Answer to Question #145913 in Complex Analysis for Apoorv Dham

Question

show that the function f(z)=1/a+z/a^2+.....can be continued analytically

Answer 1

We assume that $a\neq0$. The function $f(z)$ for $|z|<|a|$ can be written in the following way: $f(z)=\frac{1}{a}+\frac{z}{a^2}+\frac{z^2}{a^3}+...=\frac{1}{a}\frac{1}{1-\frac{z}{a}}$ . We used the formula for infinite geometric series (https://www.mathwords.com/i/infinite_geometric_series.htm). The obtained function $f(z)$ is analytic (see https://encyclopediaofmath.org/wiki/Analytic_function). For $|z|>|a|$ the series is divergent, because $\frac{|z|^n}{a^{n+1}}\rightarrow\infty$, $n\rightarrow\infty$. We point out that for $z=ae^{it}$ we receive:

$f(z)=\frac{1}{a}(1+e^{it}+e^{2it}+...)=lim_{n\rightarrow\infty}\frac{1}{a}\frac{1-e^{itn}}{1-e^{it}}.$ The latter limit does not exist,

since $e^{int}=cos\,(nt)+i\,sin\,(nt)$ and $sin(nt)$ is not convergent for $t\neq0$.

We point out that the function can be easily extended outside of the region $|z|<|a|$. We simply put $f(z)=\frac{1}{a}\frac{1}{1-\frac{z}{a}}$ and it will be analytic everywhere, except of the point $z=a$.