Theorem: If $f(z)$ is analytic in a simply connected domain $D$ containing a point $z_0$ and $C$ is a positively oriented simple closed contour in $D$ enclosing $z_0$, then:

$f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - z_0}dz$

Proof:

Step 1: Consider a small circle $C_\epsilon$ of radius $\epsilon$ centered at $z_0$. Since $f$ is analytic in the region between $C$ and $C_\epsilon$, we can apply Cauchy's Integral Theorem to the function $f(z)/(z - z_0)$, which is analytic in this annular region.

By deforming the contour from $C$ to $C_\epsilon$:

$\oint_C \frac{f(z)}{z - z_0}dz = \oint_{C_\epsilon} \frac{f(z)}{z - z_0}dz$

Step 2: Parametrize $C_\epsilon$ as $z = z_0 + \epsilon e^{i\theta}$ for $0 \leq \theta \leq 2\pi$, so $dz = i\epsilon e^{i\theta}d\theta$:

$\oint_{C_\epsilon} \frac{f(z)}{z - z_0}dz = \int_0^{2\pi} \frac{f(z_0 + \epsilon e^{i\theta})}{\epsilon e^{i\theta}} \cdot i\epsilon e^{i\theta}d\theta = i\int_0^{2\pi} f(z_0 + \epsilon e^{i\theta})d\theta$

Step 3: Since $f$ is continuous at $z_0$ (being analytic), for any $\delta > 0$ there exists $\epsilon_0 > 0$ such that $|f(z) - f(z_0)| < \delta$ whenever $|z - z_0| < \epsilon_0$.

For $\epsilon < \epsilon_0$:

$\left|i\int_0^{2\pi} f(z_0 + \epsilon e^{i\theta})d\theta - i\int_0^{2\pi} f(z_0)d\theta\right| \leq \int_0^{2\pi} |f(z_0 + \epsilon e^{i\theta}) - f(z_0)|d\theta < 2\pi\delta$

Step 4: Since $f(z_0)$ is constant:

$i\int_0^{2\pi} f(z_0)d\theta = 2\pi if(z_0)$

As $\delta$ can be arbitrarily small:

$\oint_{C_\epsilon} \frac{f(z)}{z - z_0}dz = 2\pi if(z_0)$

Combining Steps 1 and 4:

$\oint_C \frac{f(z)}{z - z_0}dz = 2\pi if(z_0)$

which gives the desired result.

Theorem: Under the same conditions, the $n$-th derivative is:

$f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z - z_0)^{n+1}}dz$

Proof:

We prove this by induction. The base case $n = 0$ is Cauchy's Integral Formula.

Assume the formula holds for $n = k$. For $n = k + 1$, consider:

$f^{(k)}(z_0) = \frac{k!}{2\pi i}\oint_C \frac{f(z)}{(z - z_0)^{k+1}}dz$

The derivative with respect to $z_0$ is:

$f^{(k+1)}(z_0) = \frac{d}{dz_0}f^{(k)}(z_0) = \frac{k!}{2\pi i}\frac{d}{dz_0}\oint_C \frac{f(z)}{(z - z_0)^{k+1}}dz$

Since $z$ on $C$ is independent of $z_0$:

$= \frac{k!}{2\pi i}\oint_C f(z)\frac{\partial}{\partial z_0}\left[(z - z_0)^{-(k+1)}\right]dz$

$= \frac{k!}{2\pi i}\oint_C f(z) \cdot (k+1)(z - z_0)^{-(k+2)}dz$

$= \frac{(k+1)!}{2\pi i}\oint_C \frac{f(z)}{(z - z_0)^{k+2}}dz$

This completes the induction.

Theorem: If $f(z)$ is entire (analytic everywhere in $\mathbb{C}$) and bounded, then $f$ is constant.

Proof:

Since $f$ is bounded, $|f(z)| \leq M$ for all $z$ and some constant $M$. By Cauchy's formula for derivatives with a circle $|z - z_0| = R$:

$|f'(z_0)| = \left|\frac{1}{2\pi i}\oint_{|z-z_0|=R} \frac{f(z)}{(z - z_0)^2}dz\right| \leq \frac{1}{2\pi}\frac{M \cdot 2\pi R}{R^2} = \frac{M}{R}$

As $R \to \infty$, $|f'(z_0)| \to 0$ for arbitrary $z_0$. Thus $f'(z) = 0$ everywhere, so $f$ is constant.