Is $d$ topologically equivalent to the usual metric on $\mathbb{C}$? 
Let $d$ be  a function  defined on $\mathbb{C} \times \mathbb{C}$ by
$$d(z,z')  = \begin{cases}  0 &\text{if} \; z= z' \\\\ |z| + |z'| &\text{if}\; z \neq z' \end{cases}$$
Is  $d$ topologically equivalent  to the usual metric on $\mathbb{C}$?

My attempt:  I know that  $d$  is  a metric on $\mathbb{C}$ but here im confusing  that "Is  $d$ topologically equivalent  to the usual metric on $\mathbb{C}$?
Any hints/solution ?

You should probably review the definition of "topologically equivalent".  But one characterization is that two metrics are topologically equivalent iff they have the same convergent sequences.
Consider the sequence $x_n = 1 + \frac{1}{n}$.  Does it converge to $1$ with respect to the usual metric on $\mathbb{C}$?  Does it converge to $1$ with respect to this new metric $d$?
For that matter, with respect to $d$, is there any sequence that converges to $1$?  For any $z_0 \ne 0$, is there any sequence that converges to $z_0$?
Another characterization of topologically equivalent metrics is that they have the same open sets. Is $\{1\}$ an open set with respect to the usual metric?  With respect to $d$?
