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Math Help - [SOLVED] Complex limit points (Take II)

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    [SOLVED] Complex limit points (Take II)

    Here's the next one I could use some help on.

    The principle argument of a complex number z, denoted by arg_P(z), is defined as the value of \theta _P = arg(z) such that -\pi < \theta _P \leq \pi. Give an example of a complex sequence ( z_n ) converging to a limit \alpha such that arg _P (z_n) fails to converge.

    I know of one such example:
    z_n = \frac{1}{n}~e^{in}
    This has a limit point of 0, but no well defined argument.

    However the text says there is at least one other series with a limit point of -1. Not only have I been unable to construct an example (please help), but I can't see the logic behind this one at all. What is so special about -1 that, say, we can't construct one with a limit point of 1?

    -Dan
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    Take -1,1/2,-1/3,1/4,-1/5,...
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    Consider z_n  = \left[ { - 1 + \frac{{\left( i \right)^{2n - 1} }}{n}} \right]
    The terms alternate between quadrants II and III; the limit is 1.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Take -1,1/2,-1/3,1/4,-1/5,...
    Quote Originally Posted by Plato View Post
    Consider z_n  = \left[ { - 1 + \frac{{\left( i \right)^{2n - 1} }}{n}} \right]
    I am having a slight problem with both TPH's and Plato's series. For both of them I am getting that \lim_{n \to \infty} arg_P (z_n) = 0 and that's just not right, since the limit point is -1 giving an argument of \pi. (Or is this the whole key to the problem?)

    Additionally I can use both sequences to construct the same situation, but with any real number as the limit point. The text disagrees with this possibility:

    Here's the whole problem statement:
    The principle argument of a complex number z, denoted by arg_P(z), is defined as the value of \theta _P = arg(z) such that -\pi < \theta _P \leq \pi. Give an example of a complex sequence ( z_n ) converging to a limit \alpha such that arg _P (z_n) fails to converge. Show that this cannot happen unless \alpha = 0 or \alpha = -1.
    My already disoriented mind is spinning...

    -Dan
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    Most complex textbooks use Arg(z) for the principal argument.
    \frac{\pi }{2} < Arg\left( { - 1 + \frac{{\left( i \right)^{2n - 1} }}{n}} \right) < \pi \,\,,n = 1,3,5 \cdots

     - \pi  < Arg\left( { - 1 + \frac{{\left( i \right)^{2n - 1} }}{n}} \right) < \frac{{ - \pi }}{2},\,\,n = 2,4,6 \cdots

    So in first case Arg\left( {z_{4n-3} } \right) \to \pi while in the second Arg\left( {z_{4n-1} } \right) \to -\pi .
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    The argument function defined by -\pi < \arg \leq \pi is continous on \mathbb{C} \setminus (-\infty,0]. This means if z_k is a sequence of points not on (-\infty,0] and if they converge to z\not \in (-\infty,0] then \lim ~ \arg (z_k) = \arg (z) by the definition of continuity. This tells us any conterexamples to this problem has to be similar to how it was done above where the sequence of points jumps between -\pi and \pi.
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    Forum Admin topsquark's Avatar
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    I see it now. Thanks to you both!

    -Dan
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    Quote Originally Posted by topsquark View Post
    I see it now. Thanks to you both!
    You started to learn complex analysis formally? I complex analysis.
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    Quote Originally Posted by ThePerfectHacker View Post
    You started to learn complex analysis formally? I complex analysis.
    I'm trying to anyway. The text I am using is more of a review text, if you know what I mean, rather than a formal textbook. ("Complex Analysis With Applications" by Silverman.) I got it at Borders with 75% off. (It cost me a whole 3 bucks. ) Anyway, I figure I get understand most of it because I have taken the "Physics" version of it before. This will fill in many of the holes in my education, I hope. This particular chapter is a pain to me, though, since I have forgotten most of the series work I did in my Intro Calc classes.

    Ah well. So I'm a sado-masochist.

    -Dan
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