partial F}{\partial x}=\frac{\partial F}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial z'}\frac{\partial z'}{\partial x}$, where $z'$ is the conjugate. Can't we cancel stuff here? That will give 2=1? How can this equation be correct? Feb 4, 2022 at 17:40