1. Radius of convergence of an analytic function

This is from a practice exam:

Let $f(z)=\sum_{k=0}^\infty a_kz^k$ have a radius of convergence 2, and let $g(z)=\sum_{k=0}^\infty b_k(z-3)^k$ have a radius of convergence 2. Let $I=\{z:f(z),g(z)\text{ both converge}\}$.

(a) If $f(z)=g(z)$ for $z\in I$, prove there is an analytic function $w$ for which $w(z)=f(z)$ when $|z|<2$ and $w(z)=g(z)$ when $|z-3|<2$.

(b) For the function $w$ of part (a), determine the radius of convergence $R$ for the power series $w(z)=\sum_{k=0}^\infty c_k\left(z-\frac{3}{2}\right)^k$.

(c) What relationship exists between the coefficient sequence $\{a_k\}$ and the coefficient sequence $\{b_k\}$?
Part (a) is fairly straightforward. Let $A=\{z:|z|<2\}$ and $B=\{z:|z-3|<2\}$. Then $A\cap B\subseteq I$, which means we can define $w$ by $w=f'\cup g'$, where $f'$ and $g'$ are the restrictions of $f$ and $g$ to $A$ and $B$, respectively. It follows immediately that $w$ is analytic on $A\cup B$.

However, I'm lost for parts (b) and (c). Any help would be much appreciated!

2. The series is convergent for the $z=-2+0.00000001$ and $z=5-0.0000000001$ for example, hence we can expect that the radius of convergence is greater than $\frac 72$. Now you have to show it's indeed $\frac 72$.

3. Hmm. I'm sorry to say I don't understand your reasoning here.

By Taylor's theorem $R\geq \sup\{r : D(3/2;r)\subseteq A\cup B\}=\frac{\sqrt{7}}{2}$. (Here I use the notation $D (z_0;r)=\{z:|z-z_0|). However, I don't see how we can get $R\geq\frac{7}{2}$, nor $R\leq\frac{7}{2}$.

4. The series is convergent on set a set which a diameter $\geq 7$, because the series converges on points of the form $-2+\delta$ end $5-\delta$ for $1>\delta>0$

5. Oh, of course... I had forgotten every power series converges precisely on a disc and a subset of its boundary. Sorry about that.