Convergence of a complex series iff statement I am trying to prove the following statement.

Let  $(z_n)\subset \mathbb{C} $ and $0\neq z\in\mathbb{C}$. Then $z_n\to z$ iff $\frac{z-z_n}{z+z_n}\to 0$.

I proved $\implies$ direction by distinguishing the cases $z>0$ and $z<0$.However, in the reverse direction I am stuck at
$$
|z-z_n|=\bigg| \frac{z-z_n}{z+z_n}\bigg| \cdot| z+z_n|\leq \bigg| \frac{z-z_n}{z+z_n}\bigg| \cdot \big( | z| +|z_n|\big)
$$
If I can show that $z_n$ is bounded I will be done. Thanks for any help!
 A: Showing the $\implies$ direction:
If $z_n\to z \neq 0$, by the reverse triangle inequality
$$ 0 \leq \left|\frac{z-z_n}{z+z_n}\right|
=\left|\frac{z-z_n}{2z-(z-z_n)}\right| \leq \frac{|z-z_n|}{\big||2z|-|z-z_n|\big|}$$
Given that $|z|\neq 0$, take any $\epsilon > 0$ then there exists some $N$ so that for all $n\geq N$,
$|z-z_n|<\epsilon$.
If we further require $\epsilon < 2|z|$ then 
$$0 \leq \left|\frac{z-z_n}{z+z_n}\right| \leq \frac{\epsilon}{2|z|-\epsilon}\quad
\text{hence}
\quad0 \leq \operatorname{lim\,sup}_{n\to\infty} \left|\frac{z-z_n}{z+z_n}\right|\leq \frac{\epsilon}{2|z|-\epsilon}$$
As this holds for every $2|z|>\epsilon > 0$, we can let $\epsilon \to 0^+$, and by the squeeze theorem
$$\operatorname{lim\,sup}_{n\to\infty} \left|\frac{z-z_n}{z+z_n}\right|
= \lim_{\epsilon\to 0^+} \frac{\epsilon}{2|z|-\epsilon} = 0
\quad \text{so} \quad\frac{z-z_n}{z+z_n}\to 0$$
In the other direction, assuming $\frac{z-z_n}{z+z_n} \to 0$,
$$ \bigg|1 - \frac{2z_n}{z+z_n}\bigg|
=\bigg|\frac{z-z_n}{z+z_n}\bigg|
$$
so $\lim_{n\to\infty} \frac{2z_n}{z+z_n} = 1$. Now if $z_n \neq 0$ for $n$ large enough (we will justify this at the end), then
$$1 = \lim_{n\to\infty} \left(\frac{2z_n}{z+z_n}\right)^{-1} = \frac{1}{2}\lim_{n\to\infty}\left(1 + \frac{z}{z_n}\right)$$
This may then be rearranged as $z\neq 0$ to $$\lim_{n\to\infty}z_n = z$$
Finally to justify the assumption $z_n\neq 0$: By definition of convergence, $\frac{z-z_n}{z+z_n} \to 0$ means that for every $\epsilon > 0$ there exists some $N_{\epsilon}$ so that for every $n\geq N_{\epsilon}$, 
$$\left|\frac{z-z_n}{z+z_n} - 0\right|<\epsilon$$
If we choose $\epsilon = 1$ then we have some $N_1$ so that for all $n\geq N_1$
$$\left|\frac{z-z_n}{z+z_n}\right|<1 = \left|\frac{z-(0)}{z+(0)}\right|$$
from which it is clear that if $z_n=0$ then $n<N_1$. Hence the assumption eventually holds.
A: I'm honestly pretty skeptical about your claim in the $\implies$ direction, because inequalities like $z > 0$ don't usually make sense for complex numbers. Positive and negative numbers are really special to $\mathbb{R}$, not $\mathbb{C}$. You could break up the proof based on the argument of $z$, but it really shouldn't come up naturally in the proof at all.

That being said, here is an idea for your converse direction. Suppose that $z_n$ is a given sequence and that $z \in \mathbb{C}$ satisfies $(z - z_n) / (z + z_n) \to 0$. Certainly $z \ne 0$, or we get a contradiction. 
Now we need to see why $z_n$ is bounded; suppose it wasn't. Then there is a subsequence $z_{n_k}$ whose moduli tend to infinity. In particular, we can have $|z_{n_k}| > 2|z|$ for all $k$. 
Try to use this to find a contradiction. The idea to have is that $z_n$ is the important term in $(z - z_n) / (z + z_n)$, while $z$ is negligible; so the sequence behaves like $-1/1 = -1$ which doesn't tend to zero.
