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$\gdef\d{\operatorname{d}}$
Let $f: U \to \CC$ be holomorphic, with $U$ open. If $\overline D$ is any closed disk inside $U$ and $z$ is any point in the interior $D$, then
$$
f(z) \spaces= \frac{1}{2\pi i} \int_{\partial D} \frac{f(w)}{w-z} \d w
$$
if we go around the boundary once counterclockwise in performing the integration.
This formula is something of a key to the notion of holomorphic function. For example, it explains why holomorphic function are so regular: the dependence on $z$ in the right-hand side of the equation is easy to control; e.g. we can easily differentiate, or even expand into a power series. Here are some corollaries:
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If $f: U \to \CC$ is holomorphic and $D(z,r)$ is any open disk contained in $U$, then the power series expansion for $f$ centered at $z$ converges in all of $D(z,r)$.
Let me pause to give a cute example application of this theorem. Consider the power series
$$ F(z) \spaces= \sum_{n \ge 0} F_n \cdot z^n $$
where $F_n$ is the $n^{th}$ Fibonacci number. Then the recurrence relation $F_{n+2} = F_{n+1} + F_n$ and normalization $F_0 = F_1 = 1$ translate into the relation
$$ F(z) \spaces= zF(z) + z^2F(z) + 1 $$
Using this, we find that $F(z)$ is the power series expansion of
$$ F(z) \spaces= \frac{1}{1-z-z^2} $$
around $0$. This function $F$ is holomorphic everywhere except when $1-z-z^2=0$ i.e. when $z=\frac{1\pm\sqrt{5}}{2}$. From this we can conclude that the radius of convergence of our series $\frac{1-\sqrt{5}}{2}$. Indeed, $F$ is holomorphic in the disk with radius $\frac{\sqrt{5}-1}{2}$, so the above theorem tells us that the radius of convergence is at least $\frac{\sqrt{5}-1}{2}$; but on the other hand, the function $F$ tends to $\infty$ as $z$ approaches $\frac{1-\sqrt{5}}{2}$, and a power series can’t tend to $\infty$ inside a region where it is defined since a power series is continuous; thus the radius of convergence is also at most $\frac{\sqrt{5}-1}{2}$, whence the claim. This tells you something about the order of growth of the $F_n$.
Another application of the Cauchy integral formula is the removable singularities theorem:
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Let $U \in \CC$ be open, and $z \in U$. Suppose $f: U \backslash \{z\} \to \CC$ is holomorphic, and is bounded in some disk around $z$. Then $f$ extends uniquely is holomorphic on all of $U$.
So merely the fact of $f$ being bounded around $z$ implies something much stronger.