If $z_{n+1}=\frac{27}{\overline{z_{n}}}+6$ and $z_1 = 3 + 6i$, then find $z_{n}$ 
Let the complex sequence $\{z_{n}\}$ satisfy $z_{1}=3+6i$, and 
  $$z_{n+1}=\dfrac{27}{\overline{z_{n}}}+6.$$
  Find the $z_{n}$.

My idea: since
$$z_{n+2}=\dfrac{27}{\overline{z_{n+1}}}+6=\dfrac{27}{\dfrac{27}{z_{n}}+6}+6?$$
So I can't. Thank you 
 A: Hint.
Clearly
$$
z_{n+2}=\dfrac{27}{\overline{z_{n+1}}}+6=\dfrac{27}{\dfrac{27}{z_{n}}+6}+6=
\frac{27z_n}{6z_n+27}+6=\frac{63z_n+162}{6z_n+27}.
$$
Setting $x=z_n=z_{n+2}$ and solving the resulting equation we obtain that $x=9$ is a solution (a fixed point). In particular, subtracting the fixed point we get
$$
z_{n+2}-9=\frac{3(z_n-9)}{2(z_n-9)+27}
$$
or
$$
\frac{1}{z_{n+2}-9}=\frac{2(z_n-9)+27}{3(z_n-9)}=\frac{2+\frac{27}{z_n-9}}{3}.
$$
Setting
$$
w_n=\frac{3}{z_n-9},
$$
you obtain that
$$
w_{n+2}=9w_n+2,
$$
and thus
$$
\frac{w_{n+2}}{3^{n+2}}=\frac{w_n}{3^n}+\frac{2}{3^{n+2}}.
$$
A: If we can solve
$$
w_{n+1}=\frac{27}{w_n}+6\tag{1}
$$
then
$$
z_n=\left\{\begin{array}{l}
w_n&\text{if $n$ is odd}\\
\overline{w}_n&\text{if $n$ is even}\\
\end{array}\right.\tag{2}
$$
$(1)$ is equivalent to
$$
w_{n+1}w_n-6w_n-27=0\tag{3}
$$
If we shift $(3)$ with $u_n=w_n+a$ to get rid of the constant, we can get an equation that we can divide by $u_{n+1}u_n$ to get a linear equation in $\frac1{u_n}$:
$$
\begin{align}
(u_{n+1}-a)(u_n-a)-6(u_n-a)-27&=0\\
u_{n+1}u_n-au_{n+1}-(a+6)u_n+a^2+6a-27&=0\tag{4}
\end{align}
$$
Letting $a=3$, we get $a^2+6a-27=0$. Then $(4)$ becomes
$$
\begin{align}
u_{n+1}u_n-3u_{n+1}-9u_n=0
&\implies\frac1{u_{n+1}}=\frac19-\frac13\frac1{u_n}\\
&\implies\frac1{u_{n+1}}-\frac1{12}=-\frac13\left(\frac1{u_n}-\frac1{12}\right)\\
&\implies\frac1{w_{n+1}+3}-\frac1{12}=-\frac13\left(\frac1{w_n+3}-\frac1{12}\right)\tag{5}
\end{align}
$$
Therefore,
$$
\frac1{w_n+3}-\frac1{12}=\left(-\frac13\right)^{n-1}\left(\frac1{w_1+3}-\frac1{12}\right)\tag{6}
$$
Using $(2)$ and $(6)$, we get $z_n$.

$w_1=3+6i$. Plugging this into $(6)$ gives
$$
w_n=\frac{12\left(1+i\left(-\frac13\right)^{n-1}\right)}{1+\left(\frac19\right)^{n-1}}-3\tag{7}
$$
Then $(2)$ gives
$$
z_n=\frac{12\left(1+i\left(\frac13\right)^{n-1}\right)}{1+\left(\frac19\right)^{n-1}}-3\tag{8}
$$
