# Liouville Theorem Related Proof.

My professor gave this question to the class for practice, to not be turned in for grading. I tried doing what he said in the hint, but am just getting nowhere. Could you please give any pointers, tips or suggestions? I want to do what the tip says, to use the power series expansion and then apply Liouville's Theorem. Thank you.

• Or just write down Cauchy's integral formula for the kth derivative of f(z) at z = 0 as a contour integral over a circle with radius R. – Count Iblis Apr 4 '17 at 23:24
• Could you explain in more detail please? – kemb Apr 4 '17 at 23:27
• I want to use the power series expansion. – kemb Apr 4 '17 at 23:27

Let us write the Taylor series of $f$ as

$$f(z) = \sum_{n=0}^{\infty} a_n z^n.$$

Define $p(z) = \sum_{k=0}^{n-1} a_k z^k$ to be the polynomial which is the sum of the first $n$ terms in the Taylor series of $f$. Then we have

$$f(z) = p(z) + a_n z^n + \sum_{k=1}^{\infty} a_{n+k} z^{n+k}.$$

Consider the function $h(z) := \frac{f(z) - p(z)}{z^n}$. This function is well-defined on $\mathbb{C} \setminus \{ 0 \}$ and in fact extends to an entire function on $\mathbb{C}$ using the power series expansion

$$h(z) = a_n + \sum_{k=1}^{\infty} a_{n+k} z^{k}.$$

$$|h(z)| = \frac{|f(z) - p(z)|}{|z|^n} \leq M + \frac{|p(z)|}{|z|^n} \leq M + 1$$
for $z$ large enough. Hence, by Liouville's theorem we have $h(z) \equiv h(0) = a_n$ and so $f(z) = p(z) + a_n z^n$ for all $z \in \mathbb{C}$.
• @BOB: I've followed the hint. The first part consists of manipulating $f(z)$ using the Taylor series. We subtract from $f$ the first $n$ terms of the series (this is $p$) and divide by $z^n$. Then we show that the resulting function is in fact entire and so is constant and then deduce that $f$ is equal to the first $n + 1$ terms of the power series - i.e it is a polynomial of degree $\leq n$. The $n$ and the $M$ are the same as in the question, and "for $z$ large enough" means $|z| > \max(R, C)$ where $C$ is a constant for which $\frac{|p(z)}{|z|^n} \leq 1$ for all $|z| > C$. – levap Apr 4 '17 at 23:45