# A complex polynomial with partial derivatives equal to zero is constant.

There is an exercise in Function Theory of One Complex Variable by Greene & Krantz that is very similar to a Proposition in the book, but I am having trouble getting to the conclusion.

Let $f : \mathbb{C} \to \mathbb{C}$ be a polynomial. Suppose further that $$\frac{\partial f}{\partial z} = 0$$
and $$\frac{\partial f}{\partial \bar z} = 0$$
for all $z \in \mathbb{C}$
Prove that $f \equiv$ constant.

Now, part of the proposition (1.3.2) proves that if $p$ is a polynomial, $$p(z, \bar z) = \sum a_{lm}z^l \bar z^m$$ with $\frac{\partial f}{\partial \bar z} = 0$, then $\frac{\partial ^{l + m}}{\partial z^l \partial \bar z^m}p$ evaluated at zero is $l!m!a_{lm}$.

Is this the constant that $f$ equals?
It doesn’t seem like I’ve used the hypothesis, or reached the conclusion!

Maybe I should use the definition of the partial… $$\frac{\partial f}{\partial z} := 1/2 (\frac{\partial }{\partial z} – i \cdot \frac{\partial }{\partial z})f$$

Any hints would be greatly appreciated.

#### Solutions Collecting From Web of "A complex polynomial with partial derivatives equal to zero is constant."

If $l\geq 1$ or $m\geq 1$, since we applied at least one time $\frac{\partial f}{\partial z}$ or $\frac{\partial f}{\partial \bar z}$, $\frac{\partial^{l+m}}{\partial z^m\partial \bar z^l}p$ must be equal to $0$. So $l!m!a_{lm}=0$ and since $l!m!\neq 0$ we have $a_{lm}=0$ if $l\geq 1$ or $m \geq 1$. It shows that the only potentially non-vanishing term is $a_{0,0}$.