### Introduction

So far we have studied the elliptic functions of real variables, i.e. we consider the $ u$ in $ \text{sn}(u, k)$ to be a real number and have so far found that they have properties similar to the circular functions (for example they are bounded and periodic). However their real nature and power is exhibited only when we go in the realm of complex numbers and study them as functions of complex variables.As an example of the power of complex variables, we can see the contrast between real variables and complex variables by looking at the behavior of circular functions and exponential functions. If we limit ourselves to the real variables the functions $ \sin x$ and $ \cos x$ have nothing to do with the exponential function $ \exp(x)$ but as soon as we meet the complex variables we have the remarkable equation $ \exp(ix) = \cos x + i \sin x$ and this puts the circular, exponential and logarithmic functions on the same footing.

We will not be discussing the deep methods of complex function theory, but we will rather find some peculiar properties of the elliptic functions when considered as functions of a complex variable. To that end we need to extend the definition of the elliptic functions to the domain of complex arguments. A simple method of defining $ \text{sn}(x + iy, k)$ could be define $ \text{sn}(iy, k)$ first and then use the addition formulas to define $ \text{sn}(x + iy, k)$. This brings us to the now famous Jacobi's Imaginary Transformation.

### Jacobi's Imaginary Transformation

We need to recast our definitions in a slightly different form as follows: \begin{align} u &= F(\phi, k) = \int_{0}^{\phi}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}}\notag\\ &= 2\int_{0}^{s}\frac{dt}{\sqrt{1 + 2(1 - 2k^{2})t^{2} + t^4}}\notag \end{align} where $ \sin\phi = 2s/(1 + s^{2})$.Thus we have $ \text{sn}(u, k) = 2s / (1 + s^{2})$ where $ s$ and $ u$ are related by the integral above. And after some manipulation we have \begin{align} \text{cn}(u, k) &= \frac{1 - s^{2}}{1 + s^{2}}\notag\\ \text{dn}(u, k) &= \frac{\sqrt{1 + 2(1 - 2k^{2})s^{2} + s^4}}{1 + s^{2}}\notag \end{align} If we substitute $ t = ix$ in the above integral formally and do some manipulation we get $$iu = 2\int_{0}^{is}\frac{dx}{\sqrt{1 + 2(1 - 2k'^{2})x^{2} + x^4}}$$ which suggests us that we should have \begin{align} \text{sn}(iu, k') &= \frac{2is}{1 + (is)^{2}}\notag\\ &= i \cdot \frac{2s}{1 - s^{2}}\notag\\ &= i \cdot \tan\phi\notag\\ &= i \cdot \frac{\sin\phi}{\cos\phi}\notag\\ &= i \cdot \frac{\text{sn}(u, k)}{\text{cn}(u, k)}\notag \end{align} and similarly \begin{align} \text{cn}(iu, k') &= \frac{1 - (is)^{2}}{1 + (is)^{2}}\notag\\ &= \frac{1 + s^{2}}{1 - s^{2}}\notag\\ &= \frac{1}{\text{cn}(u, k)}\notag\\ \text{dn}(iu, k') &= \frac{\sqrt{1 + 2(1 - 2k^{\prime 2})(is)^{2} + (is)^4}}{1 + (is)^{2}}\notag\\ &= \frac{\sqrt{1 + 2(1 - 2k^{2})s^{2} + s^4}}{1 - s^{2}}\notag\\ &= \frac{\sqrt{1 + 2(1 - 2k^{2})s^{2} + s^4} / (1 + s^{2})}{(1 - s^{2})/(1 + s^{2})}\notag\\ &= \frac{\text{dn}(u, k)}{\text{cn}(u, k)}\notag \end{align} Interchanging $ k$ and $ k^{\prime}$ we get the Jacobi's Imaginary Transformations: \begin{align} \text{sn}(iu, k) &= i\,\text{sc}(u, k')\tag{1}\\ \text{cn}(iu, k) &= \text{nc}(u, k')\tag{2}\\ \text{dn}(iu, k) &= \text{dc}(u, k')\tag{3} \end{align} We take the above as definitions for all real $ u$ for which the right hand side of the above equations is defined. It should be noted again that the right hand side of the above equations are defined only when $ \text{cn}(u, k') \neq 0$. Now we have $ \text{cn}(u, k') = 0$ when $ u = (2n + 1)K'$ where $ n$ is an integer. Therefore the elliptic functions for the purely imaginary values of the argument are defined

*except*when the argument is equal to $ (2n + 1)iK'$ where $ n$ is an integer.

