If $f(z) = \sum_{n=0}^{\infty} a_n(z-z_0)^n$ has radius convergence $ R > 0$. if $f(z) = 0$ for all $z$ $|z-z_o| < R$ show that $a_0 = a_1 = ... =0$. If $f(z) = \sum_{n=0}^{\infty} a_n(z-z_0)^n$ has a radius of convergence $R > 0$ and if $f(z) = 0$ for all $z$ $|z-z_o| < R$ show that $a_0 = a_1 = ... =0$.
Proof Attempt:
If it has a radius of convergence $R > 0$ then I know $\frac{1}{R} = \lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| > 0$.
Pick $z^*$ to be in the radius of convergence, then we know that :
$$ 0 = a_0 + a_1(z^*-z_0) + ...+a_n(z^*-z_0)^n + ...$$
Which means that for some $a_i \rightarrow 0 \leq i \leq n$ we can write:
$$a_i = (z^* - z_0)^{-i} (-a_0 - a_1(z^* - z_0) - ...-a_{i+1}(z^* - z_0)^{i+1} -....-a_n(z^*-z_0)^n) + ....$$
$$a_{i+1} = (z^* - z_0)^{-(i+1)} (-a_0 - a_1(z^* - z_0) - ...-a_{i+2}(z^* - z_0)^{(i+2} -....-a_n(z^*-z_0)^n). + ...$$
I want to setup now a contradiction, by using the fact that  as $i \rightarrow 0$ then $  | a_{i+1}/a_i | > 0 $:
$$ \lim_{i \rightarrow \infty}| \frac{a_{i+1}}{a_i} |  = (z^* - z_0)^{-1}\frac{(-a_0 - a_1(z^* - z_0) - ...-a_{i+2}(z^* - z_0)^{i+2} -....-a_n(z^*-z_0)^n). + ...)}{(-a_0 - a_1(z^* - z_0) - ...-a_{i+1}(z^* - z_0)^{i+1} -....-a_n(z^*-z_0)^n) + ....)} $$
and I am having troubles concluding anything from this ... can someone suggest maybe another way?
Attempt II
I am attempting a solution based on $f(z) = f'(z) = ... = f^k(z) = 0$
$$(1) \ a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + a_3(z-z_0)^3 + ... = 0 $$
$$(2) \ a_1 + 2a_2(z-z_0) + 3a_3(z-z_0)^2 + ... = 0 $$
$$ (3) \ 2a_2 + 3\times 2a_3(z-z_0) + ... = 0 $$
Take $(2) \times (z-z_0)$:
$$(2) \ a_1(z-z_0) + 2a_2(z-z_0)^2 + 3a_3(z-z_0)^3 + ... = 0 $$
Subtract it from (1)?
$$a_0 - [a_2(z-z)^2 + 2a_3(z-z_0)^3 + ... ] = 0 $$
Not sure where to go with this...

Hint $f^{(n)}(z_0)=n!\cdot a_n$.
Or, since $\lvert z-z_0\rvert \lt R$ has an accumulation point,  you can use the identity theorem.
The power series is unique.
