# Cauchy's Formula vs. Cauchy's Theorem confusion

I'm using Donald Sarason's book as a reference.

(One version of) Cauchy's theorem states that for a holomorphic function defined on an open, convex set, $\int_\gamma f(z)\mathrm{d}z$ = 0 for any piecewise-$C^1$ closed curve $\gamma$ in this set.

On the other hand, Cauchy's formula for a circle gives one a method to compute the value of a holomorphic $f$ for any point $z$ in the interior of the circle: $$f(z) = \frac{1}{2 \pi i} \int_C \frac{f(\xi)}{\xi - z} \mathrm{d}\xi\ .$$

My question is the following: why doesn't the integral on the right hand side of the above question vanish by Cauchy's theorem? To be painfully explicit, if we consider a circle centered around the origin, why is it not true that $$f(0) = \frac{1}{2 \pi i} \int_C \frac{f(\xi)}{\xi} \mathrm{d}\xi\ = 0$$ for all holomorphic functions $f$?

I know I'm missing something elementary; I just can't see what it is.

• Yes, but the Cauchy theorem requires the function to be holomorphic on an open convex set containing the contour, so the point $z = \xi$ will always be contained there. More particularly, in your example, if you integrate around any circle centered at the origin, the open convex set that is required to apply Cauchy's theorem will always contain the origin, since it would not be convex otherwise. I can try to elaborate this further in my answer, if you need me to. – MSobak Aug 27 '18 at 16:58
the integrand in $$f(z) = \frac{1}{2 \pi i} \int_C \frac{f(\xi)}{\xi - z} \mathrm{d}\xi\$$
is $$\frac{f(\xi)}{\xi - z}$$ which is not holomorphic within the region bounded by the boundary.
Consider the point $z=\xi$ where your function is not defined.