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