limit points of $\{1+\frac{1}{n}+i \sin\frac{1}{m}|n,m\in \mathbb{Z}\}$ I have this option in some objective question paper.

$\big\{1+\frac{1}{n}+i \sin\frac{1}{m}|n,m\in \mathbb{Z}\big\}$ has infinite number of isolated points and infinite number of limit points.

 A: Each $1+\frac{1}{n_0}$ is a limit point as $$\lim_{m\to \infty}\bigg(1+\frac{1}{n_0}+i \sin\frac{1}{m}\bigg)=1+\frac{1}{n_0}.$$Also, for each fixed $n_0,m_0\in \Bbb N$ the point, $1+\frac{1}{n_0}+i \sin\frac{1}{m_0}$ is an isolated point, as $$\lim_{n,m\to \infty}\bigg(1+\frac{1}{n}+i \sin\frac{1}{m}\bigg)=1.$$
A: You've got a lot of limit points :  
For each $m \in \mathbb{N}$, $1 + i \sin\frac{1}{m}$ is a limit point of this set.
For each $n \in \mathbb{N}$, $1 + \frac{1}{n}$ is a limit point of this set.
Isolated points :
The set is discrete, every point is isolated.
A: the set is not simply the union of those two sets, it has all elements of the form 
$$1+\frac{1}{n}+i \sin(\frac{1}{m})$$ 
Which would have for example 
$$1+\frac{1}{1}+i \sin(\frac{1}{1})$$ 
If we let n be fixed to some value k, the sequence 
$$a_m=1+\frac{1}{k}+i \sin(\frac{1}{m})$$ 
Converges to $1+\frac{1}{k}$, so $$1+\frac{1}{n}+sin(\frac{1}{m})$$  is a limit point. (To see that it is not an element of the set, note that for rational x, $\sin(x) = 0 \rightarrow x=0$, and $1/m \neq 0$)
