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$F(z)$ = $u(z)$ + $iv(z)$ where $F(z)$ is non-constant and entire. Show that $u(z)$ is not bounded.

My intuition is that Liouville's theorem should help or I should make $F(z)$ equal a non-constant function then go from there.

Any pointers would be appreciated, thanks!


marked as duplicate by Martin R, Lord Shark the Unknown, Claude Leibovici, TheSimpliFire, Parcly Taxel Feb 17 '18 at 9:53

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Consider the function

$G(z) = e^{F(z)}; \tag 1$

since $F(z)$ is entire, so is $G(z)$; furthermore, with

$F(z) = u(z) + iv(z), \tag 2$

we have

$G(z) = e^{u(z) + iv(z)} = e^{u(z)} e^{iv(z)}, \tag 3$


$\vert G(z) \vert = \vert e^{u(z)} e^{iv(z)} \vert = \vert e^{u(z)} \vert \; \vert e^{iv(z)} \vert = \vert e^{u(z)} \vert, \tag 4$


$\vert e^{iv(z)} \vert = \vert \cos v(z) + i \sin v(z) \vert = 1;\tag 5$

since $u(z)$ is bounded, (4) shows that $G(z)$ is also bounded; so $G(z)$ is a bounded entire function, hence is a constant by Liouiville's theorem. Then (1) shows that $F(z)$ must be constant well, in contradiction to the hypothesis that $F(z)$ is a non-constant entire function; hence $u(z)$ cannot be bounded.

  • 1
    $\begingroup$ I didn't even think about proof by contradiction, thanks! $\endgroup$ – Schonker Feb 16 '18 at 3:17

Assume that $u(z)$ is bounded by $M$. This means that if $m>M$, then we can find some $\epsilon>0$ so that $\epsilon< |u(z)-m|$. This gives that $\epsilon < |F(z)-m|$, which follows from the Pythagorean Theorem. Set $G(z)=\frac{1}{F(z)-m}$. We then have $|G(z)|<\frac{1}{\epsilon}$, which is bounded.


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