Some Consequences of the Cauchy theorem
(Lecture 15 of Mathematical Methods II.)
Cauchy's inequality. On a circle of radius~$r$ circling the pole $z_0$, with $M=\max|f|$ on the circle: $|f^{(n)}(z_0)|\le\frac{n!M}{r^n}\,.$
Liouville's theorem. If an holomorphic function on the whole complex plane (such a function is called ``entire) is bounded on $\mathbf{C}$, then $f=\mathrm{cste}$.
Morera's theorem. If $f$ is continuous in a simply connected domain and if $\oint_\mathcal{C}f(z)\,dz=0$ for every closed path in $D$, then $f$ is holomorphic in~$D$.
Fundamental theorem of algebra. A polynomial equation
\begin{equation} P(z)=\sum_{i=1}^N\alpha_i z^i=\alpha_0+\alpha_1 z+\alpha_2 z^2+\cdots+\alpha_Nz^N \end{equation}
with $\alpha_i\in\mathbf{C}$ and $\alpha_N\neq0$ has exactly $n$ roots in $\mathbf{C}$ (possibly degenerate).
Proof: We prove first that there is at least one root: if $P(z)$ would never be zero, then $1/P(z)$ would be entire and since $|1/P(z)|$ is bounded (it tends to zero to infinity), from Liouville's theorem, it should be constant, in contradiction with the fact that $\alpha_N\neq0$. Therefore, $P(z)$ has at least one zero.
Let's call $z_0$ this root. Then
\begin{equation} P(z)=P(z)-P(z_0)=\sum_{i=0}^N\alpha_i(z^i-z_0^i)=(z-z_0)Q(z) \end{equation}
with $Q$ a polynomial of degree $N-1$. Repeating the same reasoning shows that $Q$ has at least one zero (possibly $z_0$ again).
We have already stated the following when studying harmonic functions. We can now prove these statements:
Gauss' mean value theorem. An analytic function is equal to the average of its values on any surrounding circle.
Proof: From Cauchy's integral formula $f(a)=\frac{1}{2\pi i}\oint_Cf(z)/(z-a)\,dz$ on $C=\{z\,:\,|z-a|=r\}$, we find $f(a)=\frac{1}{2\pi i}\int_0^{2\pi}f(a+re^{i\theta})ire^{i\theta}/(r e^{i\theta})\,d\theta=\int_0^{2\pi}f(a+re^{i\theta})rd\theta/(2\pi r)$.
Extremum modulus theorem. An analytic function in a region~$\mathcal{R}$ reaches the maximum and the minimum of its modulus on the boundary of $\mathcal{R}$.
We conclude with another projection of complex results into the real space:
If $f$ is holomorphic on a circle $C$ of radius $R$ and for every $z=re^{i\theta}$ inside it:
\begin{equation} f(re^{i\theta})=\frac{1}{2\pi}\int_0^{2\pi}\frac{(R^2-rRe^{i(\phi-\theta)})f(Re^{i\phi})}{R^2-2Rr\cos(\theta-\phi)+r^2}\,d\phi\,. \end{equation}
from which we obtain, decomposing $f$ between its real and imaginary parts $f=u+iv$, the Poisson's integral formulas on the disk:
\begin{align} u(r,\theta)&=\frac{1}{2\pi}\int_0^{2\pi}\frac{(R^2-r^2)u(R,\phi)}{R^2-2Rr\cos(\theta-\phi)+r^2}\,d\phi\,, % v(r,\theta)&=\frac{1}{2\pi}\int_0^{2\pi}\frac{(R^2-r^2)v(R,\phi)}{R^2-2Rr\cos(\theta-\phi)+r^2}\,d\phi\,. \end{align}
and also the counterpart with $u\leftrightarrow v$. This specifies completely harmonic functions inside a circle from the values on its boundary. This is used for solving the two-dimensional Laplace equation with boundary conditions given on the unit disc.
The countepart on the half plane reads, for $f$ analytic on the upper half-plane $(y\ge0)$ containing a point $\zeta=\xi+i\eta$:
\begin{equation} f(\zeta)=\frac{1}{\pi}\int_{-\infty}^\infty\frac{\eta f(x)}{(x-\xi)^2+\eta^2}\,dx \end{equation}
whose real and imaginary parts $f(\xi+i\eta)=u(\xi,\eta)+iv(\xi,\eta)$ provide:
\begin{align} u(\xi,\eta)=\frac{1}{\pi}\int_{-\infty}^\infty\frac{\eta u(x,0)}{(x-\xi)^2+\eta^2}\,dx\,. v(\xi,\eta)&=\frac{1}{\pi}\int_{-\infty}^\infty\frac{\eta v(x,0)}{(x-\xi)^2+\eta^2}\,dx\,. \end{align}
And also the same relation with $u\leftrightarrow v$. These are known as Poisson's integral formulas for the half-plane.
For instance, the harmonic function $u(x,y)$ defined on the half-plane (positive $y$) such that $u(x,0)=1$ for $-1\le x\le1$ and zero otherwise is:
\begin{equation} u(x,y)=\frac{1}{\pi}\int_{-1}^1\frac{ydt}{(t-x)^2+y^2}=\frac{1}{\pi}\mathrm{arctan}\left(\frac{y}{x-t}\right)\Bigg|_{-1}^1=\frac{1}{\pi}\mathrm{arctan}\left(\frac{y}{x-1}\right)-\frac{1}{\pi}\mathrm{arctan}\left(\frac{y}{x+1}\right)\,.\\ \end{equation}
A counterpart, the Schwarz integral formula (not given) comes back to the complex plane, it allows to recover an holomorphic function, up to an imaginary constant, from the boundary values of its real part.
