The following is an irksome problem that my complex analysis class is having trouble solving:
Let $*$ be an operator that takes a function $F:\mathcal{H}\to\mathcal{H}$ to a function $F^*:\mathcal{H}\to\mathcal{H}$ defined by $$F^*(z)=z-\frac{1}{F(z)}.$$ Let $G(z)=z$ be the identity on $\mathcal{H}$ and define a sequence of functions $$G,G^*,G^{**},G^{***},...$$ The problem is: How can one show that this sequence converges to a limit function uniformly on compact sets in $\mathcal{H}$?
The problem screams out for one to use Montel's theorem, which says that asserts that every locally bounded family of holomorphic functions is normal. Thus, if we can show that the sequence of function iterates is uniformly bounded on compact sets, the result will fall out.
I have not, however, managed to find a uniform bound. On one hand, I have managed to show that the denominator of $n^{th}$ iterate, $$G^{n}(z)=z-1/G^{n-1}(z)=\frac{zG^{n-1}(z)-z}{zG^{n-1}(z)}:=\frac{A_n(z)}{B_n(z)}$$ has all of its roots of the form $\pm \sqrt{2-\Re(\zeta_n)}$ where $\zeta_n$ is a primitive $n^{th}$ root of unity. The key observation is that these roots are real, so are not in $\mathcal{H}$.
In any case, supposing there is a limit $F$, then it must be a fixed point of $\cdot^*$ so that we can solve the quadratic equation $$F^*(z)=F(z)=z-1/F(z)$$ to for a closed form expression $F(z)=\frac{z-\sqrt{z^2-4}}{2}$ (where it is minus, to make sure the mapping takes $\mathcal{H}\to\mathcal{H}$).
It also might help to see the $n^{th}$ iterate as a function defined by a continued fraction expansion $$G^n(z)=z-\cfrac{1}{z-\cfrac{1}{z-\cfrac{1}{\ddots\cfrac{1}{z-1/z}}}}$$
I'll appreciate any help!