Equivalent metrics in $\mathbb C$ (via stereographic projection) Problem statement
Let $\overline{d}$ be the distance in $\bar{\mathbb C}$ induced by the distance of $\mathbb R^3$ through the stereographic projection, i.e., if $z,z' \in \bar{\mathbb C}$, we define $\overline{d}(z,z')=d(\phi(z),\phi(z'))$ ($\phi$ is projection's inverse and $d$ is the euclidean distance). Prove that $\overline{d}$ restricted to $\mathbb C$ is equivalent to $\mathbb C$ with the usual distance.
The attempt at a solution
I am trying to prove the statement showing that $(\mathbb C,\overline{d})$ and $(\mathbb C,d')$ ($d'$ is the usual distance in $\mathbb C$) have the same convergent sequences but I got stuck.
Let $z_n \to z$ in $(\mathbb C, d')$, I want to prove that $z_n \to_{\overline{d}} z$
Let $\epsilon>0$, I know that $\sqrt{(a_n-a)^2+(b_n-b)^2}<\epsilon$ if $n \geq N$ for some $N \in \mathbb N$.
Now $\overline{d}(z_n,z)=\sqrt{(\dfrac{2a_n}{1+{a_n}^2+{b_n}^2}-\dfrac{2a}{1+a^2+b^2})^2+(\dfrac{2b_n}{1+{a_n}^2+{b_n}^2}-\dfrac{2b}{1+a^2+b^2})^2+(\dfrac{{a_n}^2+{b_n}^2-1}{1+{a_n}^2+{b_n}^2}-\dfrac{a^2+b^2-1}{1+a^2+b^2})^2}$. 
I would like to find $c>0:\overline{d}(z,z')<cd'(z,z')$, I've tried to do it but I couldn't.
I've tried to do an analogous thing for the other way round and I also got stuck.
One thing that I've used is the fact that $\phi(z_n), \phi(z) \in \mathbb S^2$, this means that $d(\phi(z_n),\phi(z))\leq 2$, which is the diameter of $S^2$.
 A: Continuous functions $\Bbb R^n\to\Bbb R$ (and those defined on subspaces of $\Bbb R^n$) are closed under the operations of addition, multiplication, and division where the denominator doesn't vanish. Since the coordinate projections $\Bbb R^n\to\Bbb R$ are continuous, and power maps $\Bbb R\to\Bbb R:x\mapsto x^k$ are continuous, we can conclude that any rational function $\Bbb R^n\supset A\to\Bbb R$ is continuous, hence any vector function $\Bbb R^n\supset X\to\Bbb R^n$ whose coordinates are rational functions is also continuous.
Since stereographic projection and its inverse are both rational functions, they are both continuous, and hence they are both homeomorphisms. In any homeomeorphism of metric spaces, pulling or pushing their respective distance functions yields an equivalent metric.
A: I'll accept blue's answer but here it is an alternative solution in the spirits of my initial approach:
I've just read in Ahlfors' textbook the equality:
$\overline{d}(z,z')=\dfrac{2|z'-z|}{(\sqrt{1+|z|^2})(\sqrt{1+|z'|^2})}$
On the other hand, if $\{z_n\}_{n \in \mathbb N}$ is a convergent sequence, then it is also bounded. So there is $k\geq 0: |z_n|\leq k$ for all $n$
Suppose $z_n \to z$ and call $|z|=c$, by the things previously said, we have
$\dfrac{2|z_n-z|}{(\sqrt{1+k^2})(\sqrt{1+c^2})}\leq \dfrac{2|z_n-z|}{(\sqrt{1+|z_n|^2})(\sqrt{1+|z|^2})}\leq 2|z_n-z|$.
From these inequalities it follows directly the equivalence of the metrics.
