<s>[INST] 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 
 [/INST] 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
$$
\frac{w_{n+2}}{3^{n+2}}=\frac{w_n}{3^n}+\frac{2}{3^{n+2}}.
$$
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