# Image of a complex function contained in a line. Prove that the function is constant.

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!

• 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. – irchans Oct 30 '18 at 0:55
• Note that the open mapping theorem for complex analysis gives this to you for free. – Theo Bendit Oct 30 '18 at 0:57

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 $$g'\equiv 0.$$

• I thought I could do this problem, but it turns out that I cannot. Could you tell me how to find A and B? – BalancedTryteOperators Oct 30 '18 at 3:27
• 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 $0.$ – DanielWainfleet Oct 30 '18 at 3:48

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.

• 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 – DanielWainfleet Oct 31 '18 at 4:07

I think you might want to consider $$\frac{1}{f(z)}$$.

• 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? – user296602 Oct 30 '18 at 0:57