Expanding product of binomials $(z^k + z^{-k})$ Suppose $z$ is a complex number, and consider the product
$$f_m(z)=\prod_{k=1}^m \left(z^k + \frac 1 {z^k} \right),$$
for $m = 1,2,\dots$ . Of course, one should be able to expand this into a sum of terms that are either powers of $z$ or of $1/z$. It is easy to see that the highest and lowest order terms will be $z^{1 + 2 + \cdots + m} = z^{\frac 1 2m(m+1)}$ and its reciprocal, respectively.
Here's the situation for low values of $m$:
$$\begin{split}
f_1(z) = z &+ \frac 1 z \\
f_2(z) = z^3 + z &+ \frac 1 z + \frac 1 {z^3} \\
f_3(z) = z^6+z^4+z^2+\ &\color{red}2 + \frac 1 {z^2}+ \frac 1 {z^4} + \frac 1 {z^6} \\
f_4(z) = z^{10} + z^8 + z^6 + \color{red}2z^4 + \color{red}2z^2 +\  &\color{red}2 + \frac {\color{red}{2}} {z^2} + \frac {\color{red}2} {z^4} + \frac 1 {z^6} + \frac 1 {z^8} + \frac{1}{z^{10}} \\
f_5(z)=z^{15} + z^{13}+ z^{11} + \color{red}2z^{9} + \color{red}2z^7 + \color{orange}3z^5 + \color{orange}3z^3 + \color{orange}3z &+ \frac{\color{orange}3} z + \frac {\color{orange}3}{z^3} + \frac {\color{orange}3} {z^5} + \frac {\color{red}2} {z^7} + \frac {\color{red}2} {z^9} + \frac 1 {z^{11}} + \frac 1 {z^{13}} + \frac 1 {z^{15}}
\end{split}$$
It seems that


*

*In each $f_m$ the exponent of $z$ falls from $\tau_m = \frac 1 2 m(m+1)$ down to $-\tau_m$, skipping every other value;

*For all $m$ the coefficient $a_{m,n}$ of the $n$-th term in the analytic part is equal to the coefficient $a_{m,-n}$ of the $n$-th term in the singular part (which makes sense intuitively).



So:
  
  
*
  
*Is there a closed-form expression for $a_{m,n}$?
  
*How can I prove fact 1. above?
  


A proof of 1. could go as follows.

We obtain for  integral $m\geq 1$
  \begin{align*}
\color{blue}{f_m(z)}&=\prod_{k=1}^m\left(z^k+\frac{1}{z^k}\right)\\
&=\prod_{k=1}^m\left(z^{-k}\left(1+z^{2k}\right)\right)\\
&=z^{-\sum_{k=1}^mz^k}\prod_{k=1}^m\left(1+z^{2k}\right)\tag{1}\\
&=z^{-\frac{1}{2}m(m+1)}\sum_{j=0}^{\sum_{k=1}^m k}a_jz^{2j}\tag{2}\\
&\,\,\color{blue}{=z^{-\frac{1}{2}m(m+1)}\sum_{j=0}^{\frac{1}{2}m(m+1)}a_jz^{2j}}\\
\end{align*}
  In the last line we observe the smallest power of the expression is $-\frac{1}{2}m(m+1)$ while the greatest power is
   $-\frac{1}{2}m(m+1)+m(m+1)=\frac{1}{2}m(m+1)$.

Comment:


*

*In (1) we have a finite product where each term is either $1$ or $z^{2k}$. We so obtain a polynomial with smallest power $0$ and greatest power of $z^2$ equal to $\sum_{k=1}^m k$. This polynomial is written more conveniently in (2).

