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?
Best Answer
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.