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I'm following the book Complex Variables and Applications by Churchill-Brown and I'm having trouble with a problem. It says the following:

  • Determine all the accumulation points of $z_{n} = (-1)^n (1+i)\frac{n-1}{n}$

The book gives the answers which are $\pm (1+i)$, and I think I understand the concept of an accumulation point. Basically, it's a point $z_0$ such that, if you take a deleted neighborhood around it, at least one point within the neighborhood lies in the original set. What I have trouble with is the fact of taking the original set $z_n$ and using the definition of the deleted neighborhood $0 < |z - z_0 | < \epsilon $, and finding those accumulation points.

At this point in the book there are no such concepts as sequences, limits or anything related to that, just basic definitions of neighborhoods and sets in the complex plane.

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3 Answers 3

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Note that $z_{2n}\to1+i$ and $z_{2n+1}\to-1-i$, and $z_{2n}\ne 1+i$ and $z_{2n+1}\ne -1-i$, so $\pm(1+i)$ are accumulation points.

Now

$$|z_{2n}-(1+i)|=\frac{1}{2n}|1+i|=\frac{\sqrt 2}{2n}$$ $$|z_{2n+1}-(-1-i)|=\frac{1}{2n+1}|1+i|=\frac{\sqrt 2}{2n+1}$$

So you might see that:

  1. Inside any deleted neigborhood of $\pm(1+i)$ there are an infinity of terms (all terms for which$\frac{\sqrt 2}{2n}<\epsilon$, or $\frac{\sqrt 2}{2n+1}<\epsilon$ respectively, so $\pm(1+i)$ are accumulation points
  2. Outside any union of two neigborhoods of $\pm(1+i)$ there are only a finite number of terms of the sequence.

Now take $w\in\mathbb C$, $w\ne\pm(1+i)$ and prove that it is not an accumulation point. Take a neighborhood $U$ of $w$ small enough so it doesn't contain any of the two points $\pm(1+i)$. If U does not contain any point of $z_n$, then we are done. Otherwise, take a deleted neigborhood $V$ of radius less than $\min_{z_n\in U, z_n\ne w}|w-z_n|$. It is easy to see that the deleted neigborhood does not contain any elements of $z_n$, which was to be proved.

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  • $\begingroup$ So, it means that those last results do the job as the $\epsilon$'s required for the definition of a deleted set? Also, how is $z_{2n} = \frac{1}{2n}$ and for the odd ones too? $\endgroup$ Commented Jan 31, 2017 at 3:41
  • $\begingroup$ I corrected that part. Yes, the last two results do the job. It takes a little bit of work to write it in detail, but I think you get the idea. $\endgroup$
    – Momo
    Commented Jan 31, 2017 at 3:53
  • $\begingroup$ Thank you very much, I'll mark the question as answered. I still have one question though, in the last paragraph you are saying that I should take a neighborhood small enough so that it doesn't contain the accumulation points, this would mean that this subset must be part of the original set? $\endgroup$ Commented Jan 31, 2017 at 4:03
  • $\begingroup$ I'm afraid I don't understand what do you mean by "subset", and "original set", so you might want to clarify a bit. $\endgroup$
    – Momo
    Commented Jan 31, 2017 at 4:07
  • $\begingroup$ Subset would be the neighborhood $U$ and the original set would be the one stated in the question, $z_n$. $\endgroup$ Commented Jan 31, 2017 at 4:11
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For large $n$, $(n-1)/n = 1 - 1/n$ is close to $1$, so $(-1)^n (1+i) (n-1)/n$ is close to $-(1+i)$ if $n$ is odd or $+(1+i)$ if $n$ is even...

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A more intuitive answer is that when n is even, because factor (n-1)/n < 1, this set S has point 1 + i as limit with n = 1,2,3,… without entering it. While each removed neighborhood of z0 = 1 + i expands from it. So however small epsilon is, it overlaps S and each deleted neighbourhood of z0 contains at least one point of S what the definition is of an accumulation point. The same reasoning for point -(1 + i) when n is odd.

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