**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 $S^2$.