Connectedness in $\mathbb{C}$

$S=\{i, \frac{i}{2}, \frac{i}{3},...\}$ then $S$ is connected? 
"My textbook says, if we join any two points of $S$ by polygonal path  , then there are points on this path which do not belongs to $S$ . Thus $S$ is not connected". 
"Yes I can see too, there are points on path( between any two points on $S$) which do not belongs to $S$. But is that mean not connected?
According to me, this only means, not polygonally connected! But how they concluded not connected? 
Is, 
not polygonally connected⇒not connected ? 
If not then, is my textbook solution is wrong? 
In perticular, how to check connectedness in $\mathbb{C}$?
Please need help.......

Your objection makes sense. Although $S$ is indeed not connected, just asserting that it is not polygonally connected isn't enough to prove it.
You can prove that it is not connected observing that the set $\{i\}$ is a subset of $S$ which is both open and closed. It is closed because it is already a closed subset of $\mathbb C$ and it is open because $B\left(i,\frac12\right)\cap S=\{i\}$. Since $\{i\}\neq\emptyset$ and $\{i\}\neq S$, this proves that $S$ is not connected.