Doubly Periodic Complex Functions

# Doubly Periodic Complex Functions

 Definition: A function complex function $f$ is said to be Doubly Periodic if $f(z + \omega) = f(z)$ for all $\omega \in \Omega$ where $\Omega = \{ m\tau_1 + n\tau_2 : m, n \in \mathbb{Z} \}$ and $\tau_1, \tau_2 \in \mathbb{C}$ are real linearly independent.
 Lemma 1: Let $\tau_1, \tau_2 \in \mathbb{C}$ be real linearly independent and let $\Omega = \{ m \tau_1 + n \tau_2 : m, n \in \mathbb{Z} \}$. Then $\displaystyle{\sum_{\omega \in \Omega, \omega \neq 0} \frac{1}{\omega^{\alpha}}}$ converges if and only if $\alpha > 2$.
• Proof: Consider the origin of the complex plane. Since $\tau_1$ and $\tau_2$ are real linearly independent complex numbers, we see that $\Omega$ is a lattice of complex numbers. There are $8$ points of $\Omega$ that are closest to the origin, call them $\omega_1, \omega_2, ..., \omega_8$. Let:
(1)
\begin{align} \quad R = \max_{1 \leq k \leq 8} |\omega_k| \quad \mathrm{and} \quad r = \min_{1 \leq k \leq 8} |\omega_k| \end{align}
• Observe that $0 < r \leq R$. Moreover, $r \leq |\omega| \leq R$ for all lattice points $\omega \in \{ \omega_1, ..., \omega_8 \}$.
• Now consider the next $16$ points closest to the origin. Again, by a similar argument we have that $2r \leq |\omega| \leq 2R$ for all lattice points $\omega$ in this "layer".
• We consider on in this process. At "layer $n$", we consider the next $8n$ closest lattice points to the origin and $nr \leq |\omega| \leq nR$ for all lattice points $\omega$ in the $n^{\mathrm{th}}$ "layer". So for lattice points $\omega$ in this layer we see that:
(2)
\begin{align} \quad \frac{1}{(nR)^{\alpha}} \leq \frac{1}{|\omega|^{\alpha}} \leq \frac{1}{(nr)^{\alpha}} \end{align}
• For each $N \in \mathbb{N}$ let $S(N)$ be the sum of all $\frac{1}{|\omega|^{\alpha}}$ for $\omega$ in layers $1$, $2$, …, $N$. Then for each $N$ we see that:
(3)
\begin{align} \quad \frac{8}{R^{\alpha}} + \frac{2 \cdot 8}{(2R)^{\alpha}} + ... + \frac{N \cdot 8}{(NR)^{\alpha}} \leq S(N) \leq \frac{8}{r^{\alpha}} + \frac{2 \cdot 8}{(2r)^{\alpha}} + ... + \frac{n \cdot 8}{(nr)^{\alpha}} \end{align}
• Thus, for all $N$:
(4)
\begin{align} \quad \frac{8}{R^{\alpha}} \sum_{n=1}^{N} \frac{1}{n^{\alpha - 1}} \leq S(N) \leq \frac{8}{r^{\alpha}} \sum_{n=1}^{N} \frac{1}{n^{\alpha - 1}} \end{align}
• Taking the limit as $N \to \infty$ and applying the comparison and $p$-tests gives us that $\displaystyle{\lim_{N \to \infty} S(N) = \sum_{\omega \in \Omega, \omega \neq 0} \frac{1}{\omega^{\alpha}}}$ converges if and only if $\alpha > 2$. $\blacksquare$
 Lemma 2: Let $\tau_1, \tau_2 \in \mathbb{C}$ be real linearly independent and let $\Omega = \{ m \tau_1 + n \tau_2 : m, n \in \mathbb{Z} \}$. If $\alpha > 2$ and $R > 0$ then $\displaystyle{\sum_{\omega \in \Omega, |\omega| > R} \frac{1}{(z - \omega)^{\alpha}}}$ converges absolutely and uniformly for $|z| \leq R$.
• Proof: Let $R > 0$. Then there exists an $\omega^* \in \Omega$ closest to the origin and such that $|\omega^*| > R$. Let $d > 0$ be such that $|\omega^*| = R + d$. Then for all $\omega \in \Omega$ with $|\omega| > R$ we have that $|\omega| \geq R + d$.
• Let $z$] be such that $|z| \leq R$. Then:
(5)
\begin{align} \quad \left | \frac{z - \omega}{\omega} \right | = \left | 1 - \frac{z}{\omega} \right | \geq 1 - \frac{|z|}{|\omega|} \geq 1 - \frac{R}{R + d} \end{align}
• Let $\alpha > 2$. Then from above we see that:
(6)
\begin{align} \quad \left | \frac{z - \omega}{\omega} \right |^{\alpha} \geq \left ( 1 - \frac{R}{R + d} \right )^{\alpha} \end{align}
• Let $M > 0$ be such that:
(7)
\begin{align} \quad \frac{1}{M} = \left ( 1 - \frac{R}{R + d} \right )^{\alpha} \end{align}
• Then we see that from above that:
(8)
• By Lemma $1$ we know that the series $\displaystyle{\sum_{\omega \in \Omega, |\omega| > R} \frac{1}{(z - \omega)^{\alpha}}}$ converges absolutely for all $|z| \leq R$. Moreover, the equality at $(*)$ and applying the Weierstrass $M$-test shows us that $\displaystyle{\sum_{\omega \in \Omega, |\omega| > R} \frac{1}{(z - \omega)^{\alpha}}}$ converges uniformly for all $|z| \leq R$. $\blacksquare$