Identity theorem for $f(z)=f(z^2)$ Given a holomorphic function $f:\mathbb{C}\backslash\{0\} \rightarrow :\mathbb{C}$, I've got to show 
that if  $f(z)=f(z^2)$ for all $z \in \mathbb{C}\backslash\{0\}$ then $f(z)$ is constant. 
I've got to use the identity theorem here.
I guess that $1$ would be an accumulation point as $f(1)=f(1^2)$. I've defined the series $z_n=1$ with the obvious convergence to $1$. But somehow I fail to show with the identity theorem that $f(0)=f(1)=f(1^2)=f(z)=..$ follows which would imply that $f$ is constant. What am I missing?
 A: Another, different solution, which uses the identity theorem (although, the exponential solution is, of course, very lovely):
Note that $f$ is analytic around $1$, so
$$
\sum_{n=0}^{\infty} a_n z^n=f(1+z)=f((1+z)^2)=f(1+z^2+2z)=\sum_{n=0}^{\infty} a_n (z^2+2z)^n,
$$
implying, by induction, that each $a_n=0$ for $n>0$ since $(z^2+2z)^n=2^n z^n+ O(z^{n+1})$ as $z\to 0$.
Now, you can apply the identity theorem, since $f$ must be equal to $a_0$ on $B(1,1)$.
A: $1$ is not an accumulation point of the (constant) sequence $z_n = 1$, so that does not work. What you need is a sequence $(z_n)$ with a limit point $z^\star$ in the domain such that $z_n \ne z^\star$ and $f(z_n) = f(z^\star)$.
A possible choice is the sequence $z_n = 2^{1/2^n} = 2, \sqrt{2}, \sqrt{\sqrt{2}}, \sqrt{\sqrt{\sqrt{2}}}, \ldots$. Then
$$
 z_{n+1}^2 = z_n \implies f(z_{n+1}) = f(z_n) 
$$
and $z_n \to 1$, so that $f(z_n) = f(1)$ for all $n$. Now you can apply the identity theorem.

Here is another solution, without the identity theorem. It shows that it is sufficient to require that $f$ is continuous:
Fix $a \in \Bbb C \setminus \{0 \}$ and define a sequence $(z_n)$ as
$$
 z_0 = a \, , \, z_{n+1} = \sqrt{z_n}
$$
where the square root is chosen such that $-\pi/2 \le \arg(z_{n+1}) < \pi/2$. Again $f(z_{n+1}) = f(z_n)$ and $z_n \to 1$, so that $f(z_n) = f(1)$ for all $n$. 
It  follows that $f(a) = f(1)$. Since $a$ was arbitrary, $f$ is constant.
A: Let $g(z)=f(\exp(z))$, so g is entire function and $g(z)=g(2z)$. Now it is easy to show that g is bounded since $g(2^kz)=g(z)$. So g is constant function since it is bounded entire function. So f is constant function too.
A: A slick solution for those that know a pinch of Algebra:
Let $f(1)=w$.  Being analytic, if $f$ were non-constant then it would map only finitely many of the points in the (compact) closed annulus $\big\{z: \frac{1}{2}\leq \vert z \vert \leq 2\big\}$ to $w$, so it suffices to show that $f$ takes on the value $w$ infinitely many times on the unit circle.  The unit circle is interesting because it contains many finite groups and the squaring map acts as a homomorphism.
In particular let $\omega_{2^k}$ be the standard generator for the (multiplicative) group $G_{2^k}$ of $2^k$th roots of unity .  $\phi:G_{2^{k+1}}\longrightarrow G_{2^{k}}$ given by $\phi(\omega_{2^{k+1}})=(\omega_{2^{k+1}})^2=\omega_{2^{k}}$ is a surjective group homomorphism. (Surjectivity isn't strictly needed but this is an easy consequence of $\ker \phi =\big\{\pm 1\big\}$ and the Counting Formula.)
Then $f\big(G_2\big) = f\big(\phi(G_2)\big)= f\big(G_1\big)=\big\{w\big\}$ and in general
$f\big(G_{2^{k+1}}\big) = f\big(\phi(G_{2^{k+1}})\big)= f\big(G_{2^{k}}\big)=\big\{w\big\}$
where the RHS follows by induction hypothesis.  Thus
$\big\{w\big\}=f\big(G_{2^k}\big)$
for $k \in \mathbb N\implies$ $f$ is constant in our domain $\mathbb C-\big\{0\big\}$
