What might I use to show that an entire function with positive real parts is constant? So the question asks me to prove that an entire function with positive real parts is constant, and I was thinking that this might somehow be related to showing an entire bounded function is constant (Liouville's theorem), but are there any other theorems that might help me prove this fact? 
 A: Well, can't we just say that, since $-f(z)$ is entire, $e^{-f(z)}$ is also entire, and if we write
$f(z) = u(z) + iv(z), \tag{1}$
where $u(z)$, $v(z)$ are the (harmonic) real and imaginary parts of $f(z)$ (so that $u(z) = Re \; f(z)$), then
$\vert e^{-f(z)} \vert = \vert e^{-u(z) - iv(z)} \vert = \vert e^{-u(z)} \vert \vert e^{-iv(z)} \vert = e^{-u(z)}, \tag{2}$
since
$e^{-u(z)} > 0 \tag{3}$
and
$\vert e^{-iv(z)} \vert = \vert \cos (-v(z)) + i\sin (-v(z)) \vert = 1; \tag{4}$
but $-u(z) < 0$ by hypothesis; thus $e^{-u(z)} < 1$, whence $e^{-f(z)}$ is a bounded entire function, hence constant; hence $f(z)$ must itself be a constant. QED.
Can't we just say that?  I think we can!
A: Both of the two other answers are already excellent, but if you really want to bring on the big guns, use Picard's Little Theorem - noting that the half plane consists of more than two points.
A: The other three answers are overkill to me.. Simply consider $e^{-f}$ if $f$ is your function. Is it bounded?
A: It isn't a nonconstant polynomial, by the fundamental theorem of algebra.  It doesn't have an essential singularity at infinity, by the Casorati-Weierstrass theorem.  What other possibilities are there?
Alternatively, if you add $1$, you get a function satisfying $|g(z)|\geq 1$ for all $z$.  What can you say about the reciprocal of $g$?
A: Have you learned the Riemann Mapping theorem?  If so, what can you do with the image of this entire function?  Remember, the composition of analytic functions is analytic.  Liousville's theorem should then finish the problem off.
