Theorem: Let $U \subset \Bbb C$ be an open setand $f \in H(U) $, i.e. $f $ is holomorphic on $U$. Let $z_0\in U$ be a point and $r>0$ such that $\bar{D}(z_0,r)\subset U $. Then $\forall z \in \bar{D}(z_0,r)$, $$f(z)=\frac{1}{2\pi i}\displaystyle \oint _{\lvert \xi-z_0=r \rvert}\frac{f(\xi)}{\xi -z}d\xi$$.
Proof:
Since $U$ is open, we can choose $\varepsilon>0$ such that $D(z_0,r+\varepsilon) \subset U$.
Fix a point $z \in D(z_0,r+\varepsilon)$. We define a function $$F(\xi)=\frac{f(\xi)-f(z)}{\xi - z} \;\;\;\;\text{if} \;\;\xi\in D(z_0,r+\varepsilon)-\{z\}$$ and $$F(\xi)=f'(z)\;\;\;\;\text{if} \;\;\xi = z$$.
Then clearly, $$F \in C(D(z_0,r+\varepsilon))\;\cap\; H(D(z_0,r+\varepsilon)-\{ z\}) \;\;\;\; \color{darkorange}{(\star)}$$
Then,$\dots$ (skip the whole proof)
My question is that why $\color{darkorange}{(\star)}$ is true?
Is subtracting $z$ a necessity so that it grantees $F$ is holomorphic on the open disk? Or $F$ is holomorphic on $D(z_0,r+\varepsilon)-\{z\}$ without a doubt. We can use the fact that $F$ is holomorphic on an open disk without finite points, so $F$ have a antiderivative on $D(z_0,r+\varepsilon)$.
We know that $F$ is holomorphic on $D(z_0,r+\varepsilon)-\{z\}$ by the quotiënt rule for differentation of holomorphic functions, since the denominator is nonzero on this open set. We do not however know whether $F$ is holomorphic at $z$, although we know that $f$ is.
It turns out that we the function $F$ as defined here is holomorphic, and this fact can be used to derive the Taylor series of holomorphic functions. But the proof of this fact relies on the theorem you are stating itself. This may be proved at a later point in your book; if not, you can read pages 124-125 of Ahlfor's Complex Analysis.
So strictly speaking, excluding $z$ isn't a necessity, but at the point in your book that hasn't been proved yet, so $z$ is excluded to avoid circular reasoning.