Show that the metric $d: \mathbb{C}^{2} \to \mathbb{R}^{\ge0}$ defined by $d(z_1,z_2)=(x_1-x_2)^2+(y_1-y_2)^2$, where $z_1=x_1+y_1i$ and $z_2=x_2+y_2i$ is unbounded. Create a new bounded metric.
$(\mathbb{C},d)$ is unbounded because $\nexists N \in \mathbb{R} \ni \forall z_1,z_2 \in \mathbb{C}$, $d(z_1,z_2) < N$.
Using linear algebra we can reframe the problem as follows:
Let $X=\{a+bi | a,b \in \mathbb{R} \ni a^2+b^2=1\}$.
First, an inner product on a vector space $X$ is a function that takes each ordered pair $(u,v)$ of elements in $X$ to a number $\langle u, v \rangle \in F$, where $F$ denotes the scalar field of real numbers or the scalar field of complex numbers.
The length of a vector is called the norm of a vector.
And the inner product induces a norm by $||x||:=\sqrt{\langle x, x \rangle}$, which in turn, induces a metric by $d(x,y):=||x-y||$.
This means that Inner Product Spaces $\subset$ Normed Vector Spaces $\subset$ Metric Spaces.
Because $a^2+b^2=1$ for $a,b \in \mathbb{R}$, we have $|z|=1$ and thus the distance of the point $z$ to the origin is $1$. Now if we think of vectors as points instead of arrows, then $||z||$ is the distance of the point $z$ to the origin, so that $||z||=1$. What this does is that it essentially puts an upper bound on the maximum possible distance between any two points in $X$.
Let $z_1,z_2 \in X$, then $z_1$ and $z_2$ lie on the circle $|z|=1$.
As the diameter of the circle is $2$, the maximum possible distance between $z_1$ and $z_2$ is $2$; that is, $||z_1-z_2|| \le 2$.
Define $d: X \times X \longrightarrow \mathbb{R}$ by $d(z_1,z_2)=||z_1-z_2||$.
$d(z_1,z_2)=||z_1-z_2||=\sqrt{\langle z_1-z_2,z_1-z_2 \rangle}$
With this new metric defined on X, we have that $ \forall z_1,z_2 \in X, \exists N \in \mathbb{R} \ni d(z_1,z_2) < N$, namely, $N=3$.
Is this the correct approach here?