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Question: Suppose that $f$ is an entire function and that $|f(z) − z + 2z^2| ≤ \arctan(|z|)$ for all $z ∈ \Bbb C$. Compute $f(2)$.

I have been trying to think about how the "entire" property would help me, but I couldn't find any theorems nor definitions to apply. Should I just unwrap the function just like the real functions since this is involves absolute value and inequalities?

I am using the Stein and Shakarchi's Complex Analysis, Chapter 5 was for entire function.

Any help would be greatly appreciated! :)

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Hints:

  1. What is $f(0)$?
  2. $\arctan{|z|}$ is bounded. What does this tell you about the left-hand side?
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    I was thinking about the Louiville's Theorem, but I couldn't find any examples using this theorem from the internet. I know that if arctan is bounded, then LHS is bounded => |f(z)| is bounded (plus since f is entire), f must be a constant. – 2017-02-25
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    oh wait, then maybe I can induce that |f(0)-0+2*0|< arctan(0) = 0 iff |f(0)|=0 iff f(0)=0 and from above what I said, f(2)=0 since f is cosntant? – 2017-02-25
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    $z-2z^2$ is not bounded. – 2017-02-25
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    Thanks for your reply Chappers. Right, z-2z^2 is not bounded but I suppose when z=0, then |f(0)-z+2z^2| = |f(0)| < arctan|z| which is bounded so f(0) is bounded? (and f(0)=0). May I use this logic? – 2017-02-25
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    Hang on. $\arctan{(|0|)}=0$, so $|f(0)-0|=0$, and so $f(0)=0$, and that's the first part. Now, $ 0 \leq \arctan{|z|} < \pi/2 $, for *any* $z$, so $|f(z)-z+2z^2| \leq \pi/2$ for any $z$. Thus $f(z)-z+2z^2$ is bounded, and entire by the given condition on $f$. – 2017-02-25