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How to show that the origin is the only accumulation point of the set $$z, = \frac{i}{n} (n = 1, 2, ...)$$(I was able to show it's a accumulation point,but was not able to show it's the only one)

def of accumulation point:A point $z$ is said to be an accumulation point of a set $S$ if each deleted neighborhood of $z$ contains at least one point of $S$.

Thanks for your help

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    What's your definition of accumulation point? ​ ​ – 2017-02-26
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    @RickyDemer A point z0 is said to be an accumulation point of a set S if each deleted neigh-borhood of z0 contains at least one point of S – 2017-02-26
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    So, you'll need to show that 0 is the only such z0. ​ (Have you showed that negative zs don't work?) ​ ​ ​ ​ – 2017-02-26
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    @RickyDemer yes i -ve z's doesn't work i have showed it. – 2017-02-26
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    How about Zs that are greater than 1? ​ ​ – 2017-02-26
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    @RickyDemer for any points in 3rd and 4th quadrant it's done and for all points in first and second quadrant where |X| >$\epsilon$ it's done we have to prove for remaining cases.(where $\epsilon$ >0 and belongs to real) – 2017-02-26
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    Oh, I missed that the numerator was the imaginary unit rather than just 1. ​ In any case, now, for nonzero X, find a positive $\epsilon$ for which it's done. ​ ​ ​ ​ – 2017-02-26
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    @RickyDemer okk i got you idea thanks. – 2017-02-26

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For all $z\in\mathbb C$, $|z-z_n|\geq |z|-|z_n|$ as a consequence of the triangle inequality. With your $z_n=\dfrac{i}{n}$ this implies $|z-z_n|\geq |z|-\frac1n$. If $z\neq 0$, then there exists $N$ such that $\frac{1}{N}<\frac12|z|$, from which it follows that for all $n\geq N$, $|z-z_n|>\frac12|z|$. Thus the disk of radius $\frac12|z|$ centered at $z$ contains none of the points $\{z_N, z_{N+1},z_{N+2},\ldots\}$. Of the remaining finitely many points $\{z_1,z_2,\ldots,z_{N-1}\}$, consider all of the distances $|z-z_k|$. There is a smallest positive one of these because the set is finite*. Call that smallest positive distance $d$. Then the open disk centered at $z$ with radius $\min\{d,\frac12|z|\}$, with $z$ removed, contains no $z_n$. This shows that $z$ is not an accumulation point.

*(It could be that $N=1$, in which case there is nothing further to show; rather than taking the minimum of the empty set you just wouldn't need $d$ at all.)

More generally, when you have a convergent sequence $\{w_n\}$ with $w=\lim\limits_{n\to\infty}w_n$, then for all $\varepsilon>0$, the disk of radius $\varepsilon$ centered at $w$ contains all but finitely many of the $w_n$s. Hence if $z\neq w$, taking $\varepsilon=\frac12|z-w|$, the open disk centered at $z$ with radius $\frac12|z-w|$ will contain only finitely many of the $z_n$s, and you can take a smaller radius if necessary to avoid the remaining finitely many points.