$1/z_n$ converges to 0 if and only if $z_n$ diverges. For complex numbers. In real analysis there was an easy property that converted limits to infinity in limits at zero. More precisely, $1/z_n$ converges to 0 if and only if $z_n$ diverges (this is converges to infinity). 
I want to apply this property to complex numbers as follows:

Let $z_n$ a complex sequence. $1/z_n$ converges
  to 0 if and only if $z_n$ diverges

I've written $1/z_n = x_n/(x_n^2+y_n^2)-i(y_n/(x_n^2+y_n^2))$ where $x_n = Re(z_n)$ and $y_n = Im(z_n)$ but I can't see either of the implications.
 A: The proposed proposition is false.  The sequence $z_n = (-1)^n$ diverges, but $1/z_n$ does not approach $0$.
Here is a somewhat modified proposition that is true:

$1/z_n$ converges to $0$ if and only if $z_n$ diverges to $\infty$.

To prove this, you can use the following facts:


*

*$w_n \to \infty$ as $n\to \infty$ means $\forall M>0\  \exists N\in\mathbb N\  \forall n\ge N\  |z_n|> M$.

*$w_n\to 0$ as $n\to\infty$ means $\forall \varepsilon>0\  \exists N\in\mathbb N\  \forall n\ge N\  |z_n| < \varepsilon$.

*For $A,B\in\mathbb C$, $|A|<|B|$ if and only if $|1/A|>|1/B|$.


You can also try to figure out how to prove the three bulleted facts.
A: I would phrase it this way:
If $z_n$ is a complex sequence
then $1/z_n\to 0$
 if and only if 
$|z_n| \to \infty$.
In terms of
standard limit nomenclature,
this becomes:
If $z_n$ is a complex sequence
then 
$\left(\forall \epsilon > 0,
\ \exists n(\epsilon)
\text{ such that }
n > n(\epsilon) 
\implies \dfrac1{|z_n|} < \epsilon
\right)$
 if and only if 
$\left(\forall v > 0,
\ \exists n(v)
\text{ such that }
n > n(v)
\implies |z_n| > v
\right)$.
The proof becomes
quite straightforward.
A: This is the way I look at it now:
If $1/z_n$ converges to 0, that implies $1/|z_n|$ converges to zero (note that the implication $|z_n|$ converges to |z| does not imply that $z_n$ converges to z). Now, using the real case |zn| converges to infinity $\iff$ zn converges to infinity.
For the other side, we just need to note that the implication $|z_n|$ converges to 0 does imply that $z_n$ converges to 0 (remember that $z_n$ converges to z $\iff \ d(z_n,z)=|z_n-z|$ converges to 0)
