What are the fixed points of $f^n$ for $f(z)=\frac{z^2-1}{2z}$? Consider the function $f(z)=\frac{z^2-1}{2z}$ on the complex plane. How can we find the fixed points of $f^n$ for fixed $n$?
For real $z$, the problem isn't too hard:

Write $z=\cot(\alpha)$ for some $\alpha\in(0,\pi)$. We can notice that, conveniently, $$f(z)=\frac{\cot(\alpha)^2-1}{2\cot(\alpha)}=\cot(2\alpha).$$ Therefore, if $z$ is a fixed point of $f^n$, $\cot(\alpha)=\cot(2^n\alpha)$, and $(2^n-1)\alpha$ must be a multiple of $\pi$. That is, $z=\cot\left(\frac{k}{2^n-1}\pi\right)$ for some $0< k<2^n-1$.

Maybe a variation on this approach works for complex numbers too, but since I can't find any info on the complex $\tan$ function, I'm kind of lost on that route.
By plugging in the expressions for $f$, $f^2$, $f^3$, $f^4$ on Wolfram|Alpha, it seems as if we only missed the points $i$ and $-i$. But how could we prove this?
 A: Let’s consider a complex number $z$ on the Argand plane. The numbers $z$ and $-\frac{1}{z}$ will create the same angles with the $y$-axis, and will lie both on the same side of the $x$-axis, and on opposite sides of the $y$-axis. In particular, except that $z$ lies on either axis, $\frac{z^2-1}{2z}=\frac{z-\frac{1}{z}}{2}$ will create a smaller angle with the $y$-axis than $z$. This means that all of the possible fixed points of $f^n$ for any $n$ are either pure real or pure imaginary.

The pure real case has already been taken care for. For the pure imaginary case, we can notice that $$\frac{ix-\frac{1}{ix}}{2}=i\cdot\frac{x+\frac{1}{x}}{2},$$ so that it suffices to find the real fixed points of $g^n$ for $g(x)=\frac{x+\frac{1}{x}}{2}$. This is easy, since:


*

*For $x\in(-\infty,-1)$, $g(x)>x$ and $g(x)\in (-\infty,-1)$.

*For $x\in(-1,0)$, $g(x)<x$ and $g(x)\in (-1,0)$.

*For $x\in(0,1)$, $g(x)>x$ and $g(x)\in (0,1)$.

*For $x\in(1,\infty)$, $g(x)<x$ and $g(x)\in (1,\infty)$.


This leaves only $-1$ and $1$ as candidate fixed points – they both work since $g(-1)=-1$, $g(1)=1$. Therefore, the only fixed points we had missed were precisely $-i$ and $i$.
