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 a0a\neq0. The function f(z)f(z) for z<a|z|<|a| can be written in the following way: f(z)=1a+za2+z2a3+...=1a11zaf(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)f(z) is analytic (see https://encyclopediaofmath.org/wiki/Analytic_function). For z>a|z|>|a| the series is divergent, because znan+1\frac{|z|^n}{a^{n+1}}\rightarrow\infty, nn\rightarrow\infty. We point out that for z=aeitz=ae^{it} we receive:

f(z)=1a(1+eit+e2it+...)=limn1a1eitn1eit.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 eint=cos(nt)+isin(nt)e^{int}=cos\,(nt)+i\,sin\,(nt) and sin(nt)sin(nt) is not convergent for t0t\neq0.

We point out that the function can be easily extended outside of the region z<a|z|<|a|. We simply put f(z)=1a11zaf(z)=\frac{1}{a}\frac{1}{1-\frac{z}{a}} and it will be analytic everywhere, except of the point z=az=a.



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