# Magnitude of differentiable complex function $f(z)$

If $f:\mathbb{C} \rightarrow \mathbb{C}$ is a differentiable function on $|f(z)|\ge 1$ everywhere, what can one conclude about $f$?

Does $|f(z)|\ge 1$ mean the magnitude of $f(z)$ is greater than 1? If so, what can I conclude about $f$?

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One can conclude that it is constant. Since $|f(z)|\geq 1$ for all $z$, then $1/f(z)$ is defined for all $z$, and is also analytic. Moreover, $|1/f(z)|\leq 1$. The conclusion then follows by Liouville's Theorem.
Thanks @CameronBuie Liouville's theorem states that every bounded entire function must be constant. Do we (did you) get that $f$ is bounded because $|f(z)|\geq 1$? As in, $f(z)$ is greater than $1$ for all $z$? – Derrick May 25 '12 at 17:18
@Derrick: For $g$ to be bounded means there is some $M>0$ such that for all $z\in\mathbb{C}$, $|g(z)|\leq M$. In this case, the function is $g=1/f$. Since $|f(z)|\geq 1$, then $|g(z)|=1/|f(z)|\leq 1$. Thus, $g$ is bounded and entire, so constant, and since $g$ is constant, so is $f=1/g$ (once we conclude that $g$ isn't identically zero, which follows from the fact that $f$ is defined everywhere). Does that clear it up? P.S.: Be careful about saying things like "$f(z)\geq 1$" when $f$ is a complex-valued function. The complex numbers are not an ordered field. – Cameron Buie May 25 '12 at 17:45
There is a result that follows immediately from liouville theorem(so essentially this solution does not differ from the above,but maybe makes just more instantaneous the understanding that the function is constant):the image of an entire noncostant function is a dense subset of $\mathbb{C}$,thus your function is constant.
Thanks @Piton , Liouville's theorem states that every bounded entire function must be constant. Do we (did you) get that $f$ is bounded because $|f(z)|\geq 1$? As in, $f(z)$ is greater than $1$ for all $z$? – Derrick May 25 '12 at 17:18