# Existence of an entire function

my Complex Analysis final is in a couple of days and i'm struggling with this question -

"Is there an entire function $$f$$ that satisfies $$|f(z)| = |z| + 1$$ for every $$z$$ in the complex plane for which $$|z| \ge 2017$$?"

I deducted if such function exists, it must be a polynomial. I tried tinkering around with $$1/f$$ and $$f(1/z)$$ but didn't really find anything useful, also tried using Rouché's theorem but I can't seem to prove/disprove the existence of a function.

I'd love a hint! :)

• Is this the actual question, or are there further conditions? There are trivial solutions such as $f$ being the zero function. Jan 26, 2019 at 6:54
• My mistake, I wrote it wrong. So sorry, I edited it. It should have been equals instead of smaller than. Jan 26, 2019 at 7:00
• Can you prove that $f$ must be a degree one polynomial? Jan 26, 2019 at 7:01
• That's what I tried to do, but for that I have to prove that f has only one root inside the circle of radius 2017, right? Also, I am not sure, how does that prove such $f$ doesn't exist? Jan 26, 2019 at 7:05
• Once you know that $f$ is a polynomial, its degree can extracted from the knowledge on its growth speed as $|z|\to\infty$. Jan 26, 2019 at 7:23

You already know that $$f$$ is a polynomial function. Let $$n$$ be its degree. Then$$\lim_{\lvert z\rvert\to\infty}\frac{\lvert f(z)\rvert}{\lvert z\rvert^n}=1.$$But then it follows from your hypothesis that $$n=1$$ and that $$f(z)=az+b$$ with $$\lvert a\rvert=1$$. However there are no such numbers $$a$$ and $$b$$ so that $$\lvert az+b\rvert=\lvert z\rvert+1$$ when $$\lvert z\rvert$$ is large enough.