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Let $f$ be a holomorphic function on a region $G \subseteq \mathbb{C}$, and suppose that the image $f(G)$ is contained in a line in $\mathbb{C}$. Prove that $f$ is constant.

There are various theorems in the field of complex analysis that prove a function to be constant. Liouville's Theorem proves a function is constant if it is bounded (but not necessarily dominated). The Cauchy-Riemann equations can be used to prove a function is constant if you can prove the derivatives are zero everywhere.

Which theorem should be used? I'm asking for a hint, not an answer!

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    $\begingroup$ Hint: You probably should not be looking for a powerful theorem to apply. Think about if $f(z)\in L$ where $L$ is the line, then what can you infer just from the definition of holomorphic. $\endgroup$ – irchans Oct 30 '18 at 0:55
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    $\begingroup$ Note that the open mapping theorem for complex analysis gives this to you for free. $\endgroup$ – Theo Bendit Oct 30 '18 at 0:57
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There exist constants $A, B$ such that the image of the holomorphic function $g(z)=Af(z) +B$ is a subset of $\Bbb R.$

Use the Cauchy-Riemann equations for $g'$ to show that $g'\equiv 0.$

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  • $\begingroup$ I thought I could do this problem, but it turns out that I cannot. Could you tell me how to find A and B? $\endgroup$ – BalancedTryteOperators Oct 30 '18 at 3:27
  • $\begingroup$ A line in $\Bbb C$ is $\{ D+rE: r\in \Bbb R\}$ for some constants $D,E\in \Bbb C$ with $E\ne 0$.... So let $A=1/E $ and $B=-D/E.$... Now since $Im(g(z))=0$ for all $z\in G,$ we have $g'(z)=\partial Re(g(z))/\partial Re(z)+i\cdot \partial Im (g(z))/\partial Re(z)=$ $=\partial Re(g(z))/\partial Re(z). $ ................................ And by the Cauchy-Riemann equations, $\partial Re(g(z))/\partial Re(z)=$ $Re (g'(z))=$ $=\partial Im(g(z))/\partial Im(z),$ which is $0$ because $Im(g(z))$ is always $0.$ $\endgroup$ – DanielWainfleet Oct 30 '18 at 3:48
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Either approach can be used.

  • With the Cauchy-Riemann equations (and composing with an affine map), you can reduce to showing that a real-valued entire function is constant. This isn't hard, but it's lengthy.

  • With Liouville's theorem, you can get a far better result (called the Casorati-Weierstrass theorem): if an entire function misses an open set, then it is constant. The proof is to consider $(a - f(z))^{-1}$ for some $a$ which is a positive distance from the image of $f$.

  • Liouville's theorem can be applied in a different way: a Mobius transformation takes the line to the boundary of the unit disk, and so the composition of $f$ and the transformation is bounded.

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  • $\begingroup$ Also the open mapping theorem , mentioned in a comment to the Q. My A, and my comment to it, is just the details of the 1st method you give. I don' think it's lengthy.............+1 $\endgroup$ – DanielWainfleet Oct 31 '18 at 4:07
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I think you might want to consider $\frac{1}{f(z)}$.

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    $\begingroup$ Although this is going in the right direction, it doesn't work. What if the image of $f$ is $\mathbb{R}$, in which case you have a non-differentiable function? $\endgroup$ – user296602 Oct 30 '18 at 0:57

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