# Thread: [SOLVED] Complex limit points (Take II)

1. ## [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

2. Take $-1,1/2,-1/3,1/4,-1/5,...$

3. 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.

4. Originally Posted by ThePerfectHacker
Take $-1,1/2,-1/3,1/4,-1/5,...$
Originally Posted by Plato
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

5. 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 $Arg\left( {z_{4n-1} } \right) \to -\pi$.

6. 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 $\pi$.

7. I see it now. Thanks to you both!

-Dan

8. Originally Posted by topsquark
I see it now. Thanks to you both!
You started to learn complex analysis formally? I complex analysis.

9. Originally Posted by ThePerfectHacker
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