If we restrict ourselves to the purely imaginary values of the argument then we can also see that the elliptic functions are periodic with an imaginary period. Thus $ \text{sn}(u, k)$ is has period $ 2iK'$, $ \text{cn}(u, k)$ has period $ 4iK'$ and $ \text{dn}(u, k)$ has a period $ 4iK'$. This is in quite contrast with the circular functions which have only one period (which is real) whether we restrict them to the real variables or not. We therefore see that the elliptic functions are

*doubly periodic*. We will have occasion to talk about this interesting feature in a later post.

We can see that the same rules of differentiation hold for the imaginary variables as in case of real variable. For example we have \begin{align} \frac{d}{d(iu)}(\text{sn}(iu, k)) &= -i \cdot \frac{d}{du}(\text{sn}(iu, k))\notag\\ &= -i \cdot \frac{d}{du}\left(i \cdot \frac{\text{sn}(u, k')}{\text{cn}(u, k')}\right)\notag\\ &= \frac{d}{du}\left(\frac{\text{sn}(u, k')}{\text{cn}(u, k')}\right)\notag\\ &= \frac{\text{cn}(u, k')\,\text{cn}(u, k')\,\text{dn}(u, k') + \text{sn}(u, k')\,\text{sn}(u, k')\,\text{dn}(u, k')}{\text{cn}^{2}(u, k')}\notag\\ &= \frac{\text{dn}(u, k')}{\text{cn}^{2}(u, k')}\notag\\ &= \text{nc}(u, k')\,\text{dc}(u, k')\notag\\ &= \text{cn}(iu, k)\,\text{dn}(iu, k)\notag \end{align}

