# Determine the accumulation points of a set of complex numbers

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.

• 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 $B(z,\epsilon)$.
– Joe
Sep 18, 2021 at 23:10
• Prove it analogous to how you would prove $\{\frac1n\}\subset\mathbb{R}$ has only one accumulation point. Sep 18, 2021 at 23:42

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 $$z=\lim_{n\to\infty}\frac{i^{a_n(z)}}{a_n(z)}$$. $$|z|=\lim_{n\to\infty}\big|\frac{i^{a_n(z)}}{a_n(z)}\big|=\lim_{n\to\infty}\frac{1}{|a_n(z)|}=0$$