Theorem (Cauchy Integral Formula): Suppose $f$ is entire, $a \in \mathbb C$, and $C$ is the curve $Re^{i\theta}, 0 \leq \theta \leq 2\pi$ with $R \gt \vert a \vert$. Then
\[f(a) = \frac{1}{2\pi i}\int_C \frac{f(z)}{z - a} dz\]Proof: TODO.
Lemma: Suppose $a$ is contained in the circle $C_\rho$. Then
\[\int_{C_\rho} \frac{dz}{z - a} = 2 \pi i\]Proof: Let $C_\rho$ have center $\alpha$ and radius $\rho$. Then
\[\begin{align*} \frac{1}{z - a} &= \frac{1}{(z - \alpha) - (a - \alpha)} \\ &= \frac{1}{(z - \alpha)(1 - \frac{a - \alpha}{z - \alpha})} \\ &= \frac{1}{(z - \alpha)} \cdot \frac{1}{(1 - \omega)} \end{align*}\]with $\omega = \frac{a - \alpha}{z - \alpha}$. Note that since $\vert \omega \vert \lt 1$ throughout $C_\rho$,
\[\frac{1}{1 - \omega} = \sum_{k = 0}^\infty \omega^k\]so
\[\begin{align*} \frac{1}{z - a} &= \frac{1}{z - \alpha}\left(\sum_{k = 0}^\infty\left(\frac{a - \alpha}{z - \alpha}\right)^k\right) \\ &= \frac{1}{z - \alpha} + \sum_{k = 1}^\infty \left(\frac{(a - \alpha)^k}{(z - \alpha)^{k + 1}}\right) \end{align*}\]By the fact that \(\int_{C_\rho} \frac{dz}{z - \alpha} = \int_0^{2\pi} \frac{i\rho e^{i\theta}}{\rho e^{i\theta}} d\theta = 2\pi i\)
and
\[\int_{C_\rho} \frac{dz}{(z - \alpha)^k} = 0\]for $k = 2, 3, \dots$ by the Closed Curve Theorem since the integrand is the derivative of the analytic function $\frac{-1}{(z - \alpha)^{k - 1}}$,
\[\begin{align*} \int_{C_\rho} \frac{1}{z - a} &= \int_{C_\rho} \frac{1}{z - \alpha} + \sum_{k = 1}^\infty \left(\frac{(a - \alpha)^k}{(z - \alpha)^{k + 1}}\right) \\ &= 2\pi i + 0 \\ &= 2\pi i \end{align*}\]$\blacksquare$
Theorem (Taylor Expansion of an Entire Function): If $f$ is entire, it has a power series representation. In fact, $f^{(k)}(0)$ exists for $k = 1, 2, 3, \dots$ and
\[f(z) = \sum_{k = 0}^\infty \frac{f^{(k)}(0)}{k!} z^k\]Proof: Let $C$ be a circle of radius $R$ centered at the origin such that $z$ is contained within $C$. Then by the Cauchy Integral Formula,
\[f(z) = \frac{1}{2\pi i} \int_{C} \frac{f(z)}{\omega - z} d\omega\]and using the power series expansion
\[\begin{align*} \frac{1}{\omega - z} &= \frac{1}{\omega(1 - \frac{z}{\omega})} \\ &= \frac{1}{\omega} + \frac{z}{\omega^2} + \dots \end{align*}\]gives (by the uniform convergence of the power series throughout $C$, since $\left\vert \frac{z}{\omega} \right\vert \lt 1$),
\[f(z) = \frac{1}{2 \pi i}\int_C \frac{f(z)}{\omega} d\omega + \left(\frac{1}{2\pi i}\int_C \frac{f(z)}{\omega^2} d\omega\right) z + \left(\frac{1}{2\pi i} \int_C \frac{f(z)}{\omega^3}d \omega\right)z^2 + \dots\]Note that since we chose $C$ to be large enough to contain $z$, it appears as though the coefficients depend on $z$. But, since $f$ is infinitely differentiable at 0 and
\[C_k = \frac{f^{(k)}(0)}{k!}\]the coefficients are independent of $z$. $\blacksquare$
Corollary: An entire function is infinitely differentiable.
Proof: This follows from the fact that every entire function has a power series expansion (from above), and an everywhere convergent power series is infinitely differentiable. $\blacksquare$
Corollary: If $f$ is entire and if $a$ is any complex number, then
\[f(z) = f(a) + f'(a)(z - a) + \frac{f''(a)}{2!}(z - a)^2 + \dots\]for all $z$.
