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