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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! :)

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  • $\begingroup$ Is this the actual question, or are there further conditions? There are trivial solutions such as $f$ being the zero function. $\endgroup$
    – Angina Seng
    Jan 26, 2019 at 6:54
  • $\begingroup$ My mistake, I wrote it wrong. So sorry, I edited it. It should have been equals instead of smaller than. $\endgroup$
    – Guy Schwartzberg
    Jan 26, 2019 at 7:00
  • $\begingroup$ Can you prove that $f$ must be a degree one polynomial? $\endgroup$
    – Angina Seng
    Jan 26, 2019 at 7:01
  • $\begingroup$ 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? $\endgroup$
    – Guy Schwartzberg
    Jan 26, 2019 at 7:05
  • $\begingroup$ Once you know that $f$ is a polynomial, its degree can extracted from the knowledge on its growth speed as $|z|\to\infty$. $\endgroup$
    – Sangchul Lee
    Jan 26, 2019 at 7:23

1 Answer 1

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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.

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