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I want to prove that if $Ref(z)$>0 for every $z$ in the complex plane, then the function is constant

I know that if $Ref(z)$>0 and $f$ is entire then $Ref(z)$<0 and $-f$ is also entire the |$e^{-f(z)}$|= $e^{Re(-f)}$ so how can I conclude that $f$ is constant?

Thanks

Matt
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    You should put the requirement that f is entire in the title. If f need not be entire this is obviously not true. – fleablood Oct 30 '15 at 05:08

2 Answers2

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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!

Robert Lewis
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Image of a non-constant entire function $f$ is dense in the plane. For say, $(\exists \alpha \in \mathbb{C})(\exists r > 0) (\forall z \in \mathbb{C}) \, \vert f(z) - \alpha \vert \geq r$, then $\frac{1}{f(z) - \alpha}$ would be a bounded entire function, which by Liouville's theorem, must be constant, contrary to our assumption. Now, if the function takes values only to the right of the imaginary axis, naturally it would not intersect an open ball lying to the left of the imaginary axis, which is impossible since the image of $f$ is dense if it non-constant.

Gautam Aishwarya
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