Find all holomorphic functions $ f:\mathbb{C}\to\mathbb{C}$ satisfying $|f(z)-3|\geq 1$ for all $z\in\mathbb C $

Please don't post a solution, I just want a little hint.

Obviously certain constant functions satisfy the condition. So for now let's assume f is not constant. We have

$1\leq |f-3|=\sqrt{(f-3)(\bar{f}-3)}$ so $1\leq(f-3)(\bar{f}-3)$

Since f is holomorphic it follows

$0\leq f'\cdot(\bar{f}-3)+(f-3)\cdot\bar{f}\\ \Leftrightarrow f'\cdot\bar{f}+f\cdot\bar{f}'\geq 3(f+\bar{f})'\\ \Leftrightarrow (f\bar{f})'\geq6\cdot\Re(f)'\\ \Leftrightarrow |f|^2\geq6\cdot\Re(f)\\ \Leftrightarrow \Im^2(f)+\Re^2(f)\geq6\cdot\Re(f)$

Since this doesn't look even close to solving the problem, I assume that there must be some theorem which says $f$ must be constant.

  • 6
    $\begingroup$ Hint: what can you say about $g(z) := \frac{1}{f(z)-3}$? $\endgroup$ – Daniel Schepler Jun 19 '17 at 21:32
  • $\begingroup$ Well, I would say this function has a singularity at $f^{-1}(3)$. But in our lecture we have not talked about singularities yet. We have just established Cauchy's integral formula. $\endgroup$ – user424862 Jun 19 '17 at 21:40
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    $\begingroup$ If $|f(z)-3| \ge 1$ then in particular $f(z)$ can never achieve the value 3, so $f^{-1}(\{ 3 \}) = \emptyset$. $\endgroup$ – Daniel Schepler Jun 19 '17 at 21:40
  • $\begingroup$ That's true. So I can conclude that for each z: $0\leq |g(z)|\leq 1$. $\endgroup$ – user424862 Jun 19 '17 at 21:46
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    $\begingroup$ Now use Liouville's theorem $\endgroup$ – Francesco Polizzi Jun 19 '17 at 21:47

Hint: $|f(z)-3|\geq 1 >0$ implies that $f(z)$ is never $3$. Now consider $g(z) := \dfrac{1}{f(z)-3}$.

  • $\begingroup$ Combined comments into an answer. $\endgroup$ – lhf Jun 19 '17 at 22:27

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