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Following Can a real power series have an uncountable number of real roots? and this essentially equivalent question about a sort of linear independence of powers of a real function, the natural thing to ask is the following.

Let $a_n \in \mathbb C$ be a sequence of complex numbers, not all zero, and consider the power series $f(z) = \sum_{n=0}^\infty a_n z^n$. Can $f$ converge to $0$ on an uncountable subset $S \subseteq \mathbb C$?

By the following argument, all but a countable number would have to lie on the boundary of the disk of convergence $D$ of $f$. Indeed, $D$ is a countable union of compact sets $K_n$, and if $S_n = S \cap K_n$ were infinite then it would have an accumulation point inside $K_n \subset D$, hence by the identity theorem $f = 0$. Therefore $S \cap D = \bigcup_n S_n$ is countable.

Olius
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    It depends on what you mean by zeroes - if its zeroes of the power series (or analytic function) itself the answer is no by the usual theorems, but if its boundary zeroes in the limit sense the answer is yes; for any compact of zero measure on the unit circle (eg $K$ Cantor set) it's easy to construct a power series with radius one that converges to zero at the boundary on $K$ as it's easy to construct an integrable non negative function $u$ on the unit circle that is infinity on $K$ and then if $u$ is its harmonic extension to the disc and $v$ its conjugate harmonic, $e^{-u-iv}$ works – Conrad Nov 24 '23 at 17:06

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