Answer to Question #145913 in Complex Analysis for Apoorv Dham

Question #145913
show that the function f(z)=1/a+z/a^2+.....can be continued analytically
1
Expert's answer
2020-11-23T19:22:41-0500

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".



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