Prove/Disprove $f(z)=0$ if and only if $z=0$ for a power series $f$ about the centre $0$ having radius of convergence $2$ Q. Let $f(z)$ be a power-series (with complex coefficients) centred at $0 \in \mathbb{C}$ and with a radius of convergence $2$ . Suppose that $f(0)=0$. Choose the correct statement(s) from below:
(A) $f^{-1}(0)=\{0\}$
(B) If $f$ is a non-constant function on $\{z \in \Bbb C~:~|z|<2\}$, then $f^{-1}(0)=\{0\}$;
(C) If $f$ is a non-constant function, then for all $\zeta \in \Bbb C$ with sufficiently small $|\zeta|$, the equation $f(z)=\zeta$ has a solution;
(D) $\int_\gamma f^{(n)}(z) d z=0$ for every $n \geq 1$, where $\gamma$ is a unit circle centred at $0$ , oriented clockwise, and $f^{(n)}$ is the $n^{\text{th}}$ derivative of $f(z)$.
How can I confirm $f(z)=0$ if and only if $z=0$, in the geometric series $\sum_{n=1}^\infty 2^{-n}z^n$, I got first option is true, but how to generalize. Option D says all the coefficients of the terms is zero, I feel it is impossible.
 A: A and B: These are both false. We can use the same counterexample for both claims. Let $f(z) = \frac{z(z-1)}{z-2}$. Then $f(0) = f(1) = 0$. Also $f$ is holomorphic on the disc $|z| < 2$, but it has a pole at $z=2$, so the power series expansion around $z=0$ has radius of convergence = 2.
EDIT: A comment asked why it follows that the power series for $f$ around $z=0$ has RoC=2. It's because of the fact: "The radius of convergence is always the distance from the center to the nearest non-removable singularity; if there are no singularities (i.e., if $f$ is an entire function), then the radius of convergence is infinite." The proof is included in this Wikipedia article.
C: This is true. It's the Open Mapping Theorem. $f$ is holomorphic on a neighborhood of $z=0$, so the range of $f$ is an open set containing $0$ and therefore the range of $f$ includes all points that are sufficiently close to $0$.
D: This is true. It's Cauchy's theorem. We know $f$ is holomorphic on $|z|<2$, so $f^{(n)}$ is also defined and holomorphic on that disc, so this is exactly the situation covered by that theorem.
