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.
What more can I say about this function, for part(a)?
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.
Thanks in advance,
Best Answer
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…)