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Thread: [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 $\displaystyle arg_P(z)$, is defined as the value of $\displaystyle \theta _P = arg(z)$ such that $\displaystyle -\pi < \theta _P \leq \pi$. Give an example of a complex sequence $\displaystyle ( z_n )$ converging to a limit $\displaystyle \alpha$ such that $\displaystyle arg _P (z_n)$ fails to converge.

    I know of one such example:
    $\displaystyle z_n = \frac{1}{n}~e^{in}$
    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 $\displaystyle -1,1/2,-1/3,1/4,-1/5,...$
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    Consider $\displaystyle z_n = \left[ { - 1 + \frac{{\left( i \right)^{2n - 1} }}{n}} \right]$
    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 $\displaystyle -1,1/2,-1/3,1/4,-1/5,...$
    Quote Originally Posted by Plato View Post
    Consider $\displaystyle z_n = \left[ { - 1 + \frac{{\left( i \right)^{2n - 1} }}{n}} \right]$
    I am having a slight problem with both TPH's and Plato's series. For both of them I am getting that $\displaystyle \lim_{n \to \infty} arg_P (z_n) = 0$ and that's just not right, since the limit point is -1 giving an argument of $\displaystyle \pi$. (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 $\displaystyle arg_P(z)$, is defined as the value of $\displaystyle \theta _P = arg(z)$ such that $\displaystyle -\pi < \theta _P \leq \pi$. Give an example of a complex sequence $\displaystyle ( z_n )$ converging to a limit $\displaystyle \alpha$ such that $\displaystyle arg _P (z_n)$ fails to converge. Show that this cannot happen unless $\displaystyle \alpha = 0$ or $\displaystyle \alpha = -1$.
    My already disoriented mind is spinning...

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

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

    So in first case $\displaystyle Arg\left( {z_{4n-3} } \right) \to \pi $ while in the second $\displaystyle Arg\left( {z_{4n-1} } \right) \to -\pi $.
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    The argument function defined by $\displaystyle -\pi < \arg \leq \pi$ is continous on $\displaystyle \mathbb{C} \setminus (-\infty,0]$. This means if $\displaystyle z_k$ is a sequence of points not on $\displaystyle (-\infty,0]$ and if they converge to $\displaystyle z\not \in (-\infty,0]$ then $\displaystyle \lim ~ \arg (z_k) = \arg (z)$ 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 $\displaystyle -\pi$ and $\displaystyle \pi$.
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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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