# Prove analyticity by Morera's theorem

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 Im not sure how to do that. Any help please?

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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: math.stackexchange.com/questions/159659/… –  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

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 $$\lim\limits_{\text{something}\to\text{something}} 0=\text{what}?$$