### Extension to Complex Variables

Since the addition formulas in the previous post were established using the derivatives of the elliptic functions, it is obvious that the same formulas hold for the case when both the arguments $ u_{1}$ and $ u_{2}$ are purely imaginary. It would now be a great idea to assume that the addition formulas hold generally and define for a complex number $ z = x + iy$ \begin{align} \text{sn}(z, k) &= \text{sn}(x + iy, k)\notag\\ &= \frac{\text{sn}^{2}(x, k) - \text{sn}^{2}(iy, k)}{\text{sn}(x, k)\,\text{cn}(iy, k)\,\text{dn}(iy, k) - \text{sn}(iy, k)\,\text{cn}(x, k)\,\text{dn}(x, k)}\notag\\ \text{cn}(z, k) &= \text{cn}(x + iy, k)\notag\\ &= \frac{\text{sn}(x, k)\,\text{cn}(x, k)\,\text{dn}(iy, k) - \text{sn}(iy, k)\,\text{cn}(iy, k)\,\text{dn}(x, k)}{\text{sn}(x, k)\,\text{cn}(iy, k)\,\text{dn}(iy, k) - \text{sn}(iy, k)\,\text{cn}(x, k)\,\text{dn}(x, k)}\notag\\ \text{dn}(z, k) &= \text{dn}(x + iy, k)\notag\\ &= \frac{\text{sn}(x, k)\,\text{cn}(iy, k)\,\text{dn}(x, k) - \text{sn}(iy, k)\,\text{cn}(x, k)\,\text{dn}(iy, k)}{\text{sn}(x, k)\,\text{cn}(iy, k)\,\text{dn}(iy, k) - \text{sn}(iy, k)\,\text{cn}(x, k)\,\text{dn}(x, k)}\notag \end{align} We shall keep the above definition as provisional and would replace them by a different set of formulas. Here we like to demonstrate that defining the elliptic functions for complex values using the addition formula makes an analytic function. Now these definitions were derived from the addition formulas for elliptic functions for real variables and since we have for real values of $ u, v$ $$\frac{\partial}{\partial u}(\text{sn}(u + v, k)) = \frac{\partial}{\partial v}(\text{sn}(u + v, k))$$ it follows that if we replace $ \text{sn}(u + v, k)$ in the above equation with the corresponding expression consisting of the the functions of $ u$ and $ v$ separately using addition formula, the above equation remains valid and can be independently verified by using the rules of differentiation of elliptic functions.It also follows that since the same rules of differentiation applying for purely imaginary argument, therefore we may replace the variables $ u = x$ and $ v = iy$ to get $$\frac{\partial}{\partial x}(\text{sn}(x + iy, k)) = \frac{1}{i}\cdot\frac{\partial}{\partial y}(\text{sn}(x + iy, k))$$ and this means that if we could express $ \text{sn}(x + iy, k) = f(x, y) + ig(x, y)$ in real and imaginary parts separately then we will have \begin{align} \frac{\partial}{\partial x}(f(x, y) + ig(x, y)) &= \frac{1}{i}\cdot\frac{\partial}{\partial y}(f(x, y) + ig(x, y))\notag\\ \Rightarrow \frac{\partial f}{\partial x} + i \cdot \frac{\partial g}{\partial x} &= -i \left(\frac{\partial f}{\partial y} + i\cdot \frac{\partial g}{\partial y}\right)\notag\\ \Rightarrow \frac{\partial f}{\partial x} &= \frac{\partial g}{\partial y}\text{ and}\notag\\ \frac{\partial g}{\partial x} &= -\frac{\partial f}{\partial y}\notag \end{align} We thus get the Cauchy Riemann Equations and it follows that the definition we have adopted for $ \text{sn}(z, k)$ (and correspondingly for $ \text{cn}(z, k)$ and $ \text{dn}(z, k)$) makes it an analytic function of $ z$ at all points in the complex plane where the functions are defined.

With somewhat reasonable amount of algebraical manipulation it can be shown that the differentiation formulas also hold for the complex variable $ z$ and therefore the same addition formulas apply when the variables involved are arbitrary complex numbers.

Its now easy to see how we can obtain the addition formulas in their traditional look. We have \begin{align} \text{sn}(iu + iK', k) &= \text{sn}(i(u + K'), k)\notag\\ &= i \frac{\text{sn}(u + K', k')}{\text{cn}(u + K', k')}\notag\\ &= \frac{-i}{k} \frac{\text{cn}(u, k')}{\text{dn}(u, k')}\frac{\text{dn}(u, k')}{\text{sn}(u, k')}\notag\\ &= \frac{-i}{k}\frac{1}{\text{sc}(u, k')} = \frac{1}{k\,\text{sn}(iu, k)}\notag \end{align} By analytical nature of the elliptic functions the above formula holds for all values of $ u$ whenever the right side is defined and hence we can write \begin{align} \text{sn}(u + iK', k) &= \frac{1}{k\,\text{sn}(u, k)} = \frac{1}{k}\,\text{ns}(u, k)\tag{4}\\ \text{cn}(u + iK', k) &= \frac{-i}{k}\frac{\text{dn}(u, k)}{\text{sn}(u, k)} = \frac{-i}{k}\,\text{ds}(u, k)\tag{5}\\ \text{dn}(u + iK', k) &= \frac{-i\,\text{cn}(u, k)}{\text{sn}(u, k)} = -i\,\text{cs}(u, k)\tag{6} \end{align} From the above we can also see that the above functions are not defined when the argument is of the form $ 2mK + (2n + 1)iK'$ where $ m, n$ are integers. We will later see that these are the only values where the elliptic functions are not defined.

