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Math Help - Complex Series

  1. #1
    Member mohammadfawaz's Avatar
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    Complex Series

    Hello!
    Suppose that \{z_n\} is a sequence of complex numbers satisfying Re\{z_n\}\geq0 and that the two series \sum_{n=1}^\infty z_n and \sum_{n=1}^\infty z_n^2 are convergent. Prove that the series \sum_{n=1}^\infty |z_n|^2 is also convergent.

    Well, assuming z_n=x_n+iy_n, we can deduce from the given that the following series converge:
    \sum_{n=1}^\infty x_n, \sum_{n=1}^\infty |x_n|,
    \sum_{n=1}^\infty y_n,
    \sum_{n=1}^\infty x_n^2-y_n^2,
    \sum_{n=1}^\infty 2x_n y_n.

    The sequence \sum_{n=1}^\infty |z_n|^2 can be written as:
    \sum_{n=1}^\infty x_n^2+y_n^2 or \sum_{n=1}^\infty |x_n^2-y_n^2+2ix_ny_n|

    I am stuck here for now. Any ideas?

    Thanx

    Mohammad
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by mohammadfawaz View Post
    Suppose that \{z_n\} is a sequence of complex numbers satisfying Re\{z_n\}\geq0 and that the two series \sum_{n=1}^\infty z_n and \sum_{n=1}^\infty z_n^2 are convergent. Prove that the series \sum_{n=1}^\infty |z_n|^2 is also convergent.
    If the series of positive terms \sum x_n converges, then so does \sum x_n^2 (comparison test, because x_n<1 for all sufficiently large n). If you also know that \sum(x_n^2-y_n^2) converges, then (by subtraction) so does \sum y_n^2. Hence so does \sum(x_n^2+y_n^2).
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