How find this $a_{n},b_{n}$ let sequence $\{a_{n}\},\{b_{n}\}$ such 
$$a_{1}=1,b_{1}=3$$
and
$$\begin{cases}
a_{n+1}=2+\dfrac{27a_{n}}{9a^2_{n}+4b^2_{n}}\\
b_{n+1}=\dfrac{27b_{n}}{9a^2_{n}+4b^2_{n}}
\end{cases}$$
Find $a_{n},b_{n}$
My idea: since
$$\dfrac{a_{n+1}-2}{b_{n+1}}=\dfrac{a_{n}}{b_{n}}$$
then I can't.Thank you

As stated in my previous comments, by setting $z_n=3a_n+2i b_n$ we get:
$$ z_{n+1} = 6+\frac{27}{\overline{z_n}}\tag{1} $$
with $z_1 = 3+6i$. By setting $z_n = 3\sqrt{3}\, w_n$ we get:
$$w_{n+1} = \frac{2}{\sqrt{3}}+\frac{1}{\overline{w_n}}\tag{2} $$
with $w_1=\frac{1}{\sqrt{3}}(1+2i)$. $(2)$ gives that the sequence $\{w_n\}_{n\in\mathbb{N}}$ is the orbit of an element in $PSL(2,\mathbb{C})$. Iterating twice we can remove the complex conjugation and get:
$$w_{n+2}=\frac{2}{\sqrt{3}}+\frac{1}{\frac{2}{\sqrt{3}}+\frac{1}{w_n}}.\tag{3}$$
In order to have $a_n$ and $b_n$ we only need to find a closed form for the sequence given by:
$$t_{n+1} = \frac{2}{\sqrt{3}}+\frac{1}{t_n}.\tag{4}$$
Since the roots of the polynomial $x^2-\frac{2}{\sqrt{3}}x-1$ are $-\frac{1}{\sqrt{3}}$ and $\frac{3}{\sqrt{3}}$, with the choice
$$ t_n = \frac{A\left(-\frac{1}{\sqrt{3}}\right)^{n+1}+B\left(\frac{3}{\sqrt{3}}\right)^{n+1}}{A\left(-\frac{1}{\sqrt{3}}\right)^{n}+B\left(\frac{3}{\sqrt{3}}\right)^{n}}$$
equation $(4)$ is fulfilled, and we only need to choice $A$ and $B$ such that $t_1$ and $t_2$ fit with our starting values. 
