# Analytic function on the whole plane, positive imaginary part, what can it be?

Part (a): The function f is analytic in the whole plane with positive imaginary part. What can it be?

Part (b): What if all you know is that the imaginary part of f tends to 0 at infinity?

what we know:

For part(a): Write f(z) = u(x,y) + iv(x,y)

u(x,y) and v(x,y) are both harmonic; in particular, they are harmonic conjugates to each other. So, Uxx+Uyy=0, and similarly for v.

The Cauchy-Riemann equations hold: $$Ux = Vy$$ $$Uy=-Vx$$

Since f has positive imaginary part, then v(x,y) $>/=$ 0, and its partial derivatives, Vx and Vy are both non-negative.

Some natural (or maybe not so natural?) guesses would be that f is a constant, or a polynomial, or an exponential function, since we know that f is entire. But, at the moment, I can't seem to extract any more information from the question itself.

• Positive imaginary part - that means the image is contained in the upper half-plane. But the upper half-plane is conformally equivalent to an open disk. So after composing with a suitable mapping, $f$ takes its values in a disk. What does that tell you?
– Mike
Commented Nov 18, 2014 at 22:35
• Or, if you seem to know only the very definition of complex analytic function (i.e. real and imaginary part are harmonic conjugates), what do you know about non-negative harmonic functions in $\mathbb{R}^n$? Commented Nov 18, 2014 at 22:38
• @Mike, applying your hint and making my best guess, I'd say that, consider the standard mapping phi: upper half plane -> unit disk, compose it with f, then phi(f(z)) takes values in the open disk, which is bounded. I'd now like to argue that since f is entire and bounded, and by Liouville's theorem, we conclude that f is a constant. But, that may not be correct, since, it's really phi(f(z)) that is bounded - not f(z), which lives in the upper half space. Also, f is entire, but I don't think phi(f(z)) is entire. Am I sort of on the right track? Commented Nov 18, 2014 at 23:09
• "Lebron", what is the domain of $\phi(f(z))$?
– Mike
Commented Nov 18, 2014 at 23:16
• The upper half plane... Commented Nov 18, 2014 at 23:24

Part (a) can be done as Mike suggested in the comments (or using the Liouville theorem for harmonic functions), or considering the function $e^{-if}$ which is bounded by $1$ on the whole plane, hence constant.
For part (b), if $\mathsf{Im}(f(z))$ tends to $0$ as $|z|\to\infty$, then there is $R>0$ such that, for $|z|\geq R$, $|\mathsf{Im}(f(z))|\leq 1$.
Take $K=\{z\in\mathbb{C}\ :\ |z|\leq 2R\}$. This is a compact set, hence $|\mathsf{Im}(f(z))|$ achieves a maximum on $K$, say $M>0$, but outside $K$, $|\mathsf{Im}(f(z))|\leq 1$, so $|\mathsf{Im}(f(z))|\leq\max\{M,1\}$ on the whole plane, hence it is constant.
Equivalently, set $g=e^{-if}$, then $|g|\to1$ when $|z|\to\infty$, meaning that $g$ is bounded around $\infty$. By Riemann's extension theorem, the function $h(z)=g(1/z)$, bounded around $0$, can be extended holomorphically to $z=0$; therefore the function $g$ extends holomorphically (or at least continuously if you have problems dealing with Riemann surfaces) to a function $\tilde{g}:\mathbb{C}\cup\{\infty\}\to\mathbb{C}$, i.e. $\tilde{g}:\mathbb{S}^2\to\mathbb{C}$. But as $\mathbb{S}^2$ is compact, $\tilde{g}$ is bounded and so is $g$, which is then constant and so is $f$.
(or if you know enough about holomorphic functions, on compact Riemann surfaces like $\mathbb{CP}^1=\mathbb{C}\cup\{\infty\}$ there are no non-constant holomorphic functions…)