Proof of holomorphism for Cauchy Integral Theorem

Question

Show that any continuously differentiable function f in an open disc D satisfying $$\int_{\gamma} f(z) \, dz =0 $$ for any closed, smooth, simple curve γ ⊂ D, is holomorphic in D. (Hint: use the result for shrinking circles around any given point, $z_0$, in D.)

My approach:

I know that Cauchy Integral Theorem can be proved from Green's theorem. And from the Green theorem, I know that Cauchy Riemann Equation holds to have $\int_{\gamma} f(z) \, dz =0 $, but apparently I was wrong about using Cauchy Riemann equation to prove that it is holomorphic. Can anyone help me with this proof?

Answer 1

You could deduce from this hypothesis that the line integral of $f$ is path-independent and then, as with Morera's Theorem, deduce that $f$ has a holomorphic primitive, from which it follows that $f$ is holomorphic.

If you know about the $\partial/\partial\bar z$ operator, you can rewrite the Cauchy-Riemann equations as $\partial f/\partial\bar z = 0$. Applying Green's/Stokes's Theorem to $\gamma = \partial R$, you get $$\int_\gamma f(z)\,dz = \iint_R \frac{\partial f}{\partial\bar z}d\bar z\wedge dz = 2i\iint_R \frac{\partial f}{\partial\bar z} dA.$$ If $f$ failed to be holomorphic at some $z_0$, you'd have $\frac{\partial f}{\partial\bar z}(z_0)\ne 0$, and by continuity, you could choose $R$ to make this area integral nonzero, contradicting the hypothesis for some $\gamma$.