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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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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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    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? – 2017-01-31
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    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. – 2017-01-31
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    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? – 2017-01-31
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    I'm afraid I don't understand what do you mean by "subset", and "original set", so you might want to clarify a bit. – 2017-01-31
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    Subset would be the neighborhood $U$ and the original set would be the one stated in the question, $z_n$. – 2017-01-31
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    I took a disk $U$ centered in $w$ of radius small enough so the two accumulation points are outside $U$. Then I can construct two neigborhoods of the accumulation points that are disjoint with $U$. These two neigborhoods contain all the terms of $z_n$, except for a finite number. So $U$ can contain at most a finite number of terms of $z_n$ (possibly zero). Then I shrink the radius of $U$ until all the terms of $z_n$ are left outside of $U-\{w\}$. Try to draw a picture to see what I mean. – 2017-01-31
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    If by "original set" you mean $\{z_n: n\in\mathbb N\}$, then the answer is no, it is not a subset. – 2017-01-31
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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...