Doubt in Partial derivative of complex variables

1. May 14, 2015

smart_worker

Today, I had a class on Complex analysis and my professor wrote this on the board :

The Laplacian satisfies this equation :

where,

So, how did he arrive at that equation?

2. May 14, 2015

ShayanJ

$\triangle=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$. Now you should just use chain rule to write derivatives w.r.t. x and y, in terms of derivative w.r.t. $z$ and $\bar z$.

3. May 14, 2015

smart_worker

x = (z + z¯)/2
and
y = (z - z¯)/2i

How do I solve this further?

EDIT:Sorry, I don't know how to write the latex code to represent the "Bar" above "z"

4. May 14, 2015

smart_worker

I guess, I'll have to use Wirtinger derivatives.

5. May 14, 2015

Hawkeye18

By the definition $$\frac{\partial}{\partial z}=\frac12\left(\frac{\partial}{\partial x} -i \frac{\partial}{\partial y} \right), \qquad \frac{\partial}{\partial\overline z}=\frac12\left(\frac{\partial}{\partial x} +i \frac{\partial}{\partial y} \right).$$ So if you multiply these two differential operators and use the fact that $$\frac{\partial}{\partial x}\frac{\partial}{\partial y} = \frac{\partial}{\partial y}\frac{\partial}{\partial x}$$ (equality of mixed partial derivatives), you get exactly the Laplacian.

6. May 14, 2015

Hawkeye18

I meant $1/4$ of the Laplacian, i.e. $$\frac14\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right).$$