We assume that a=0. The function f(z) for ∣z∣<∣a∣ can be written in the following way: f(z)=a1+a2z+a3z2+...=a11−az1 . 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 an+1∣z∣n→∞, n→∞. We point out that for z=aeit we receive:
f(z)=a1(1+eit+e2it+...)=limn→∞a11−eit1−eitn. The latter limit does not exist,
since eint=cos(nt)+isin(nt) and sin(nt) is not convergent for t=0.
We point out that the function can be easily extended outside of the region ∣z∣<∣a∣. We simply put f(z)=a11−az1 and it will be analytic everywhere, except of the point z=a.
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