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.
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,
 A: 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…)
