# Estimating $\int_{|z|=R} \frac{f(z)}{(z-a)(z-b)} dz$

Suppose $f$ is an entire function, and take $a,b\in\mathbb{C}$. For $R>\max\{|a|,|b|\},$ estimate $$\int_{|z|=R} \frac{f(z)}{(z-a)(z-b)} dz.$$
Assuming $f$ is bounded, let $R\rightarrow\infty$ and show $f$ is constant.

I'm not sure what it means by "estimate", are we supposed to take an upper and lower bound?

Also, after reading the second part, I suspect that it's actually a derivation of Liouville's Theorem, as the only condition to apply the theorem is that $f$ is an entire bounded function...

• Are you assume $f$ is bounded and entire or boundedness is a result of integration? Commented Sep 28, 2017 at 6:47
• I think $f$ is entire and bounded in the second part. The question was given as is. Commented Sep 28, 2017 at 6:50
• If I'm not mistaken, due to residue theorem this should be equal to $2\pi i (f(a) + f(b))$ no? Commented Sep 28, 2017 at 6:59
• $2\pi i\dfrac{f(b)-f(a)}{b-a}$ I think! Commented Sep 28, 2017 at 7:15
• That's correct. Commented Sep 28, 2017 at 7:50

## 1 Answer

The point is that if $|f| \le B$, your integrand is bounded by $\dfrac{B}{(R-|a|)(R-|b|)}$ so the integral is bounded by $\dfrac{2\pi R B}{(R-|a|)(R-|b|)}$. What happens to this as $R \to \infty$? Compare to the result of the residue theorem.