$\displaystyle (X,d)$ is the metric space $\displaystyle \mathbb{C}$ with metric $\displaystyle d(z,w) = |z-w|$. Prove:

$\displaystyle d(z,e^{i \theta}) \leq d(z,e^{i \alpha}) \leq d(z,-e^{i \theta})$ , for all $\displaystyle \alpha \in \mathbb{R}$ , where $\displaystyle z=|z|e^{i \theta}$

Not sure what's going on here so I started going backwards for the hell of it:

$\displaystyle |z|e^{i \theta} - e^{i \theta} \leq |z|e^{i \theta} - e^{i \alpha} \leq |z|e^{i \theta} + e^{i \theta}$

$\displaystyle -e^{i \theta} \leq -e^{i \alpha} \leq e^{i \theta}$

$\displaystyle |-e^{i \alpha}| \leq e^{i \theta}$

Is this correct or useful in any way? Can somebody explain this geometrically?