# How does someone topologically prove that a function is analytic on some curve

Let us say for the sake of an example, that I want to evaluate $\int_\gamma \! \frac{1}{z-5} \,$ where $|\gamma| = 2$. I know that this integral gives me zero which means that the function $f(z) = \frac{1}{z-5}$ is analytic on the circle $\gamma$ but how do you prove this fact? I am boggled down by the geometry a little bit, $z=5$ lies outside of the circle $\gamma$.

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Are you referring to Morera's theorem? –  Alex Youcis Mar 9 '12 at 3:48

Do you mean that the function $f(z)$ is analytic in the circle $|z|=2$? We normally say that a function is holomorphic in a domain, rather than over a curve or path.

To prove that a function is holomorphic in a domain (here we are looking at $\mathbb{C}-{5}$), sometimes it helps to look at first principles and to calculate the derivative $$f'(z) = \lim_{z\rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0}$$

Domains where the derivative exists are areas where the function is holomorphic (and therefore analytic) over them.

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Thanks again Calvin! –  CalculusStudent Mar 9 '12 at 4:49

In general, polynomials, rational functions away from their zeros, compositions of elementary functions and analytic-looking functions are analytic. For a proof, certainly taking the derivative may be the best way to go; you can also claim that since the second partials of the function $f(z)$ are continuous away from the point $z = 5$, and since the Cauchy-Riemann equations are satisfied, it must be analytic (i.e. holomorphic). Note that neither of these conditions alone implies analyticity, though they can be relaxed slightly (I think you only really need continuity of the function, and I think there's a proof in the case where $f$ is only bounded).

What I wanted to say was that $\oint f dz = 0$ does not imply analyticity! It must be true on every closed (Jordan) curve, or at least on triangles or rectangles (as Alex said, Morera's theorem).

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