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**[SOLVED] Finding an Antiderivative in an Annulus**

**Homework Statement**

Let A be the annulus, A = {z : 0 <= r < |z| < R}. Suppose that g is an analytic function on A (and that g' is continuous) with the property that

[tex]\int_{|z| = s} g(z) \, dz = 0[/tex]

where r < s < R. Show that there is an analytic function G on A with G on A with G' = g throught A.

**The attempt at a solution**

My plan of attack is as follows:

1. Show that for any piecewise smooth simple closed curve C in A, [itex]\int_C g(z) \, dz = 0[/itex].

2. Conclude that g(z) is path-independent on A and so there is a function G on A such that if g(z = x + iy) = u(x, y) + iv(x, y), then G

_{x}= u and G

_{y}= y or more succinctly that G' = g.

I'm having trouble with 1. If the interior of C is entirely contained in A, then by Cauchy's Theorem, the integral is 0. But what if C goes around the annulus? I was thinking of using Green's Theorem but I can't because A is open and so I can't integrate around its boundaries. But what if I apply Green's Theorem to the portion of A contained within |z| = s and C? Hmm...

If |z| = s is contained inside C, then

[tex]\int_C g(z) \, dz - \int_{|z|=s} g(z) \, dz = i\iint_B (g_x + ig_y) \, dxdy[/tex]

The second term on the LHS is 0. The term on the RHS is 0 because g is analytic and by Cauchy-Riemann equations. That means the first term on the LHS is 0. Aha! Is correct?