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I am reading Complex Variables and Applications by Brown and Churchill. On page 35, exercise 7(b) asks the reader to determine the accumulation points of the following set:

$$S=\left\{\frac{i^n}{n}:n\in\mathbb{N}\right\}$$

After calculating a few elements of $S$ for small values of $n$ by hand and plotting the points in the complex plane, it is obvious to me that $z=0$ is the only accumulation point of $S$. This is confirmed by the answer given in the textbook, and I was easily able to prove that $z=0$ is indeed an accumulation point of $S$. However, the exercise asks the reader to "determine" the accumulation points of $S$, which seems to imply that I also need to prove that there are no other accumulation points of $S$. However, I am struggling to prove that if $z\neq0$, then $z$ is not an accumulation point of $S$. Here is what I have so far:

Let $z\in\mathbb{C}\setminus\{0\}$ and let $z_1\in S$

Then there exists $n\in\mathbb{N}$ such that $z_1=\frac{i^n}{n}$

Let $\varepsilon=\frac{1}{n+1}$

Then $|z_1-z|\geq||z_1|-|z||=\left|\left|\frac{i^n}{n}\right|-|z|\right|=\left|\frac{1}{n}-|z|\right|$

I know that I need to show somehow that $\left|\frac{1}{n}-|z|\right|\geq\varepsilon$, but I am not sure how. I chose $\varepsilon=\frac{1}{n+1}$ based on the fact that:

$\left|\frac{1}{n}-|z|\right|\geq\varepsilon\implies\frac{1}{n}-|z|\geq\varepsilon$ or $\frac{1}{n}-|z|\leq-\varepsilon$

$\frac{1}{n}-|z|\geq\varepsilon\implies\frac{1}{n}\geq\varepsilon+|z|\implies\frac{1}{n}>\varepsilon$ since $|z|>0$, which is true for all $n\in\mathbb{N}$ if $\varepsilon=\frac{1}{n+1}$

However, substituting $\varepsilon=\frac{1}{n+1}$ into the second inequality above yields:

$\frac{1}{n}-|z|\leq-\varepsilon\implies\frac{1}{n}-|z|\leq-\frac{1}{n+1}\implies\frac{1}{n}+\frac{1}{n+1}\leq|z|$

which does not appear to be true for all $n\in\mathbb{N}$ since $|z|$ can be arbitrarily small. However, I am not sure how to choose $\varepsilon$ such that both inequalities are satisfied. $\varepsilon=-\frac{1}{n}$ would work, but $\varepsilon$ must be positive. Will my choice of $\varepsilon=\frac{1}{n+1}$ work?

I would appreciate any hints that might help me complete the proof.

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    $\begingroup$ For any $z\ne 0$, we have $|z|>0$. Let $\epsilon=|z|/2$, and show that there are only finitely many points in $S$ with magnitude greater than $\epsilon$, hence only finitely many (possibly zero) points in $B(z,\epsilon)$. $\endgroup$
    – Joe
    Sep 18, 2021 at 23:10
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    $\begingroup$ Prove it analogous to how you would prove $\{\frac1n\}\subset\mathbb{R}$ has only one accumulation point. $\endgroup$
    – Adam Rubinson
    Sep 18, 2021 at 23:42

1 Answer 1

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Given an $S$ accumulation point $z$ choose a convergent one-to-one $S\setminus\{z\}$ sequence $(\frac{i^{A_n(z)}}{A_n(z)})_{n=1}^\infty$ and a monotonic subsequence $(a_n(z))_{n=1}^\infty$ of $(A_n(z))_{n=1}^\infty$ so that $z=\lim_{n\to\infty}\frac{i^{a_n(z)}}{a_n(z)}$. $$|z|=\lim_{n\to\infty}\big|\frac{i^{a_n(z)}}{a_n(z)}\big|=\lim_{n\to\infty}\frac{1}{|a_n(z)|}=0$$

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  • $\begingroup$ Thank you, Oliver! I admit that this is a bit over my head (I have only learned the basic definitions of accumulation point, open set, etc. and nothing about complex limits or monotonic subsequences), but I was able to figure it out based on the comments above, so I will mark this as the accepted answer. $\endgroup$
    – user772759
    Sep 19, 2021 at 2:20

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