What is $f(z)$?

Question

Let $0<a<b<c<d$ and $0<x<y$. Given a meromorphic function $p(z)=\frac{f(z)}{g(z)}$, where $f(z)$ and $g(z)$ are polynomials, such that $p(z)$ has:

1. poles only at $0,\pm ix,\pm iy$ (and $\bar{ \infty}$)
2. roots only at $\pm b, \pm d$
3. and values of $1$ only at $\pm a,\pm c$.

From 1. I get $g(z)=z(z^2+x^2)(z^2+y^2)$. From 2. I get that $f(z)$ is polynomial of order $4$ and $f(\pm b)=f(\pm d)=0$. From 3. I get $f(\pm a)=g(\pm a)$ and $f(\pm c)=g(\pm c)$. Further to exclude other roots and values of $1$, let's force $f'(\pm a)=f'(\pm b)=f'(\pm c)=0$.

So I end up with $11$ variables in $14$ equation. Is there a chance to find an answer to the question:

What is $f(z)$?

... and $g(z)$...

The Orginal values, Jacky Chong mentions are: $$ a=1, b=2, c=3, d=4, x=1, y=2 $$

Answer 1

This is not a solution but too long for a comment.

Based on the original information, we can write \begin{align} p(z) = C\frac{(z-2)^{n_1}(z+2)^{n_2}(z-4)^{n_3}(z+4)^{n_4}}{z^{d_1}(z-i)^{d_2}(z+i)^{d_3}(z-2i)^{d_4}(z+2i)^{d_5}} \end{align} with $n_1+n_2+n_3+n_4> d_1+d_2 +d_3+d_4+d_5$ since $\lim_{z\rightarrow \infty}p(z) = \infty$.

Moreover, since \begin{align} 1=p(1) = C\frac{(-1)^{n_1}3^{n_2}(-3)^{n_3}5^{n_4}}{(1-i)^{d_2}(1+i)^{d_3}(1-2i)^{d_4}(1+2i)^{d_5}} \end{align} then it follows if $d_2 = d_3$ and $d_4=d_5$, we then have \begin{align} 1 = C \frac{(-1)^{n_1+n_3} 3^{n_2+n_3}5^{n_4}}{2^{d_2}5^{d_4}} \ \implies \ \ C= \frac{(-1)^{n_1+n_3}2^{d_2}5^{d_4}}{3^{n_2+n_3}5^{n_4}}. \end{align} On the other hand, we have \begin{align} 1 = p(-1) = \frac{(-1)^{n_1+n_3}2^{d_2}5^{d_4}}{3^{n_2+n_3}5^{n_4}} \frac{(-3)^{n_1}(-5)^{n_3}3^{n_4}}{(-1)^{d_1}2^{d_2}5^{d_4}} = (-1)^{d_1}3^{n_1+n_4-n_2-n_3} 5^{n_3-n_4}=1 \end{align} which means $d_1$ is even, $n_1=n_2$ and $n_3=n_4$.

This means \begin{align} p(z) = C \frac{(z^2-4)^{m_1}(z^2-16)^{m_2}}{z^{2k_1}(z^2+1)^{k_2}(z^2+4)^{k_3}}. \end{align} Next, since \begin{align} 1=p(\pm 3) = C \frac{5^{m_1}(-7)^{m_2}}{3^{2k_1}10^{k_2}13^{k_3}}=p(\pm 1)=C\frac{(-3)^{m_1}(-15)^{m_2}}{2^{k_2}5^{k_3}} \implies p(z) = \text{const}. \end{align} Thus, it seems when we assume $d_2=d_3$ and $d_4=d_5$ we get that $p(z)$ must be constant.