We can now start with the formula $$\text{sn}(u_{1} + u_{2}, k) = \frac{s_{1}^{2} - s_{2}^{2}}{s_{1}c_{2}d_{2} - s_{2}c_{1}d_{1}}$$ and replace $ u_{1}$ by $ u_{1} + iK'$ to get $$\text{sn}(u_{1} + iK' + u_{2}, k) = \dfrac{\dfrac{1}{k^{2}s_{1}^{2}} - s_{2}^{2}}{\dfrac{c_{2}d_{2}}{ks_{1}} + \dfrac{s_{2}c_{1}d_{1}}{ks_{1}^{2}}}$$ and since $$\text{sn}(u_{1} + u_{2}, k) = \frac{1}{k\,\text{sn}(u_{1} + iK' + u_{2}, k)}$$ we obtain the addition formulas in their traditional form (which is similar to those of circular functions) \begin{align} \text{sn}(u_{1} + u_{2}, k) &= \frac{s_{1}c_{2}d_{2} + s_{2}c_{1}d_{1}}{1 - k^{2}s_{1}^{2}s_{2}^{2}}\tag{7}\\ \text{cn}(u_{1} + u_{2}, k) &= \frac{c_{1}c_{2} - s_{1}s_{2}d_{1}d_{2}}{1 - k^{2}s_{1}^{2}s_{2}^{2}}\tag{8}\\ \text{dn}(u_{1} + u_{2}, k) &= \frac{d_{1}d_{2} - k^{2}s_{1}s_{2}c_{1}c_{2}}{1 - k^{2}s_{1}^{2}s_{2}^{2}}\tag{9} \end{align} The equivalence of these forms with the previous forms of addition formula can also be seen algebraically. By replacing $ u_{1}$ by $ u_{1} + K$ in the above formulas we can obtain the following new form of the addition formulas: \begin{align} \text{sn}(u_{1} + u_{2}, k) &= \frac{s_{1}c_{2}d_{1} + s_{2}c_{1}d_{2}}{d_{1}d_{2} + k^{2}s_{1}s_{2}c_{1}c_{2}}\tag{10}\\ \text{cn}(u_{1} + u_{2}, k) &= \frac{c_{1}c_{2}d_{1}d_{2} - k^{\prime 2}s_{1}s_{2}}{d_{1}d_{2} + k^{2}s_{1}s_{2}c_{1}c_{2}}\tag{11}\\ \text{dn}(u_{1} + u_{2}, k) &= \frac{1 - k^{2}s_{1}^{2} - k^{2}s_{2}^{2} + k^{2}s_{1}^{2}s_{2}^{2}}{d_{1}d_{2} + k^{2}s_{1}s_{2}c_{1}c_{2}}\tag{12} \end{align} Again the algebraic equivalence of these formulas with the previous ones can be verified. We shall now use the standard addition formulas ($(7), (8), (9)$ above) to define the elliptic functions of a complex variable as follows: $$\text{sn}(z, k) = \text{sn}(x + iy, k) = \frac{\text{sn}(x, k)\,\text{cn}(iy, k)\,\text{dn}(iy, k) + \text{sn}(iy, k)\,\text{cn}(x, k)\,\text{dn}(x, k)}{1 - k^{2}\,\text{sn}^{2}(x, k)\,\text{sn}^{2}(iy, k)}$$ or \begin{align} \text{sn}(z, k) &= \text{sn}(x + iy, k)\notag\\ &= \frac{\text{sn}(x, k)\,\text{dn}(y, k') + i\,\text{sn}(y, k')\,\text{cn}(y, k')\,\text{cn}(x, k)\,\text{dn}(x, k)}{\text{cn}^{2}(y, k') + k^{2}\,\text{sn}^{2}(x, k)\,\text{sn}^{2}(y, k')}\tag{13}\\ \text{cn}(z, k) &= \text{cn}(x + iy, k)\notag\\ &= \frac{\text{cn}(x, k)\,\text{cn}(y, k') -i\,\text{sn}(x, k)\,\text{dn}(x, k)\,\text{sn}(y, k')\,\text{dn}(y, k')}{\text{cn}^{2}(y, k') + k^{2}\,\text{sn}^{2}(x, k)\,\text{sn}^{2}(y, k')}\tag{14}\\ \text{dn}(z, k) &= \text{dn}(x + iy, k)\notag\\ &= \frac{\text{dn}(x, k)\,\text{cn}(y, k')\,\text{dn}(y, k^{\prime}) -i\,k^{2}\,\text{sn}(x, k)\,\text{cn}(x, k)\,\text{sn}(y, k')}{\text{cn}^{2}(y, k') + k^{2}\,\text{sn}^{2}(x, k)\,\text{sn}^{2}(y, k')}\tag{15} \end{align} The above formulas will be taken as standard definitions when both of $ x, y$ are non-zero and we can prove (same as before) that these formulas make elliptic functions analytic in $ z$ for all values of $ z$ for which the formulas are defined. Also the same rules of differentiation hold and therefore the addition formulas are extended to the complex arguments. Note that the above definitions fail only when $ z = 2mK + i(2n + 1)K'$ where $ m, n$ are integers and at these exceptional points the elliptic functions are not defined.

