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?
What might I use to show that an entire function with positive real parts is constant?
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4Jonas Meyer's alternative answer, and Soarer's answer, both explain how to reduce your problem to an application of Liouville's theorem. This is typically how such questions are expected to be solved in a first course in complex analysis (which I'm guessing is where your question comes from). – 2011-03-16
5 Answers
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.
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2More elementary would be to use an explicit Möbius transformation sending the right half plane to the unit disk. – 2011-03-15
The other three answers are overkill to me.. Simply consider $e^{-f}$ if $f$ is your function. Is it bounded?
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1my bad, I paid attention to your Casorati-Weierstrass proof :P – 2011-03-16
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 $g$?
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!
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.