If $(z_{n}) \in \overline{ \mathbb{C}}$, $z_{n} \to \infty$ as $n \to \infty$, what happens to $|z_{n}|$, $Re(z_{n})$, $Im(z_{n})$, $Arg(z_{n})$? Suppose the sequence $(z_{n}) \in \overline{\mathbb{C}}$ (where $\overline{\mathbb{C}}$ is the extended complex plane) converges to infinity as $n \to \infty$.  I need to determine what this implies about $|z_{n}|$, $Re(z_{n})$, $Im(z_{n})$, $Arg(z_{n})$.
I know that a sequence $(z_{n})$ converges to $z=x+iy$ if and only if $(x_{n}) \to x$ and $(y_{n}) \to y$. I suppose, therefore, that when $(z_{n}) \to infty$ it does so because either $Re(z_{n})$, $Im(z_{n})$, or both must converge to $\infty$. But, I am not 100 percent sure about this, or how to prove it.
As for $|z_{n}|$, I suppose it would be the same thing.
Finally, for $Arg(z_{n})$, not sure what this would mean...perhaps that it converges to $\pi/2$? 
Could somebody please explain this to me?
Thanks.
 A: Complex $\infty$ is not in $\mathbb C$.  Intuitively, to go to $\infty$ in $\mathbb C$ you can go in any direction.  $|z_n| \to \infty$ (that's really the definition of $z_n \to \infty$).  But $\text{Arg}(z_n)$ could do anything at all (within the interval $(-\pi, \pi]$).  At least one of $\text{Re}(z_n)$ and $\text{Im}(z_n)$ must be unbounded (in particular they don't both have finite limits), but it could be that neither goes to $\pm\infty$.  For example, you might have $z_n = (2i)^n$, where $\text{Im}(z_n) = 0$ if $n$ is even and $\text{Re}(z_n) = 0$ if $n$ is odd.
EDIT: In the above example, $\text{Arg}(z_n)$ cycles through $0$, $\pi/2$, $\pi$, $-\pi/2$.  It is $0$ for $n = 0, 4, 8, \ldots$, $\pi/2$ for $n=1,5,9,\ldots$, $\pi$ for $n = 2, 6, 10, \ldots$, and $-\pi/2$ for $n=3,7,11,\ldots$.
For other examples, you could take any sequence of arguments $\theta_n$ with $-\pi < \theta_n \le \pi$, and let 
$z_n = 2^n \exp(i \theta_n)$.  Then $\text{Arg}(z_n) = \theta_n$, while
$|z_n| = 2^n \to \infty$.
