as part of a project that I'm doing on fractals I was watching this video, in which it was stated that the fixed points of the function
$$z\mapsto f(z)=\frac{(7z+2)-e^{i\pi z} (5z+2)}{4}$$
were given by
$$ \lim_{n \rightarrow \infty } (z_n) =2n+\frac{i}{\pi}\ln{\frac{5n+1}{3n+1}}$$
but deriving this was left as an exercise. When I try I get this:
$\frac{\left(7z+2\right)-e^{i\pi z}(5z+2)}{4}-z=0$
$\left(3z+2\right)-e^{i\pi z}(5z+2)=0$
$1=e^{i\pi z}\frac{\left(5z+2\right)}{\left(3z+2\right)}$
$0=\ln{e^{i\pi z}\frac{\left(5z+2\right)}{\left(3z+2\right)}}$
$0=i\pi z+\ln{\frac{\left(5z+2\right)}{\left(3z+2\right)}}$
which seems close, but I don't understand how to get from here to the given result. Any help is appreciated, thanks!
In the logarithm of $$ 1=e^{i\pi z}\frac{5z+2}{3z+2}\implies 2i\pi\,n=i\pi z +\ln\frac{5z+2}{3z+2}. $$ there is a phase correction depending on the real part of $z$. This suggests the decomposition $z=2n+w$ with $Re(w)\in (-1,1]$. For large values of $|z|$ and thus $|n$ the fraction $\frac{5z+2}{3z+2}$ is close to $\frac53$ so that the solutions are close to $$ z_n^0=2n+\frac{i}{\pi}\ln\frac53\approx z_n^1= 2n+\frac{i}{\pi}\ln\frac{5n+1}{3n+1}. $$
Let's extract this form from the exact equation and apply Taylor expansions to the remainder terms. \begin{align} -i\pi w=i\pi(2n-z) &= \ln\frac{5(2n+w)+2}{3(2n+w)+2} \\[1em] &=\ln\frac{5n+1}{3n+1} + \ln\left(1+\frac{w}{2(n+\frac15)}\right) - \ln\left(1+\frac{w}{2(n+\frac13)}\right) \\[1em] &=\ln\frac{5n+1}{3n+1} + \left(\frac{w}{2(n+\frac15)}\right) - \left(\frac{w}{2(n+\frac13)}\right) - \frac12 \left(\frac{w}{2(n+\frac15)}\right)^2 + \frac12 \left(\frac{w}{2(n+\frac13)}\right)^2+\dots \\[1em] &=\ln\frac{5n+1}{3n+1} + \frac{w}{15(n+\frac15)(n+\frac13)}+O(n^{-3}) \end{align}
One could solve this for $w$ to get an expression with a better asymptotic approximation order, but already in this form it is visible that $z_n^1$ is correct with asymptotic error $O(n^{-2})$, as $|w|\le 1$.
But in fact this approximation is also good for small $n$, as the table of numerical results shows \begin{array}{l|l|l|} n& z_n^1 & z_n^* & |z_n^1-z_n^*|\cdot n^2 \\ \hline 0 & 0j & 0j & 0.0\\ 1 & (2+0.129063552413j) & (1.99828953966+0.129128337607j) & 0.00171168680229\\ 2 & (4+0.143871333295j) & (3.99940562717+0.143887760063j) & 0.00237839910788\\ 3 & (6+0.149606801731j) & (5.99970248341+0.149613026078j) & 0.00267823526895\\ 4 & (8+0.152652852595j) & (7.9998220574+0.152655829237j) & 0.00284747997595\\ 5 & (10+0.154541937583j) & (9.999881771+0.154543582053j) & 0.00295601083948\\ 6 & (12+0.155828039883j) & (11.9999157981+0.1558290414j) & 0.0030314830087\\ 7 & (14+0.15676013392j) & (13.9999370036+0.156760788158j) & 0.00308698922499\\ 8 & (16+0.157466704434j) & (15.9999511033+0.157467154984j) & 0.00312952065372\\ 9 & (18+0.15802076878j) & (17.9999609501+0.158021092099j) & 0.00316314778196\\ 10 & (20+0.158466893431j) & (19.9999680969+0.158467133232j) & 0.00319040052269\\ 11 & (22+0.158833821294j) & (21.9999734475+0.158834004027j) & 0.00321293371277\\ 12 & (24+0.159140922028j) & (23.9999775569+0.15914106445j) & 0.00323187520713\\ 13 & (26+0.159401724899j) & (25.9999807813+0.159401838043j) & 0.00324802011499\\ 14 & (28+0.159625965758j) & (27.9999833577+0.159626057128j) & 0.00326194504545\\ 15 & (30+0.159820829484j) & (29.9999854487+0.159820904326j) & 0.00327407832235\\ 20 & (40+0.160506694619j) & (39.9999917075+0.160506727028j) & 0.00331702567588\\ 30 & (60+0.161198597698j) & (59.9999962659+0.161198607556j) & 0.00336073346512\\ 40 & (80+0.161546838952j) & (79.9999978857+0.161546843166j) & 0.00338287769512\\ 50 & (100+0.16175652236j) & (99.9999986415+0.161756524535j) & 0.00339625833067\\ 60 & (120+0.161896620462j) & (119.999999054+0.161896621727j) & 0.00340521827801\\ 70 & (140+0.1619968424j) & (139.999999304+0.1619968432j) & 0.00341163768546\\ \end{array} which confirms the that the error is inverse quadratic in $n$ as the estimate $z_n^*-z_n^1\approx \dfrac{\ln\frac{5n+1}{3n+1}}{\pi^2(3n+1)(5n+1)}$ predicts.