The periodic nature of these functions for complex values is quite obvious from the above definitions. Thus $ \text{sn}(z, k)$ has periods $ 4K$ and $ 2iK'$, $ \text{cn}(z, k)$ has periods $ 4K$ and $ 4iK'$ and $ \text{dn}(z, k)$ has periods $ 2K$ and $ 4iK'$. We will focus more on the double periodicity of elliptic functions in the next post.

### Residues of Elliptic Functions

Since we have the rules of differentiation available for the elliptic functions we can expand them in form of Taylor's series around the point $ u = 0$ as follows: \begin{align} \text{sn}(u, k) &= u - \frac{1}{6}(1 + k^{2})u^{3} + O(u^{5})\notag\\ \text{cn}(u, k) &= 1 - \frac{1}{2}u^{2} + O(u^{4})\notag\\ \text{dn}(u, k) &= 1 - \frac{1}{2}k^{2}u^{2} + O(u^{4})\notag \end{align} and therefore \begin{align} \text{sn}(u + iK') &= \frac{1}{k\,\text{sn}(u)} = \frac{1}{ku}\left(1 - \frac{1}{6}(1 + k^{2})u^{2} + O(u^{4})\right)^{-1}\notag\\ \Rightarrow \,\text{sn}(u + iK') &= \frac{1}{ku} + \frac{1 + k^{2}}{6k}u + O(u^{3})\notag \end{align} similarly \begin{align} \text{cn}(u + iK') &= \frac{-i}{k}\frac{\text{dn}(u)}{\text{sn}(u)} = \frac{-i}{ku} + \frac{2k^{2} - 1}{6k}iu + O(u^{3})\notag\\ \text{dn}(u + iK') &= -i\frac{\text{cn}(u)}{\text{sn}(u)} = \frac{-i}{u} + \frac{2 - k^{2}}{6}iu + O(u^{3})\notag \end{align} It thus follows that the residues of $ \text{sn}(u, k),\,\text{cn}(u, k),\,\text{dn}(u, k)$ at the point $ u = iK'$ are $ 1 / k, -i / k, -i$ respectively.Again we have \begin{align} \text{sn}(u + 2K + iK') &= -\text{sn}(u + iK')\notag\\ \text{cn}(u + 2K + iK') &= -\text{cn}(u + iK')\notag\\ \text{dn}(u + 3iK') &= -\text{dn}(u + iK')\notag \end{align} it follows that the residues of $ \text{sn}(u, k)$ and $ \text{cn}(u, k)$ at point $ u = 2K + iK'$ are $ -1 / k$ and $ i / k$ respectively while the residue of $ \text{dn}(u, k)$ at point $ u = 3iK'$ is $ i$.

By double periodicity we can see that these are only residues for the elliptic functions. From the signs of residues it is easily seen that the sum of residues in any period parallelogram is zero.

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