Given the following definition $d(z , w) = \begin{cases}0 & z=w \\ |z|+ |w| & z\neq w \end{cases}$
I have to prove that $d(z,w)= 0\Rightarrow z = w$ Which is in part of checking that $d$ is a metric on $\Bbb C$
Given what I know about complex numbers , if $z= a+ bi$ and $w = c +di $ then $|z|= \sqrt{a^2+b^2}$ and $ |w|= \sqrt{c^2+d^2}$ so $|z|+|w| >0 \forall z,w \in \Bbb C $ and so $d(x,y)=0$ only if $z=w$
I am worried that this is not an eloquent proof. I am prone to being pedantic on a daily basis though.
So if this is pedantic and a waste of your time I apologise
Assume $d(z,w)=0$ and $z\not=w$. Then
$$0=d(z,w)=|z|+|w|$$
If the sum of two non-negative numbers is $0$, then they are both $0$. That is $|z|=|w|=0$, i.e. $z=w=0$. Contradiction.