Take the 2-minute tour ×

I ask for the proof of the following:

If $p$ is a polynomial with degree $n\ge 1$ and zeros in $A\subseteq \mathbb C$ whose order (multiplicity) is given by $n:A\to \mathbb N^*$ then $A$ is finite and non empty and there exists a $c\in \mathbb C^*$ so that \begin{equation}p(z)=c\prod_{a\in A}(z-a)^{n_a}\end{equation}

using only tools from Complex Analysis, like Cauchy's Theorem and Formula, Liouvile's Theorem...

Here is what I have come up with so far: By the Fundamental Theorem of Algebra, $A$ is finite and \begin{equation}n=\sum_{a\in A}n_a\end{equation} We can easily show that the product $\prod_{a\in A}(z-a)^{n_a}$ has the same zeros of the same order as $p$ and so $$p(z)\left(\prod_{a\in A}(z-a)^{n_a}\right)^{-1}=p(z)\prod_{a\in A}(z-a)^{-n_a}$$ is entire. It remains to show that either it is bounded (and then use Liouville) or that it is constant (for example that it has derivative $0$). How do we prove either of those?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

From $\deg p = n$ we obtain an estimate

$$\lvert p(z)\rvert \leqslant C\cdot \lvert z\rvert^n$$

for $\lvert z\rvert \geqslant R$. Similarly, we obtain an estimate

$$\left\lvert \prod_{a\in A} (z-a)^{n_a}\right\rvert \geqslant c\cdot \lvert z\rvert^n$$

with a $c > 0$ for $\lvert z\rvert \geqslant S$ (we can choose for example $c = 2^{-n}$ and $S = 2 \cdot \max \{ \lvert a\rvert : a \in A\}$).

Together that yields

$$\left\lvert \frac{p(z)}{\prod_{a\in A} (z-a)^{n_a}}\right\rvert \leqslant \frac{C}{c}$$

for $\lvert z\rvert \geqslant T := \max \{ R,S\}$. Since the quotient is also bounded on the compact set $\{ z : \lvert z\rvert \leqslant T\}$, it is a bounded entire function.

share|improve this answer
Why not take $R=1$? –  Optional Dec 5 '13 at 16:22
We could take $R = 1$, or any $R > 0$, if we wish. That would only lead to different constants $C$. I wrote $R$ out of habit, since usually I want both an upper and a lower bound, and for the lower bound, you can't generally pick $R = 1$. –  Daniel Fischer Dec 5 '13 at 16:27
Nevermind. Thanks for the answer. –  Optional Dec 5 '13 at 16:32

Your Answer


