Complex Analysis: Liouville's theorem Proof

Question

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?

Answer 1

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.

Answer 2

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

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.