Cauchy's Integral Formula implies Holomorphicity? Is the converse direction of the Cauchy Integral Formula true? Meaning,

if $f:\mathbb{C}\supseteq U\rightarrow\mathbb{C}$, and

$$\forall a \in U \space \space f(a) = \frac{1}{2\pi i}\int_{|z-a|=R}\frac{f(z)}{z-a}dz$$

does it mean that f is complex-differenatiable?
If you parametrize your integral, you see that your equation is equivalent to

$$ f(a) = \dfrac{1}{2\pi} \int_0^{2\pi} f(a + r e^{i\theta})\; d\theta$$

and this is true for harmonic functions, not just holomorphic ones.
Well...yes, because then you would have an expression for the nth derivative of f, making f infinitely differentiable/analytic in U and thus holomorphic.

https://en.wikipedia.org/wiki/Analytic_function