Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.