show that holomorpic function f such that f(1/2n) = f(1/2n+1) are constant this is from my homework assignment:
let $f$ be holomorphic on the unit ball such that $f(\frac{1}{2n}) = f(\frac{1}{2n+1})$ for all $n \in \mathbb{N}$
show that f is constant.
 A: Uniqueness Theorem. Let $G$ be a connect domain where $\{x_n\}\subseteq G$ and $x_n\to x_0\in G$. Suppose $f,g$ are analytic and $f(x_n)=g(x_n)$, then $f=g$ on all of $G$. 
Proof. 
We will assume that $f$ is entire and $f(x_n)=0$ where $x_n\to 0$. We wish to show that $f$ is the constant zero function. This special case will establish the full theorem. 
Since $f$ is continuous, $\lim f(x_n) = 0$ and so $f(0)=0$. We now have 
$\displaystyle \left|f'(0)\right|=\lim_{R\to 0}\left|\frac{1}{2\pi i} \cdot \int_{B(0,R)} \frac{f(z)}{z} dz\right|\le \lim_{R\to 0} \frac{1}{2\pi} \cdot \frac{max_{B(0,R)}|f(z)|}{R} \cdot 2\pi R=0$
So, $f'(0)=0$. $f'$ is continuous at zero, so we now have 
$\displaystyle \left|f''(0)\right|=\lim_{R\to 0}\left|\frac{1}{2\pi i} \cdot \int_{B(0,R)} \frac{f'(z)}{z} dz\right|\le \lim_{R\to 0} \frac{1}{2\pi} \cdot \frac{max_{B(0,R)}|f'(z)|}{R} \cdot 2\pi R=0$
So, $f''(0)=0$. In general if $f^{(k)}(0)=0$, then 
$\displaystyle \left|f^{(k+1)}(0)\right|=\lim_{R\to 0}\left|\frac{1}{2\pi i} \cdot \int_{B(0,R)} \frac{f^{(k)}(z)}{z} dz\right|\le \lim_{R\to 0} \frac{1}{2\pi} \cdot \frac{max_{B(0,R)}|f^{(k)}(z)|}{R} \cdot 2\pi R=0$
Thus, $f^{(k)}(0)=0$ for every $k$. Thus, the taylor series coefficients of $f$ are zero and so $f$ is the zero function. 
