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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?

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    $\begingroup$ I'd use that one! There are definitely others, but they would all be considered "overkill" and probably use Liouville's theorem in their proof. $\endgroup$ – Matt Mar 15 '11 at 21:16
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    $\begingroup$ Jonas 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). $\endgroup$ – Matt E Mar 16 '11 at 3:30
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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 $g$?

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  • $\begingroup$ How do we continue along your first line of reasoning? I guess if we show that $f$ has a removable singularity at infinity we are done, but how do we exclude a pole? $\endgroup$ – Ovi Apr 13 at 3:52
  • $\begingroup$ @Ovi: A pole at infinity implies it is a polynomial, and being a nonconstant polynomial implies it is surjective, by the fundamental theorem of algebra. $\endgroup$ – Jonas Meyer Apr 16 at 1:11
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The other three answers are overkill to me.. Simply consider $e^{-f}$ if $f$ is your function. Is it bounded?

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    $\begingroup$ I don't think that $1/(1+f)$ is overkill. $\endgroup$ – Jonas Meyer Mar 16 '11 at 1:37
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    $\begingroup$ my bad, I paid attention to your Casorati-Weierstrass proof :P $\endgroup$ – Soarer Mar 16 '11 at 4:13
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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!

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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.

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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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    $\begingroup$ More elementary would be to use an explicit Möbius transformation sending the right half plane to the unit disk. $\endgroup$ – Dan Petersen Mar 15 '11 at 21:39

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