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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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On any circle inside the domain of convergence it's a Fourier series, hence its coefficients can be obtained by integration. Or equivalently, use the Cauchy integral formula. And by the way, sums of Laurent series are analytic too. You don't need to bother rewriting as Taylor series, and actually you can't do that, since domains of convergence would be disjoint. –  Alexander Shamov Aug 22 '12 at 2:32
    
Oh I can't use the Cauchy integral formula, because that's from chapter 4 of my book, and this is a problem of the chapter 3. The only things that I can use. Is that a power series is $C^{\infty}$, and the principle of analytic continuation :/ –  Daniel Aug 22 '12 at 2:35
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Do you know the identity theorem for holomorphic functions? –  EuYu Aug 22 '12 at 3:40
    
Nop :/,only for series that are locally a power series ( I know that are equivalent, but this will be proved later in my book). –  Daniel Aug 22 '12 at 3:58
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1 Answer

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 :( –  Daniel Aug 28 '12 at 1:24
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