# If we have a non-constant entire function $f(z)$ show that $u(z)$ is not bounded. [duplicate]

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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 TaxelFeb 17 '18 at 9:53

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

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

whence

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

since

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

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.

• I didn't even think about proof by contradiction, thanks! – 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.