State and prove Cauchy integral theorem

Statement (Cauchy's integral theorem). If f(z) is analytic at all points inside and on a simple closed contour C, then

∮_C f(z) dz = 0.

Proof (using Green's theorem)

Assume, as in Cauchy's original proof, that f′(z) is continuous in the region R bounded by C. Let f(z) = u(x, y) + i v(x, y) and dz = dx + i dy. Then

∮ f(z) dz = ∮ (u + iv)(dx + i dy)
          = ∮ (u dx − v dy) + i ∮ (v dx + u dy)

Green's theorem: ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy.

Apply it to each integral.

∮ (u dx − v dy) = ∬_R ( −∂v/∂x − ∂u/∂y ) dx dy
∮ (v dx + u dy) = ∬_R (  ∂u/∂x − ∂v/∂y ) dx dy

Since f(z) is analytic, u and v satisfy the Cauchy–Riemann equations in R:

∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x.

So both integrands vanish identically: −v_x − u_y = −v_x + v_x = 0 and u_x − v_y = 0.

Hence ∮_C f(z) dz = 0 + i·0 = 0. ∎

Remark. Goursat later proved the theorem without assuming that f′ is continuous. The theorem also gives path independence: if f is analytic in a simply connected region, ∫ f(z) dz between two points does not depend on the path joining them.