Proof: Let $g(\zeta) = f(\zeta + a)$. Then
\[g(\zeta) = \sum_{k = 0}^\infty \frac{g^{(k)}(0)}{k!} \zeta^k = \sum_{k = 0}^\infty \frac{f^{(k)}(a)}{k!} \zeta^k = f(\zeta + a)\]from which the statement follows with $\zeta = z - a$. $\blacksquare$
Proposition: If $f$ is entire and if
\[g(z) = \begin{cases} \frac{f(z) - f(a)}{z - a}, &z \neq a \\ f'(a), & z = a \end{cases}\]then $g$ is entire.
Proof: If $z = a$ then $g$ is entire since the derivative of an entire function is itself entire. So suppose otherwise. Then
\[\begin{align*} \frac{f(z) - f(a)}{z - a} &= \frac{f(a) + f'(a)(z - a) + \frac{f''(a)}{2!}(z - a)^2 + \dots - f(a)}{z - a} \\ &= f'(a) + \frac{f''(a)}{2!}(z - a) + \dots \end{align*}\]so it is an everywhere convergent power series, so $g$ is entire. $\blacksquare$
Corollary: Suppose $f$ is entire with zeroes at $a_1, a_2, \dots, a_N$. Then if $g$ is defined by
\[g(z) = \frac{f(z)}{(z - a_1)\dots(z - a_N)}\]for $z \neq a_k$, $\lim_{z \to a_k}$ exists for $k = 1, 2, \dots, N$ and if $g(a_k)$ is defined by these limits, then $g$ is entire.
Proof: Define $f_0(z)$ as $f(z)$ and $f_k(z)$ as
\[f_k(z) = \frac{f_{k - 1}(z) - f_{k - 1}(a)}{z - a_k} = \frac{f_{k - 1}(z)}{z - a_k}.\]Then by the above proposition, $f_k$ is entire if $f_{k - 1}$ is entire. Since $f_0 = f$, the proof follows by induction. $\blacksquare$
Theorem (Liouville’s Theorem): A bounded entire function is constant.
Proof: Let $a, b \in \mathbb C$ and $C$ a circle centered at the origin containing both $a$ and $b$. Then by the Cauchy Integral Formula,
\[\begin{align*} f(b) - f(a) &= \frac{1}{2\pi i}\int_C \frac{f(z)}{z - b} - \frac{f(z)}{z - a} dz \\ &= \frac{1}{2\pi i}\int_C \frac{f(z)(b - a)}{(z - b)(z - a)} dz \\ &\ll \frac{M \vert b - a \vert}{(R - \vert b \vert)(R - \vert a \vert)} \end{align*}\]which approaches 0 as $R \to \infty$. $\blacksquare$
Theorem (Extended Liouville Theorem): If $f$ is entire and if, for some integer $k \geq 0$, there exist positive constants $A$ and $B$ such that
\[\vert f(z) \vert \leq A + B\vert z \vert^K\]then $f$ is a polynomial of degree at most $k$.
Proof: Define
\[g(z) = \begin{cases} \frac{f(z) - f(0)}{z}, & z \neq 0 \\ f'(0), & z = 0 \end{cases}\]then $g$ is entire, so it is a polynomial. By the hypothesis on $f$, $\vert g(z) \vert \leq C + D\vert z \vert^{k - 1}$, so $g$ is a polynomial of degree at most $k - 1$, so $f$ is a polynomial of degree at most $k$. $\blacksquare$
Theorem (Fundamental Theorem of Algebra): Every non-constant polynomial with complex coefficients has a zero in $\mathbb C$.
Proof: Let $P(z)$ be a non-constant polynomial with $P(z) \neq 0$. Then $f(z) = \frac{1}{P(z)}$ is entire, and since $P(z)$ is non-constant, $\lim_{z \to \infty}P(z) = \infty$. So $f(z)$ is bounded. But then $f$ is constant, so $P(z)$ is constant; a contradiction. $\blacksquare$
Definition: The convex hull if a set $S$ of complex numbers is the smallest convex set containing $S$.
Theorem (Gauss-Lucas): The zeroes of the derivative of any polynomial lie within the convex hull of the zeroes of the polynomial.
Proof: TODO.