(complex variables) Show that a convergent sequence in the plane means that the corresponding points of the sequence on the Riemann sphere converges I'm just learning about the Riemann Sphere in complex variables and I've been posed the following problem. How do I begin to solve it?

Given a sequence $\{z_n\}$ of complex numbers where $\left| z_n - z \right| \rightarrow 0$, show that the sequence $\{ Z_n \}$ of the corresponding points on the unit Riemann sphere also converges. Is the converse also true?

We've defined the Riemann Sphere as the stereographic projection that projects a unit sphere onto the complex plane.
 A: The Riemann sphere minus the point at infinity is homeomorphic to the complex plane, so any sequence converging in the plane must converge on the sphere.
The converse is not true. Pick a sequence converging to infinity on the Riemann sphere. This clearly doesn't converge to a complex number. 
A: Here is my solution to the question I've posed above. Comments would be appreciated:
Let $\{z_n\}$ be a convergent sequence of complex numbers that corresponds to a sequence of points $\{Z_n\}$ on the unit sphere of the stereographic projection. We then have $\lim_{n \to \infty} z_n = z$ and $\left| z_n - z \right| \to 0$. Let $b = z_n - z $. Then $b \in \mathbb{C}$ corresponds to the point $B \in \mathbb{R}^3$ on the unit sphere at $(x_1,x_2,x_3)$ such that
\begin{align*}
B = (x_1,x_2,x_3) = \left( \frac{2\operatorname{Re} b}{1+ \left| b \right|^2}, \frac{2\operatorname{Im} b}{1 + \left| b \right|^2}, \frac{\left| b \right|^2 - 1}{\left| b \right|^2 + 1} \right).
\end{align*}
We see that in the limit $\left| b\right| \to 0$, $\operatorname{Re}b \to 0$, and $\operatorname{Im}b \to 0$. Thus
\begin{align*}
\lim_{n \to \infty} B = (x_1,x_2,x_3) = \left( \frac{2 \cdot 0}{1+ 0}, \frac{2 \cdot 0}{1 + 0}, \frac{0 - 1}{0 + 1} \right) = \left( 0, 0, -1 \right).
\end{align*}
Thus, the limit exists on the unit sphere. This means a convergent sequence of complex numbers $\{z_n\}$ implies that the corresponding points $\{Z_n\}$ on the unit sphere will converge.
However, it is not true that $\{ Z_n\} \to Z$ necessarily implies the convergence of the corresponding sequence $\{z_n\}$ in $\mathbb{C}$. To prove this, we show that $\lim_{n\to\infty}z_n=z$ does not exist. Let $\{ Z_n\} \to Z$ and $Z = (0,0,1)$. In the limit, we have
\begin{align*}
x_1 = \frac{2 \operatorname{Re} z}{1 + \left| z \right|^2} &= 0 \\
2 \operatorname{Re} z &= 0\\
\operatorname{Re} z &= 0,
\end{align*}
\begin{align*}
x_2 = \frac{2 \operatorname{Im} z}{1 + \left| z \right|^2} &= 0 \\
2 \operatorname{Im} z &= 0\\
\operatorname{Im} z &= 0,
\end{align*}
and
\begin{align*}
x_3 = \frac{ \left| z \right|^2 -1 }{ \left| z \right|^2 + 1} &= 1 \\
\left| z \right|^2 -1 &= \left| z \right|^2 + 1 \\
-1 &= 1
\end{align*}
Since $-1 \neq 1$, there is no such point $z \in \mathbb{C}$ in the limit that corresponds to $Z = (0,0,1)$.
