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Let be a series $ \sum\limits_{ - \infty }^\infty {a_jz^j} $. Convergent on $1<|z|<4$. Such that vanishes on $2<|z|<3$ It's true that all the coefficients $a_j$ are zero?

I know by the principle of analytic continuation, that this would be true, if I know that the series has the form $ \sum\limits_{ 0 }^\infty {b_jz^j} $. Here I don't know how to do. Maybe in the convergence zone, I can write the series on that form, but I don't know how :/

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    Nop :/,only for series that are locally a power series ( I know that are equivalent, but this will be proved later in my book).2012-08-22

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If you multiply your series by $z^{-1 - n}$ and integrate it around the circle $|z| = 2.5$, you'll get $2\pi i a_n$. Thus each $a_n$ is zero. Note that to rigorously justify this, you use that the the integral of the (infinite) sum of the functions is the sum of the integrals, which will hold here since your series converges uniformly on any subannulus $a < |z| < b$ where $1 < a < b < 4$.

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    I can't use integrals :(2012-08-28