TITLE: If $z_{n+1}=\frac{27}{\overline{z_{n}}}+6$ and $z_1 = 3 + 6i$, then find $z_{n}$
QUESTION [5 upvotes]: 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

REPLY [4 votes]: 1. First terms of the sequence:
$$
z_1 = 3+6i;
$$
$$
z_2 = \dfrac{39}{5}+\dfrac{18}{5}i = {\large 9}-\dfrac{6}{5}+\dfrac{18}{5}i;
$$
$$
z_3 = \dfrac{363}{41}+\dfrac{54}{41}i = {\large 9} - \dfrac{6}{41}+\dfrac{54}{41}i;
$$
$$
z_4 = \dfrac{3279}{365}+\dfrac{162}{365}i = {\large 9} - \dfrac{6}{365}+\dfrac{162}{365}i;
$$
$$
z_5 = \dfrac{29523}{3281}+\dfrac{486}{3281}i = {\large 9} - \dfrac{6}{3281}+\dfrac{486}{3281}i;
$$
$$
...
$$
2. 
Now we can fill the pattern:
$$
z_n = {\large 9} - \dfrac{12}{9^{n-1}+1} + \dfrac{4\cdot 3^n}{9^{n-1}+1} i.\tag{1}
$$
3.
It remains to apply math. induction to prove $(1)$.
Denote 
$a_n = \mathbf{Re} z_n, ~~~ b_n = \mathbf{Im} z_n$.
Show that
$$
a_n = 9-\dfrac{12}{9^{n-1}+1} =  \dfrac{9^n-3}{9^{n-1}+1}; \tag{2}
$$
$$
b_n = \dfrac{4\cdot 3^n}{9^{n-1}+1}. \tag{3}
$$
If it holds for some $n$, then
$$
z_{n+1} = \dfrac{27 (a_n+ib_n)}{a_n^2+b_n^2}+6;
$$
$$
a_{n+1} = \mathbf{Re} z_{n+1} = \dfrac{27 a_n}{a_n^2+b_n^2}+6;
$$
$$
b_{n+1} = \mathbf{Im} z_{n+1} = \dfrac{27 b_n}{a_n^2+b_n^2};
$$
$$
a_n^2+b_n^2 = \dfrac{(9^n-3)^2+ 16\cdot 9^n}{(9^{n-1}+1)^2} = \dfrac{9^{2n}+10\cdot 9^n+9}{(9^{n-1}+1)^2} = \dfrac{(9^n+1)(9^n+9)}{(9^{n-1}+1)^2} = \dfrac{9(9^n+1)}{9^{n-1}+1};
$$
$$
a_{n+1} = \dfrac{27 a_n}{a_n^2+b_n^2}+6 = \dfrac{27(9^n-3)}{9^{n-1}+1} \cdot \dfrac{9^{n-1}+1}{9(9^n+1)} +6 =
\dfrac{3(9^n-3)+6(9^n+1)}{9^n+1} = \dfrac{9^{n+1}-3}{9^n+1};
$$
$$
b_{n+1} = \dfrac{27 b_n}{a_n^2+b_n^2} = 27 \cdot \dfrac{4\cdot 3^n}{9^{n-1}+1} \cdot \dfrac{9^{n-1}+1}{9(9^n+1)} = \dfrac{4\cdot 3^{n+1}}{9^n+1}.
$$
That's all.