I left you last month with a collection of necessary conditions on any elliptic function, that is to say a doubly periodic meromorphic function in the complex plane. This month I shall demonstrate that such functions actually exist. But first I need another result from complex analysis, the {\it Schwarz Reflection Principle\/}. This says that if $f$ is analytic in a region $R$ in the upper half pane part of whose boundary is a segment $L$ of the real number line and if $f$ is also analytic at all points on this segment, then we can reflect $R$ across the real line to get a region $R^*$ in the lower half plane and extend $f$ to an analytic function in $R \cup L \cup R^*$. Specifically, using $\bar{z}$ to denote the complex conjugate of $z$ ($\overline{a+bi} = a-bi$), then the extension of $f$ is given by $$ g(z) = \left\{ \begin{array}{ll} f(z),& z \in R \cup L, \\[5pt] \overline{f(\overline{z})},& z \in R^*. \end{array} \right. $$This is easily checked. If $z \in R^*$ and we limit $h$ to be small enough that $z+h \in R^*$, then $$ \frac{g(z+h)-g(z)}{h} = \frac{\overline{ f(\bar{z} + \bar{h})-f(\bar{z})}}{h} = \overline{\left(\frac{ f(\bar{z} + \bar{h})-f(\bar{z})}{\bar{h}}\right)}$$ which approaches $\overline{f'(\bar{z})} = g'(z)$ as $h$ approaches 0. This can be applied to any situation where part of the boundary is a straight line. Simply rotate the function so that the straight line is horizontal, then add or subtract a constant to put that line on the real axis, reflect, then shift and rotate back. Since rotation is multiplication by a constant and shifting is addition of a constant, these actions do not affect analyticity.

Consider an arbitrary ellipse centered at the origin, $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\ a > b > 0. $$ The length of the upper arc from $x=0$ to $x = \alpha$ equals $$ a \times \int_0^{\alpha/a} \frac{1-k^2 t^2}{\sqrt{(1-t^2)(1-k^2t^2)}}\,dt, $$ where $k^2 = 1-b^2/a^2$. This is an example of what is called an $\textit{elliptic integral}$ for obvious reasons. In general, an elliptic integral is any definite integral with a polynomial in the numerator and the square root of a cubic or quartic polynomial without repeated roots in the denominator. The starting point for studying elliptic integrals is the $\textit{incomplete elliptic integral of the first kind}$: $$ F(x) := \int_0^x \frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}\,dt, \ 0 < k < 1.$$ For $0 \leq x \leq 1$, the value of this integral is real, Set $$ K = F(1) = \int_0^1 \frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}\,dt. $$ Once $t$ is larger than 1, $1-t^2$ is negative. For $1 < x < 1/k$, the integral contributes an amount that is pure imaginary. We can write $$ F(1/k) = \int_0^{1/k} \frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}\,dt = K + iK'.$$ For $x=1/k$ to infinity, the contribution is negative and real. Making the substitution $t = 1/(ku)$, we see that $$ \int_{1/k}^{\infty} \frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}\,dt = \int_1^0 \frac{1}{\sqrt{(1-u^2)(1-k^2u^2)}}\,du = -K.$$ If we integrate from $-\infty$ to 0, we continue from $iK'$ to $-K+iK'$ to $-K$ and back to 0. The net effect is that our elliptic function $E$ maps the entire upper half plane to the interior of a rectangle $2K$ units wide and $K'$ units high. See Figure 1.

At this point, it is instructive to think about what happens when the denominator is the square root of a quadratic polynomial such as $$ \int_0^x \frac{1}{\sqrt{1-t^2}}\,dt = \arcsin x.$$ No one really wants to work with the arcsine. Better to work with its inverse, the sine. The same is true for functions definid by elliptic integrals. We want to work with the inverse function of $F$ that maps the area inside the rectangle to the entire upper half plane, call it $E$ for elliptic function. Now we get to invoke the Schwarz reflection principle. This principle enables us to extend $E$ to the mirror image of our rectangle below the real axis, mapping that rectangle to the lower half plane, so the whole complex plan is the image under $E$ of a rectangle $2K$ units wide and $2K'$ units high. But now we can extend $E$ across the line segment from $K-iK'$ to $K+iK'$. The domain is now a rectangle of wide $4K$ and height $2K'$ mapping the rectangle to two complete copies of the complex plane. At this point we can continue to extend our function both horizontally and vertically until it is defined on the entire complex plane, but notice what happens. Horizontally, our function is periodic with period $4K$. Vertically it is periodic with period $2K'$. We have created a doubly periodic function.