The trigonometric functions are periodic, and once we move into the complex plane so is the exponential function, $e^z = e^{z+2\pi i}$. But in the complex plane we also have functions with two independent periods. Subject to a constraint I will explain later, these are the $\textit{Elliptic functions}$. The name comes from their association with elliptic integrals which are generalizations of the integral encountered when computing the arc length of a portion of an ellipse. I will explain this connection next month. Usually elliptic functions are normalized so that one of the periods is the real number 1 and other has positive imaginary part. The effect, as shown in Figure 1, is to tile the plane with parallelograms. Once we know the values in one parallelogram, we know the values over the entire complex plane.

One of the immediate consequences of Cauchy's theorem is that any function $f$ that is analytic at $a$ has a power series expansion at $a$. We can see this by recentering, setting $g(\omega) = f(\omega + a)$, and letting $C$ be a simply closed curve encircling $a$ within the region of analyticity: \begin{eqnarray*} f(\omega) = g(\omega - a) & = & \frac{1}{2\pi i}\int_C \frac{g(z)}{z-(\omega-a)} dz\\ & = & \frac{1}{2\pi i}\int_C \frac{g(z)}{z} \frac{1}{1-(\omega-a)/z} dz \\ & = & \frac{1}{2\pi i}\int_C \frac{g(z)}{z} \sum_{j=0}^{\infty} \frac{(\omega-a)^j}{z^j} dz \\ & = & \sum_{j=0}^{\infty} (\omega-a)^j \frac{1}{2\pi i} \int_C \frac{g(z)}{z^{j+1}}. \end{eqnarray*} This is a power series in $\omega - a$ with explicit coefficients given by the integrals.

Furthermore, the radius of convergence is the distance from $a$ to the nearest singularity, that is the nearest point where $f$ is not analytic. It clearly cannot be greater than this distance. If it were less then analyticity on the boundary guarantees that we can push the region of convergence out a little bit farther. This involves a nice application of completeness. Let $R$ be the candidate for the radius of convergence, a positive value less than the distance to the nearest singularity. Let $C_R$ be the set of points of distance less than or equal to $R$ from $a$. If our function is analytic at each point on the boundary, then each boundary point is the center of a small disc where the function remains analytic. If we take the union of all of these discs plus the interior of $C_R$, this union of open sets contains $C_R$. But $C_R$ is closed and bounded. Completeness of the complex plane means that there is a finite subcollection of these open discs that contains all of $C_R$. This implies that we can push the radius out a little farther.

The proof of Liouville's Theorem is fairly simple. Assume that we do have a bound, call it $M$. Take any two points in the plane, $a$ and $b$, and consider a circle $C$ centered at the origin with radius $R$ for which $a$ and $b$ lie in the interior of the circle. By Cauchy's theorem, \begin{eqnarray*} f(b) - f(a) & = & \frac{1}{2\pi i} \int_C \frac{f(z)}{z-b}dz - \frac{1}{2\pi i} \int_C \frac{f(z)}{z-a}dz \\ & = & \frac{1}{2\pi i} \int_C \frac{f(z)(b-a)}{(z-a)(z-b)} dz. \end{eqnarray*} This tells us that $|f(b)-f(a)|$ is bounded above by the circumference of the circle ($2\pi R$) divided by $2\pi$ times the maximum possible value of the integrand, $$ |f(b) - f(a)| \leq R \frac{M|b-a|}{(R-|a|)(R-|b|)}. $$ We have a single power of $R$ in the numerator and a quadratic power of $R$ in the denominator and no limit on the size of $R$. That means that the upper bound can be brought as close to zero as we wish, so that $ |f(b) - f(a)| = 0$. We say that $f$ $\textit{has a pole of order $k \geq 1$ at $a$}$ if there exists an integer $k$ for which $$ \lim_{z\to a} (z-a)^k f(z) \neq 0\quad {\rm but} \quad \lim_{z\to a} (z-a)^{k+1} f(z) = 0. $$ If $f$ has an isolated pole of order $k$ at $a$, then $(z-a)^kf(z)$ is analytic at $a$ and we can write $f$ as $$ f(z) = \sum_{j=-k}^{\infty} c_j (z-a)^j. $$ This power series that can include negative powers is called a $\textit{Laurent series}$. If $f$ has a pole at $a$ and $C$ is a simply closed curve with $a$ in its interior, then $$ \int_C f(z) dz = \sum_{j=-k}^{\infty} \int_C c_j (z-a)^j dz = (2\pi i) c_{-1}.$$ We call the coefficient of $(z-a)^{-1}$ the $\textit{residue}$ of $f$ at $a$. If there is more than one pole encircled by $C$, then $\int_Cf(z) dz$ equals $2 \pi i$ times the sum of the residues.

If $f(a)=0$, we say that $a$ is a $\textit{zero}$ of $f$, and its order is the smallest positive power $k$ so that $$ \lim_{z\to a} \frac{f(z)}{(z-a)^k} \neq 0. $$ Now an amazing result is that if $C$ is a simply closed curve with no zeroes or poles of $f$ on the curve, then in the interior of the region bounded by $C$ the number of zeros, denoted $\mathbb{Z}$, minus the number of poles, denoted $\mathbb{P}$, is given by $$ \mathbb{Z} - \mathbb{P} = \frac{1}{2\pi i} \int_C \frac{f'(z)}{f(z)} dz, $$ where each zero or pole is counted according to its order. If $f$ is elliptic, then the number of zeros in the period parallelogram must equal the number of poles. Here also the proof is simple. If $f$ has a zero or pole at $a$ then we can write $f(z) = (z-a)^{k_a} g_a(z)$ where $g_a$ is analytic at $a$ and we have a zero of order $k_a$ if $k_a$ is positive and a pole of order $-k_a$ if $k_a$ is negative. Now, \begin{eqnarray*} f'(z) & = & k_a(z-a)^{k_a-1}g_a(z) + (z-a)^k g_a'(z) \\ \frac{f'(z)}{f(z)} & = & \frac{k_a}{z-a} + \frac{g_a'(z)}{g_a(z)}. \end{eqnarray*} Therefore, $$ \frac{1}{2\pi i} \int_C \frac{f'(z)}{f(z)} dz = \sum k_a, $$ where the sum is over all points encircled by $C$ and $k_a = 0$ if $a$ is neither a pole nor a zero. There are a lot of conditions on elliptic functions. How do we know such functions even exist? That will be next month's topic.