# Complex analysis

1. May 24, 2017

### Silviu

1. The problem statement, all variables and given/known data
Suppose f is entire and there exist constants a and b such that $|f(z)| \le a|z|+b$ for all $z \in C$. Prove that f is a polynomial of degree at most 1.

2. Relevant equations

3. The attempt at a solution
We have that for any $z \neq 0$, $\frac{|f(z)|}{a|z|} \le b$. So if we take the limit as $z \to \infty$ it is obvious that if f is polynomial, it can't have a degree greater than 1. However I am not sure why it must be polynomial.

2. May 24, 2017

### stevendaryl

Staff Emeritus
I'm not sure what you are supposed to be able to use in your proof. One fact about entire functions is the Cauchy estimate formula, which says:
• Take a circle in the complex plane of radius $R$ centered on $z=0$.
• Let $M_R$ be the largest value of $|f(z)|$ on the circle.
• Then $|f^{(n)}(0)| \leq \frac{n! M_R}{R^n}$ (where $f^{(n)}$ means the $n^{th}$ derivative).
See if you can use this to prove $f^{(n)} = 0$ for $n > 1$.

3. May 24, 2017

### Staff: Mentor

This doesn't follow from $|f(z) | \le a|z| + b$