Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $f$ be continuous on the complex plane and analytic on the complement of the coordinate axes. Show that $f$ is analytic everywhere. Hint: Morera's theorem. I think that I need to show that the integral is zero almost everywhere so by Moreras theorem its analytic, but I`m not sure how to do that. Any help please?

share|cite|improve this question
It's certainly 0 on the complement of the coordinate axes by analyticity. Now, Morera's theorem needs the integral over every triangle to be 0. The only case where there's any trouble is a triangle with border on the axes. Argue that this case is also fine by taking the limit of a triangle off the axis. – Euler....IS_ALIVE Sep 23 '12 at 19:53
See the second half of the answer here:… – user31373 Sep 23 '12 at 20:29
What about the absolute value function, $|z|$? Surely it's continuous everywhere and analytic on the complex plane minus the origin ($ \mathbb C \setminus\{0\}$). Why does the same strategy with Morera's theorem not imply that $|z|$ is entire? – user58875 Jan 19 '13 at 12:10
@Nick: The absolute value function is not analytic anywhere. Check the Cauchy-Riemann equations. It maps the entire complex plane to the real line, so it cannot be conformal. There are many reasons. – robjohn Jan 19 '13 at 13:38
up vote 4 down vote accepted

It's enough to show that the integral of $f$ along every circle is $0$. If the circle does not cross or touch one of the axes, you've got it. If crosses an axis, approximate it by two simple closed curves: one of the follows an arc of the circle until it's very close to the coordinate axis, then moves along a line close to the axis until it reaches the arc, then moves along that arc. The other does the same on the other side of the axis. The integrals along those two lines approximately cancel each other, because you assumed continuity, and the approximation can be made as close as you want by making the lines close enough to the axis, again because you assumed continuity. So in the limit, they cancel each other and the whole thing approaches the integral along the circle. So $$\lim\limits_{\text{something}\to\text{something}} 0=\text{what}?$$

If it merely touches an axis without crossing, the problem is simpler. If it crosses both axes, it's more complicated in details, but conceptually pretty much the same.

share|cite|improve this answer
yes, thank you. I understand the idea now and I`ll try to write the details and see how that works. – Danny Sep 25 '12 at 4:00
Just a question: is that will make any difference if I choose circle, rectangle, or anything else? How many cases do I have here? Is it different when I take the curve intersect both axes or I deal with it as you explained above? I do think it`s similar but I just need to break it down to the one axes case, right? – Danny Sep 25 '12 at 4:34
It shouldn't matter whether it's a circle or a rectangle. – Michael Hardy Sep 25 '12 at 5:05

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.