Question: Suppose $f(z) = \sum_{n=1}^{\infty}a_{n}z^{n}$ is a power series with finite radius of convergence. Show that $f$ cannot be analytically continued along the path $\gamma(t) = te^{i\alpha}$ , $0 \leq t \leq R$ for some angle $\alpha$ - where $R$ is the radius of convergence of $f(z)$. Also, determine the radius of convergence of the power series of $f(z)$ about $te^{i\alpha}$ for $0 \leq t \lt R$.

What I was thinking :

to use the definition mentioned in the book that: $f(z)$ is said to admit an analytic continuation along $\gamma$ if for each $t$ the power series $f_{t} (z) = \sum_{n=0}^{\infty} a_{n}(t) (z-\gamma (t))^{n}$ is convergent such that: $f_{a} (z)$ is the power series representation of $f(z)$ at $z_{0}$ & such that when $s$ is near $t$ ; then $f_{s}(z) = f_{t}(z)$ in the intersection of the discs of convergence.where $\gamma(t)$; $a \leq t \leq b$ is the curve with $\gamma(a) = z_{0}$ .

Now, if possible, assuming contrary; if $f(z)$ has analytic continuation along the given path then $\sum_{n=0}^{\infty} a_{n}(t) (z-te^{i\alpha})^{n}$ is convergent for all $\alpha$ . Then $f_{R}(z) = \sum_{n=0}^{\infty} a_{n}(R) (z-Re^{i\alpha}))^{n}$ denotes the valoue of the function in the disc centred at $R$. Now, I want to derive some contradiction from it!! Ami I on the right track?? If yes, then what to do next??


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