example of analytic continuation

The function defined by

$$f(z):=\sum _{n=0}^{\mathrm{\infty }}{z}^n=1+z+z^2+\mathrm{\dots }$$

is, as a sum of power series, analytic in the disc of convergence .  The function

$$g(z):=\frac{1}{1-i}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{z-i}{1-i}\right)}^n=\frac{1}{1-i}+\frac{z-i}{{(1-i)}^2}+\frac{{(z-i)}^2}{{(1-i)}^3}+\mathrm{\dots }$$

similarly is analytic in the bigger disc .  But we have

$$f(z)=\frac{1}{1-z}\mathit{\hspace{1em}}\mathrm{in}D,$$

$$g(z)=\frac{1}{1-i}\cdot \frac{1}{1-\frac{z-i}{1-i}}=\frac{1}{1-z}\mathit{\hspace{1em}}\mathrm{in}E;$$

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  $E\setminus D$.  It is clear that $\frac{1}{1-z}$ is the analytic continuation of $f(z)$ to the domain $ℂ\setminus \{1\}$.