Stuck on a complex analysis proof:

Question

I am trying to prove that $f$ entire, one-to-one, is a linear polynomial.

I classified $f$'s singularities at infinity -- by Liouville's Theorem and the one-to-one assumption, $f$ can and must have a pole of order $k$ at infinity.

Then $f(1/z)$ has a pole of order $k$ at $z = 0$.

Consider $g:= z^k \cdot f(1/z)$. Then $g$ is entire and bounded, hence constant.

So I have $C = z^k \cdot f(1/z)$, which is

$C/z^k = f(1/z)$.

I'm not sure how to finish and show that $f(z) = C\cdot z^k$, i.e., $f$ must be a polynomial, which then it follows that $k$ must be $1$ for the one-to-one assumption.

I tried inverting the arguments, but that seems faulty.

Thanks,

Maybe this could be useful: http://math.stackexchange.com/questions/29758/entire-1-1-function — Olivier Oloa, Jun 02 '15 at 06:59

score 0 · Accepted Answer · answered Jun 02 '15 at 06:58

0

If you already have $f\left(\frac1z\right) = \frac C{z^k}$, you can now use this to prove that:

$$f(z) = f\left(\frac{1}{\frac1z}\right) = \frac C {\frac1z^k} = Cz^k$$

answered Jun 02 '15 at 06:58

5xum

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Ah, very nice @5xum. I was making the mistake of thinking that 1/f(1/z) = f(z). Thanks so much. – user245178 Jun 02 '15 at 07:20

Stuck on a complex analysis proof:

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