If $u(z)$ is non-constant and harmomic in the plane, then show that $u(z)$ comes arbitrarily close to every real value

complex-analysisharmonic functions

This question was asked in my complex analysis assignment and I am not able to solve it.

If $u(z)$ is non constant and harmonic in the plane, then show that $u(z)$ comes arbitrarily close to every real value.

Every harmonic function can be proved both real part and imaginary part of an aanalytic function. So, $u(z)$ itself would be analytic.
So function can't be bounded and can be defined for all $z\in \mathbb{C}$. Now there is a theorem that an entire function comes in $\epsilon$ neighbourhood of every complex number (I forgot the theorem's name). So, I am done.

Am i right?

Is the proof correct?

Best Answer

Your argument is correct and the theorem you want to use is Picard's Theorem. But you to complete it by observing that if $u+iv$ attains some value in the disk of radius $\epsilon$ around a real number $a$ then $u$ takes a value in $(a-\epsilon, a+\epsilon)$.

Actually $u$ attains very real value (by continuity). To see this consider $e^{u+iv}$ and $e^{-(u+iv)}$ Using Liouville's Theorem show that $u$ can neither be bounded above nor be bounded below.

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