Theorem: Cauchy's Integral Formula

Let $D = \set {z \in \C: \cmod z \le r}$ be the closed disk of radius $r$ in $\C$.
Let $f: U \to \C$ be holomorphic on some open set containing $D$.
Then for each $a$ in the interior of $D$:
:$\ds \map f a = \frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f z} {z - a} \rd z$
where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.

Proof:
Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $\map f z$ is holomorphic on $R$.
Let $z_0$ be any point in the region $R$ such that:
:$\dfrac {\map f z} {z - z_0}$ is holomorphic on $R \setminus \set {z_0}$
We draw a circle $C_r$ with center at $z_0$ and radius $r$ such that $r \to 0$.
This makes $C$ and $C_r$ a multiply connected region for a sufficiently small $r > 0$.
{{explain|multiply connected region}}
{{DiagramWanted}}
According to Cauchy's Integral Theorem for a multiply connected region:
{{begin-eqn}}
{{eqn | l = I
      | o = :=
      | r = \oint_C \frac {\map f z} {z - z_0} \rd z
      | c = 
}}
{{eqn | r = \oint_{C_r} \frac {\map f z} {z - z_0} \rd z
      | c = 
}}
{{eqn | r = \oint_{C_r} \frac {\map f {z_0} + \paren {\map f z - \map f {z_0} } } {z - z_0} \rd z
      | c = 
}}
{{eqn | r = \map f {z_0} \oint_{C_r} \frac {\rd z} {z - z_0} + \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z
      | c = 
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = z - z_0
      | r = r e^{i \theta}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \d z
      | r = i r e^{i \theta} \rd \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \oint_{C_r} \frac {\rd z} {z - z_0}
      | r = \int_0^{2 \pi} \frac {i r e^{i \theta} } {r e^{i \theta} } \rd \theta
      | c = 
}}
{{eqn | r = i \int_0^{2 \pi} \rd \theta
      | c = 
}}
{{eqn | r = 2 \pi i
      | c = 
}}
{{end-eqn}}
Now:
:$\ds I = 2 \pi i \map f {z_0} + \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z$
From Complex-Differentiable Function is Continuous, $f$ is continous.
According to Epsilon-Delta definition, for every $\epsilon \in \R_{>0}$ there exists a $\delta \in \openint 0 r$ such that:
:$\forall z \in \C: \cmod {z - z_0} < \delta \implies \cmod {\map f z - \map f {z_0} } < \epsilon$
Hence:
{{begin-eqn}}
{{eqn | l = \cmod {\oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z}
      | r = \cmod {\oint_{C_\delta} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z}
}}
{{eqn | o = \le
      | r = \oint_{C_\delta} \frac {\cmod {\map f z - \map f {z_0} } } {\cmod {z - z_0} } \cmod {\d z}
}}
{{eqn | o = \le
      | r = \frac \epsilon \delta \oint_{C_\delta} \cmod {\d z}
      | c = 
}}
{{eqn | r = 2 \pi \epsilon
      | c = 
}}
{{end-eqn}}
{{explain|The precise meaning of $\cmod {\d z}$}}
As $\epsilon \to 0$:
:$\ds \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z = 0$
So:
{{begin-eqn}}
{{eqn | l = I
      | r = \map f {z_0} \oint_{C_r} \frac {\rd z} {z - z_0} + \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z
      | c = 
}}
{{eqn | r = 2 \pi i \map f {z_0} + 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \oint_C \frac {\map f z} {z - z_0} \rd z
      | r = 2 \pi i \, \map f {z_0}
      | c = 
}}
{{end-eqn}}
{{qed}}