simplification of $z_2\frac{\partial}{\partial z_1}\circ z_1\frac{\partial}{\partial z_2}$

Question

Let $P(z_1,z_2)\in\mathbb{C}[z_1,z_2]$ with homogeneous degree $n$. How do we simplify the expression $z_2\frac{\partial}{\partial z_1}\circ z_1\frac{\partial}{\partial z_2}-z_1\frac{\partial}{\partial z_2}\circ z_2\frac{\partial}{\partial z_1}$ to $z_2\frac{\partial}{\partial z_2}-z_1\frac{\partial}{\partial z_1}$ ?. I want to see the all steps. Thanks!

Answer 1

Notice that using Leibniz's formula,

$$z_2 \frac {\partial} {\partial z_1} \circ z_1 \frac {\partial} {\partial z_2} = z_2 \left( \frac {\partial} {\partial z_1} z_1 \right) \frac {\partial} {\partial z_2} + z_2 z_1 \frac {\partial} {\partial z_1} \circ \frac {\partial} {\partial z_2} = z_2 \frac {\partial} {\partial z_2} + z_2 z_1 \frac {\partial^2} {\partial z_1 \partial z_2} .$$

Similarly (just exchange $z_1$ with $z_2$),

$$z_1 \frac {\partial} {\partial z_2} \circ z_2 \frac {\partial} {\partial z_1} = z_1 \frac {\partial} {\partial z_1} + z_1 z_2 \frac {\partial^2} {\partial z_2 \partial z_1} .$$

Now just subtract the second from the first in order to obtain the desired result.

Notice that this is true for complex-differentiable functions of two complex variables, not just for polynomials. The homogeneity condition plays no role.