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I'm being asked to find an alternate proof for the one commonly given for Liouville's Theorem in complex analysis by evaluating the following given an entire function $f$, and two distinct, arbitrary complex numbers $a$ and $b$: $\lim_{R\to\infty}\oint_{|z|=R} {f(z)\over(z-a)(z-b)} dz $

What I've done so far is I've tried to apply the cauchy integral formula, since there are two singularities in the integrand, which will fall in the contour for $R$ approaches infinity. So I got:

$2{\pi}i\biggl({f(a)\over a-b}+{f(b)\over b-a}\biggr)$

Which equals $2{\pi}i\biggl({f(a)-f(b)\over a-b}\biggr)$

and I got stuck here I don't quite see how I can get from this, plus $f(z)$ being bounded and analytic, that can tell me that $f(z)$ is a constant function. Ugh, the more well known proof is so much simpler -.- Any suggestions/hints? Am I at least on the right track?

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    One of the hypotheses of Liouville's theorem is that f(z) is bounded, as well as entire. Sorry I didn't state that in the question.2012-04-02

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You can use the $ML$ inequality (with boundedness of $f$) to show $\displaystyle \lim_{R\rightarrow \infty} \oint_{|z|=R} \frac{f(z)}{(z-a)(z-b)}dz = 0$.

Combining this with your formula using the Cauchy integral formula, you get $ 0 = 2\pi i\bigg(\frac{f(b)-f(a)}{b-a}\bigg)$ from which you immediately conclude $f(b) = f(a)$. Since $a$ and $b$ are arbitrary, this means $f$ is constant.

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    @JasonDeVito, its modulus must be $\leq 0$, but a modulus cannot be negative, so it must be $0$. I see...2016-04-05
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\lim_{R\to\infty}\oint_{|z|=R} {f(z)\over(z-a)(z-b)} \; dz=2{\pi}i\biggl({f(a)-f(b)\over a-b}\biggr) \to 2\pi if'(b)\text{ as }a\to b.

If one could somehow use boundedness of $f$ to show that $ \lim_{R\to\infty}\oint_{|z|=R} {f(z)\over(z-a)(z-b)} \;dz \to 0\text{ as }a\to b, $ then one would have shown that f'(b)=0. Since $b$ was arbitrary, one would have f'=0 everywhere.

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    @MichaelHardy, no. Actually, $b$ is not completely arbitrary, $a \neq b$, so you have to show that $f(a)=f(b)$. I would have liked to have seen more steps in the solution above where they showed that, though.2016-04-05