The function defined by $$f(z) := \sum_{n=0}^\infty z^n = 1+z+z^2+\ldots$$ is, as a sum of power series, analytic in the disc of convergence $D = \{z\in\mathbb{C}\,\vdots\;\; |z| < 1\}$ The function $$g(z) := \frac{1}{1-i}\sum_{n=0}^\infty\left(\frac{z-i}{1-i}\right)^n = \frac{1}{1-i}+\frac{z-i}{(1-i)^2}+\frac{(z-i)^2}{(1-i)^3}+\ldots$$ similarly is analytic in the bigger disc $E = \{z\in\mathbb{C}\,\vdots\;\; |z-i| < \sqrt{2}\}$ But we have $$f(z) = \frac{1}{1-z}\quad\mathrm{in}\; D,$$ $$g(z) = \frac{1}{1-i}\cdot\frac{1}{1-\frac{z-i}{1-i}} = \frac{1}{1-z} \quad\mathrm{in}\; E;$$ thus $f(z)$ and $g(z)$ coincide in the intersection domain $D \cap E$ So we can say that $g(z)$ is the analytic continuation of $f(z)$ to the domain $E\!\smallsetminus\!D$ It is clear that $\frac{1}{1-z}$ is the analytic continuation of $f(z)$ to the domain $\mathbb{C}\!\smallsetminus\!\{1\}$