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Math Help - [SOLVED] Convergence of a series

  1. #1
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    [SOLVED] Convergence of a series

    I have a question that I can't do...can anybody help me?

    Investigate the convergence of the series (in the complex plane) for difference values of a and b:

    Generalise your solution by determining what constraints are required on the values of a and b, so as to guarantee the convergence of the series?

     S(a,b) = \sum_{n=0}^{\infty}(a+bi)^n

    Thanks if you can help.
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  2. #2
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     a^2 + b^2 < 1 assuminf a and b are both real
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  3. #3
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    Sorry, still trying to understand

    But I thought a + bi is a complex number.
    i.e. b is not real.
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  4. #4
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    Quote Originally Posted by Bucephalus View Post
    But I thought a + bi is a complex number.
    i.e. b is not real.
    in the form a + bi, both a and b are real. that is generally how complex numbers are defined. we get a complex number because there is an "i" attached to the real number b
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  5. #5
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    so b could be a complex number?

    So the assumption was that a and b are real numbers i.e.
    it is not of the form a + (c +di)i
    that's what I now understand of that that assumption is assuming.
    Is this correct?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Bucephalus View Post
    So the assumption was that a and b are real numbers i.e.
    it is not of the form a + (c +di)i
    that's what I now understand of that that assumption is assuming.
    Is this correct?
    correct. a and b are considered to be real unless otherwise stated.

    and it would not be otherwise stated, since even a + (c + di)i can be rewritten in the form a + bi
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  7. #7
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    well why isn't it....

    \sqrt{a^2 + b^2} < 1
    Because I thought that the condition for geometric series convergence is

    |r| < 1 for geomtric series

     S = \sum_{n=0}^{\infty}ar^n

    thanks.
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  8. #8
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    Quote Originally Posted by Bucephalus View Post
    \sqrt{a^2 + b^2} < 1
    Because I thought that the condition for geometric series convergence is

    |r| < 1 for geomtric series

     S = \sum_{n=0}^{\infty}ar^n

    thanks.
    this you should answer !!!
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  9. #9
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    you're right, that was a stupid question.

    Thanks.
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