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 Morera's theorem it's analytic, but I'm not sure how to do that. Any help please?

  • $\begingroup$ 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. $\endgroup$ Sep 23, 2012 at 19:53
  • $\begingroup$ See the second half of the answer here: math.stackexchange.com/questions/159659/… $\endgroup$
    – user31373
    Sep 23, 2012 at 20:29
  • $\begingroup$ 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? $\endgroup$
    – user58875
    Jan 19, 2013 at 12:10
  • $\begingroup$ @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. $\endgroup$
    – robjohn
    Jan 19, 2013 at 13:38

1 Answer 1


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.

  • $\begingroup$ yes, thank you. I understand the idea now and I`ll try to write the details and see how that works. $\endgroup$
    – Danny
    Sep 25, 2012 at 4:00
  • $\begingroup$ 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? $\endgroup$
    – Danny
    Sep 25, 2012 at 4:34
  • $\begingroup$ It shouldn't matter whether it's a circle or a rectangle. $\endgroup$ Sep 25, 2012 at 5:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .