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?

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