Show that $\log(z)=\sum_{n=1}^\infty \frac{(-1)^{n-1}(z-1)^{n}}{n}$ can be analytically continued along the curve $\gamma(t)=e^{it}$. Compute $f_t$ and their radii of convergence. What is $f_{2\pi}$?
My attempt: I'm looking for a sequence of functions $f_t$ and radii $r_t$ s.t $f_t$ is analytic in $\mathbb{D}_{r_t}(\gamma(t)), t\in[0,2\pi]$ and $f_t=f_s$ on their intersection.
For every $t$, I first expand $\frac{1}{z}$ around $e^{it}$. The power series is:$$\frac{1}{z}=\frac{1}{z-e^{it}+e^{it}}=\frac{1}{e^{it}}\frac{1}{\frac{z-e^{it}}{e^{it}}+1}\stackrel{|z-e^{it}|<1}{=}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(e^{it})^{n+1}}(z-e^{it})^{n}$$ So by integration by parts, I define:$$f_t(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(e^{it})^{n+1}}\frac{(z-e^{it})^{n+1}}{n+1}$$ The radius of convergence for each $f_t$ is $1$ by the Hadamard theorem. They're also analytic. Only thing is I'm not sure how to show that if $\mathbb{D}_{r_t}(\gamma(t))\cap\mathbb{D}_{r_s}(\gamma(s))\neq\emptyset$ then $f_t=f_s$ on the intersection.
Any help would be appreciated.