Change of variables - Wirtinger derivatives

Question

This is from my class. How did the lecturer get the last two equalities?

Answer 1

In the expression $df = \cfrac{\partial f}{\partial x}dx+\cfrac{\partial f}{\partial y}dy$, just substitute for $dx$ and $dy$. As you've put it, $dx = \cfrac{dz+d\bar{z}}{2}$ and $dy = \cfrac{dz-d\bar{z}}{2i}$. So, $df$ becomes:

$$df= \cfrac{\partial f}{\partial x}\bigg{(}\cfrac{dz+d\bar{z}}{2}\bigg{)}+\cfrac{\partial f}{\partial y}\bigg{(}\cfrac{dz-d\bar{z}}{2i}\bigg{)}$$

Regroup like terms and arrive at:

$$df= \cfrac{1}{2}\bigg{(}\cfrac{\partial f}{\partial x}+\cfrac{\partial f}{i\partial y}\bigg{)}dz+\cfrac{1}{2}\bigg{(}\cfrac{\partial f}{\partial x}-\cfrac{\partial f}{i\partial y}\bigg{)}d\bar{z}$$

Note: $\cfrac{1}{i}=-i$

Also, $df$ is defined as $df = \cfrac{\partial f}{\partial z}dz+\cfrac{\partial f}{\partial \bar{z}}d\bar{z}$. This follows from the total differential for $f(z,\bar{z})$. Additionally, note that the so called Cauchy Riemman (CR) conditions really equal $\cfrac{\partial f}{\partial \bar{z}}=0$. This is a "one-step" method to check whether a function $f(z,\bar{z})$ does satisfy the CR conditions.

Answer 2

It is just the multi-variable chain rule. You have $$(x(z,\bar{z}),y(z,\bar{z})) := (\frac{z+\bar{z}}{2},\frac{z-\bar{z}}{2i}) $$ so $$\frac{\partial }{\partial z} f(x(z,\bar{z}),y(z,\bar{z})) = \frac{\partial f}{\partial x}\frac{dx}{dz} + \frac{\partial f}{\partial y}\frac{dy}{dz} = \frac{\partial f}{\partial x}\frac{1}{2} + \frac{\partial f}{\partial y}\frac{-i}{2}$$ and similarily for $\frac{\partial}{\partial\bar{z}}f$.