Solution: Consider the sequence $(z_n)$ defined as follows: \[ z_1 = \frac{1}{2}, \quad z_{n+1} = \sqrt{z_n}, \quad n \geq 1, \] where we are consider the real square root. We claim that $(z_n)$ is a converging sequence with limit equals to $1$. It is clear that $(z_n)$ is a increasing sequence with an upper bound $1$. Thus, by the monotone convergence theorem, $(z_n)$ must converge. Let the limit be $l$. By uniqueness of limit, we have \[ \lim_{n \to \infty} z_{n+1} = \lim_{n \to \infty} \sqrt{z_n} = \sqrt{\lim_{n \to \infty} z_n} \implies l = \sqrt{l}\implies l = 0 \text{ or } 1. \] Note that $l\neq 0$ as each term of the sequence is bigger than $\frac{1}{2}$. Thus $\displaystyle \lim_{n \to \infty} z_n = 1$.

Now as per the definition of $f$, we have \begin{align*} f(z_2) & = f(\sqrt{z_1} ) = f(z_1),\\ f(z_3) & = f(\sqrt{z_2} ) = f(z_2) = f(z_1), \\ f(z_n) & = f(\sqrt{z_{n-1} } ) = f(z_{n-1}) = \dots = f(z_1). \end{align*} Since $f$ is an analytic function, so it must be continuous. Therefore, \[ \lim_{n \to \infty} f(z_n) = f\left( \lim_{n \to \infty} z_n \right) = f(1) = 1. \] We now define a function \[ h(z) = f(z) - 1. \] Since $f$ is analytic, so is $h$. We know that zeros of a non-constant analytic function are isolated. But we have \[ h(z_n) = 0,\quad z_n \implies 1. \] This implies, $h$ must be a constant function. Since $h(1) = 0$, \[ h \equiv 0 \implies f(z) - 1 = 0,\ \forall\ z \implies f\equiv 1. \] Therefore, \[ \textcolor{teal}{\boxed{ f(2\pi \iota ) = 1. }} \]