# dz and dz bar: How to derive the Wirtinger derivatives

Something interesting in Complex Analysis is the Wirtinger derivatives: $\displaystyle\boxed{\frac{\partial}{\partial z}:=\frac 12(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})}$ $\displaystyle\boxed{\frac{\partial}{\partial \bar z}:=\frac 12(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})}$

They are often simply defined as such, but one would be curious how to derive them, at least heuristically.

## How to derive Wirtinger derivatives

It turns out we can derive them as such. Any complex function $f(z)$ can be viewed as a function $f(x,y)$ by considering $z=x+iy$. Since $x=\frac 12 (z+\bar z), y=-\frac 12 i(z-\bar z)$, we can also view $f(x,y)$ as $f(z,\bar z)$.

Then by the Chain Rule (for multivariable calculus), we have $\displaystyle\frac{\partial}{\partial x}=\frac{\partial z}{\partial x}\frac{\partial}{\partial z}+\frac{\partial\bar z}{\partial x}\frac{\partial}{\partial\bar z}=\frac{\partial}{\partial z}+\frac{\partial}{\partial\bar z}$.

Similarly, we get $\displaystyle\frac{\partial}{\partial y}=i(\frac{\partial}{\partial z}-\frac{\partial}{\partial\bar z})$.

Then, solving the simultaneous equations we get the Wirtinger derivatives. $\displaystyle i\frac{\partial}{\partial x}+\frac{\partial}{\partial y}=2i\frac{\partial}{\partial z}$. Thus, $\displaystyle\frac{\partial}{\partial z}=\frac 12(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})$.

Similarly, we can get that $\displaystyle\boxed{\frac{\partial}{\partial \bar z}:=\frac 12(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})}$.

Using Wirtinger derivatives, we can express the Cauchy-Riemann equations in a succinct manner: A function satisfies the Cauchy-Riemann equations iff $\displaystyle\frac{\partial f}{\partial\bar z}=0$. ## Author: mathtuition88

https://mathtuition88.com/

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