Cauchy's Integral Formula

Lemma For \(z_0 \in \mathbb C, C_r\) be the circle of radius \(r\). Then, \(\oint_{C_r}\frac{1}{z-z_0}dz = 2\pi i\).

proof. We parameterize the circle by \(c(t) = re^{it} + z_0\)

\[\oint_{C_r}\frac{1}{z-z_0}dz = \int_0^{2\pi}\frac{1}{re^{it}-z_0+z_0}rie^{it}dt =i \int_0^{2\pi}dt = 2\pi i\]

Theorem Let \(f:D\rightarrow\mathbb C\) analytic and \(D\) simply connected. Then for any Jordon contour \(C\subset D\). For \(z\in C_{int}\),

\[f(z) = \frac{1}{2\pi i}\oint_C \frac{f(w)}{w-z}dw\]

proof. Let \(z\in D\), define \(C_\delta\) be a circle of radius \(\delta\) centered at \(z\), we choose \(\delta\) small enough so that \(C_\delta \subset C_{int}\). Then, we deform the integral over \(C\) to \(C_\delta\).

\[\begin{align*} \frac{1}{2\pi i}\oint_C \frac{f(w)}{w-z}dw &= \frac{1}{2\pi i}\oint_{C_\delta} \frac{f(w) -f(z)}{w-z} + \frac{f(z)}{w-z}dw\\ &= \frac{1}{2\pi i}\oint_{C_\delta} \frac{f(w) -f(z)}{w-z}dw + \frac{f(z)}{2\pi i}\oint_C\frac{1}{w-z}dw \\ &= \frac{1}{2\pi i}\oint_{C_\delta} \frac{f(w) -f(z)}{w-z}dw + f(z) &\text{lemma} \end{align*}\]

Thus, we need to show that \(\oint_{C_\delta} \frac{f(w) -f(z)}{w-z}dw = 0\).
Since \(C_\delta \subset C\cup C_{int}\) and it is compact. We can bound \(\frac{f(w) -f(z)}{w-z}\) on \(C_\delta - \{z\}\), in addition, since \(f\) is analytic, \(\lim_{w\rightarrow z} \frac{f(w) -f(z)}{w-z} = f'(z)\). Therefore, \(\frac{f(w) -f(z)}{w-z} \leq M\) for all \(w\in C_\delta\). By ML inequality,

\[|\frac{1}{2\pi i}\oint_{C_\delta} \frac{f(w) -f(z)}{w-z}dw| \leq |\frac1{2\pi}||(2\pi\delta)M| = \delta M\]

Note that we can pick \(\delta\) arbitrarily small so that the supremum of the integral approaches 0.

Holomorphic from Analyticity

Theorem (Analytic functions are infinitely complex-differentiable) If \(f\) in analytic on \(C\cup C_{int}\) for some Jordon contour \(C\), then

\[f^{(k)}(z) = \frac{k!}{2\pi i} \oint_C \frac{f(w)}{(w-z)^{k+1}}dw\]

proof. First, consider \(k = 1\), we have

\[\begin{align*} \frac{f(z+h) - f(z)}h &= \frac{1}{h2\pi i}\oint_C f(w)(\frac{1}{w-(z+h)} - \frac 1{w-z})dw\\ &= \frac{1}{2\pi i}\oint_C \frac{f(w)}{(w-(z+h))(w-z)}dw\\ &= \frac{1}{2\pi i}\oint_C \frac{f(w)}{(w-z)^2}dw + \frac{h}{2\pi i}\oint_C \frac{f(w)}{(w-z)^2(w-z-h)}dw \end{align*}\]

Then, let \(I = \oint_C \frac{f(w)}{(w-z)^2(w-z-h)}dw\), we need the \(\lim_{h\rightarrow 0} \frac{h}{2\pi i}I = 0\).

Similar to the CIF proof, note that \(f(w)\) is bounded on \(C\), and since \(C\) can be taken arbitrarily small, \(|I|\) is then bounded by \(\frac M{(\delta + h)^2\delta} 2\pi\delta\) so that

\[\lim_{h\rightarrow 0} |\frac{h}{2\pi i}I| = \frac{|h|M}{(\delta + |h|)^2} = 0\]

Then, we can repeat this process in our inductive step and prove this statement.

Liouville's Theorem

Theorem If \(f:\mathbb C\rightarrow\mathbb C\) is entire (analytic on \(\mathbb C\)) and bounded, then \(f\) is a constant.

proof. First note that \(f\) is analytic, take \(R\) be the radius of circle \(C\) around some \(z\), and \(M\) so that \(|f(z)| < M\) for \(z \in C\), using ML inequality and the theorem above, we can easily have

\[\begin{align*} |f^{(k)}(z)| &= \frac{k!}{2\pi} |\oint_C \frac{f(w)}{(w-z)^{k+1}}dw| \\ &= \frac{k!}{2\pi} \frac{M}{R^{k+1}} 2\pi R\\ &= \frac{k!M}{R^n} \end{align*}\]

Take \(k=1\), we have

\[\forall R > 0, |f'(z)|\leq \frac{M}{R}\]

Take \(R\) arbitrarily large, we have \(|f'(z)| = 0\) so that for any point \(z\in\mathbb C\)

\[f(z) - f(0) = \int_0^z f'(z)dz = 0\implies f(z) = f(0)\]

Fundamental Theorem of Algebra

Theorem Let \(p(z)\) be some complex polynomials. Then \(\exists z\in \mathbb C\) s.t. \(p(z) = 0\).

proof. Let \(p(z):= \sum_{j=0}^n c_j z^j\) be some n-degree complex polynomial. Assume that \(\forall z\in\mathbb C, p(z) \neq 0\). Then, \(\frac{1}{p(z)}\) is entire and bounded. By Liouville's Theorem, it is a constant. This causes contradiction as \(p\) must be in degree of \(n\).

Maximum Principles

Lemma If \(f\) is analytic on some domain \(D\), then \(|f(z)|\) is unbounded unless \(f(z)\) is a constant.

proof. Let \(z\in \mathbb D\), take some ball \(B_{\delta}(z) \subset D\). Then, we can establish

\[\begin{align*} f(z)\frac{\delta^2}{2} &= \int_0^\delta f(z)\tau d\tau\\ &= \frac{1}{2\pi}\int_0^\pi \oint_{C_\tau}\frac{f(w)}{w-z}\tau dw d\tau\\ &= \frac1{2\pi}\int_0^\pi \oint_{C_\tau}f(w)dwd\tau &\forall w\in C_\tau.(w-z)=\tau \\ &= \frac1{2\pi}\iint_{B_{\delta}(z)}fdA \end{align*}\]

Suppose \(z\) is a maximum, then

\[|f(z)|\frac{\delta^2}{2} = |\frac1{2\pi}\iint_{B_{\delta}(z)}fdA| \leq \frac{1}{2\pi}|f(z)|A = \frac{1}{2\pi}|f(z)|\pi\delta^2 = \frac{|f(z)|\delta^2}{2}\]

Implies that for all \(z\), \(f(z)\) is constant within its neighborhood, implies that \(f\) is a constant.

Theorem If \(f\) is analytic and bounded on region \(D\) and \(|f(z)|\) is continuous in the closure \(\bar D\), then \(|f(z)|\) must has its maximum on \(\partial D\). Note that first, by extreme value theorem, a closed and bounded region must have a maximum, then by lemma. Any open region cannot have a maximum, hence the maximum must be on some closure.