Complex Analysis: Liouville's theorem Proof

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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Your choice of $a$ and $b$ was arbitrary --- if you can show that $f(a)-f(b) = 0$, you're done. So how can you argue that your expressions are all equal to zero? – Neal Apr 1 '12 at 21:59
I noticed that what you said would prove it but I'm really not sure how I would go about arguing it. Any theorems/lemmas I should go review? I've looked into using cauchy integral theorem, ML formula, and Cauchy Estimates. Whenever I try using Cauchy estimates, I can't help but just end up with the normal proof of Liouville (as seen on proofwiki, wikipedia, etc.). Are one of those what I ought to be using or am I completely overlooking something important? – calvin Apr 1 '12 at 22:26
How do you know that f(z) is of power one? Don't you just know that f(z) is entire? We don't know specifically what f(z) is. Couldn't it have a power greater than that of the denominator, so the denominator wouldn't dominate? – user28118 Apr 2 '12 at 3:36
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. – calvin Apr 2 '12 at 4:10

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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I've never heard it called "the ML inequality". What does that stand for - "Maximum of function times Length of contour," perhaps? – Gerry Myerson Apr 2 '12 at 0:37
@Gerry: Exactly. I don't remember where I read it. A quick google search shows it's not an uncommon thing to call it. What do you call it? – Jason DeVito Apr 2 '12 at 0:38
I don't quite understand this method. Wouldn't the L in this case be approaching infinity since the contour is |z|=R? How would you use ML? How would you use ML in this case? – calvin Apr 2 '12 at 2:03
ahh, I see. is it because f(z) is always finite, but the (z-a)(z-b) is gonna put an R^2 on the bottom no matter what, So the ML is gonna have a first power of R in the numerator, and an R^2 in the denominator, and so it's all controlled by the R^2, making the ML approach zero (all stated informally, of course. – calvin Apr 2 '12 at 2:57
@JasonDeVito, its modulus must be $\leq 0$, but a modulus cannot be negative, so it must be $0$. I see... – Jessy Cat Apr 5 at 2:04

$$\lim_{R\to\infty}\oint_{|z|=R} {f(z)\over(z-a)(z-b)} \; dz=2{\pi}i\biggl({f(a)-f(b)\over a-b}\biggr) \to 2\pi if'(b)\text{ as }a\to b.$$

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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To put it another way: calvin, you have evaluated the integral; now estimate the integral. – Gerry Myerson Apr 2 '12 at 0:33
@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. – Jessy Cat Apr 5 at 1:36