Let $ D := \{ z \in \mathbb{C} : \left | z \right | \le 1 \} $ be a unit disk in the complex plane and define $ d(z,w) := \begin{cases} \left| z-w \right| & \text{ if } \arg (z) =\arg (w) \text{ or if one of } z \text{ and } w \text{ is zero } \\ \left| z \right|+\left| w \right| & \text{ otherwise } \\ \end{cases} $
Verify that $d$ defines a metric on $D$.
SOLUTION: Checking the metric axioms
$M_1: d(z,w) \ge 0$:
$d(z,w) \ge 0$ holds from the definition.
$M_2:d(z,w)=0 \Leftrightarrow z=w $:
$d(z,w)=0= \begin{cases} \left| z-w \right| & \text{ if } \arg (z)=\arg(w) \text{ or if one of } z \text{ and } w \text{ is zero } \\ \left| z \right|+\left| w \right| & \text{ otherwise } \end{cases} $
We get $z=w$ on the other hand if $z=w$, $d(z,w)= \begin{cases} \left| z-z \right| & \text{ if } \arg (z)=\arg (w) \text{ or if one of } z \text{ and } w \text{ is zero} \\ \left| z \right|+\left| z \right| & \text{ otherwise } \\ \end{cases} $
$ |z-z|=0 \\ \text{so } d(z,w)=0 \Leftrightarrow z=w \text{ holds} \\ $
${{M}_{3}};\,d(z,w)=d(w,z)$
$d\left( z,w \right)=\begin{cases} \left| z-w \right| & \text{ if } \arg (z) =\arg (w) \text{ or if one of } z \text{ and } w \text{ is zero } \\ \left| z \right|+\left| w \right| & \text{ otherwise } \\ \end{cases}$ $ =d\left( w,z \right)=\begin{cases} \left| w-z \right| & \text{ if } \arg (z) =\arg (w) \text{ or if one of } z \text{ and } w \text{ is zero } \\ \left| w \right|+\left| z \right| & \text{ otherwise } \\ \end{cases}$ $\text{so }d(z,w)=d(w,z) \text{ holds}$
${{M}_{4}};\text{ Triangle Inequality}: d(z,w)\le d(z,x)+d\left( x,w \right)$
Somebody to assist me in axioms $M_2$ , and $M_4$. Regards.