a general continued fraction satisfying $\frac{(i+\Theta\sqrt{z})^m}{(i-\Theta\sqrt{z})^m}=\frac{(ik+\sqrt{z})^{m+1}}{(ik-\sqrt{z})^{m+1}}$ Given any natural number $m\gt2$, let $z$,$k$ be complex numbers, where $i=\sqrt{-1}$ and consider the general continued fraction 
$$\Theta(k,z,m)=\cfrac{(m+1)}{km+\cfrac{z(0m-1)(2m+1)} {3km+\cfrac{z(m-1)(3m+1)}{5km +\cfrac{z(2m-1)(4m+1)}{7km+\cfrac{z(3m-1)(5m+1)}{9km+\ddots}}}}}$$
The variable $\Theta$ has the property of satisfying the polynomial equation of $m$th degree
$$\frac{(i+\Theta\sqrt{z})^m}{(i-\Theta\sqrt{z})^m}=\frac{(ik+\sqrt{z})^{m+1}}{(ik-\sqrt{z})^{m+1}}\tag1$$
$k\neq\pm i$ , $k\neq 0$ ,$z\neq-k^2$,$z\neq 0$
Both continued fractions in this post are just the special cases $z =\pm 1$. 
It also obeys a very fundamental transformation property
$$\Theta(k,z,m)=\frac{1}{\sqrt{z}}\Theta(\frac{k}{\sqrt{z}},1,m)\tag2$$ 
P.S.:After applying property $(1)$ and $(2)$,we arrive at the elegant solution
$$\Theta(k,z,m)=\frac{1}{\sqrt{z}}\dfrac{-\frac{ik}{\sqrt{z}}\left(\dfrac{\frac{k}{\sqrt{z}}+i}{\frac{k}{\sqrt{z}}-i}\right)^{1/m}+\left(\dfrac{\frac{k}{\sqrt{z}}+i}{\frac{k}{\sqrt{z}}-i}\right)^{1/m}+\frac{ik}{\sqrt{z}}+1}{\frac{k}{\sqrt{z}}\left(\dfrac{\frac{k}{\sqrt{z}}+i}{\frac{k}{\sqrt{z}}-i}\right)^{1/m}+i\left(\dfrac{\frac{k}{\sqrt{z}}+i}{\frac{k}{\sqrt{z}}-i}\right)^{1/m}+\frac{k}{\sqrt{z}}-i}$$

Q: How do we prove rigorously that the continued fraction does satisfy the polynomial equation? 

 A: The case with $z \neq 1$ can be obtained from the case $z=1$, because you can just factor them out of the infinite fraction while changing $k$ accordingly.
Consider $k,m$ fixed, and let $(f_n)$ be the sequence of homographies $f_n(x) = a_n + b_n / x$, with $a_n = (2n-1)km$ and $b_n = ((n-1)m-1)((n+1)m+1)$ so that
$\Theta = \lim_{n \to \infty} (m+1)/ f_0 \circ f_1 \circ f_2 \circ f_3 \dots f_n(0)$" (well I guess this is where you put the $0$ but it shouldn't really matter)
(I think you should be able to define this more purely, if you call $x_n$ the attractive fixpoint of $f_0 \circ \dots \circ f_n$, then $\Theta = \lim (m+1)/x_n$, but leaving a zero is fine).
Now consider a nonzero complex number $z$ and let $g(x) = x/z$. After conjugation by $g$, you get $g^{-1} \circ f_n \circ g (x) = za_n + (z^2b_n)/x$, and so $\lim_{n \to \infty} g^{-1} \circ f_0 \circ f_1 \circ \dots f_n (0)$ is given by the infinite fraction where we replace $a_n$ and $b_n$ with $za_n$ and $z^2 b_n$. This amounts to replacing $k$ with $zk$ and adding a $z^2$ factor in front of every numerator.
Thus, $g^{-1}((m+1)/\Theta(k,1)) = (m+1)/\Theta(kz,z^2)$, so $z\Theta(kz,z^2) = \Theta(k,1)$
If we know the conjecture for $\Theta(k,1)$, we have $\frac{(i+\Theta(k,1))^m}{(i-\Theta(k,1))^m}=\frac{(ik+1)^{m+1}}{(ik-1)^{m+1}}$
And $\frac{(i+z\Theta(kz,z^2))^m}{(i-z\Theta(kz,z^2))^m}=\frac{(ik+1)^{m+1}}{(ik-1)^{m+1}} = \frac{(ikz+z)^{m+1}}{(ikz-z)^{m+1}}$. So letting $w =z^2$ and $k' = kz$ we obtain $\frac{(i+\sqrt w \Theta(k',w))^m}{(i-\sqrt w\Theta(k',w))^m} = \frac{(ik'+\sqrt w)^{m+1}}{(ik'- \sqrt w)^{m+1}}$ (the choice of square root doesn't matter as long as we pick the same on both side)
