Complex number problem using property

Question

If $ \ |Z| \le 1$ , $ \ |W| \le 1$, then prove that $ \ |Z-W|^2 \le (|Z| - |W| )^2 +(\arg Z -\arg W )^2$

Both Z and W represent circles with radius less than 1 and center at origin. From here I am not able to proceed.

Answer 1

In polar coordinates $(r,\theta)$ the length of a path $\gamma$ from $Z$ to $W$ is given by $$l(\gamma) = \int_\gamma \sqrt{dr^2 + r^2\,d\theta^2} = \int_0^1 \sqrt{r'(t)^2 + r(t)^2 \, \theta'(t)^2} \, dt,$$ where $(r(t), \theta(t)) := p(t)$ is some parameterization of $\gamma$ with $p(0)=Z$ and $p(1)=W.$

Now let $\lambda$ be the line segment between $(|Z|, \arg Z)$ and $(|W|, \arg W)$ in the polar coordinates, i.e. $(r(t), \theta(t)) = (1-t)(|Z|, \arg Z) + t(|W|, \arg W).$ Since $|Z-W|$ is the length of the shortest path from $Z$ to $W$ we must then have $$|Z-W| \leq \int_\lambda \sqrt{dr^2 + r^2 \, d\theta^2}.$$

But, since $r \leq 1$ along $\lambda$ (since both $|Z| \leq 1$ and $|W| \leq 1$) we have $$\int_\lambda \sqrt{dr^2 + r^2 \, d\theta^2} \leq \int_\lambda \sqrt{dr^2 + d\theta^2} \\ = \{ \text{ by Pythagorean Theorem } \} \\ = \sqrt{\Delta r^2 + \Delta\theta^2} = \sqrt{|Z-W|^2 + |\arg Z - \arg W|^2}.$$

Thus, $$|Z-W|^2 \leq |Z-W|^2 + |\arg Z - \arg W|^2.$$