Question

If $\left|z\right|\le 1,\left|w\right|\le 1$ show that
${|z-w|}^2\le {(\left|z\right|-\left|w\right|)}^2={(Argz-Argw)}^2$

Answer 1

Let $z=r{e}^{i\theta }$, $w=R{e}^{i\varphi }\therefore r\le 1$, $R\le 1$,
$Argz=\theta ,Argw=\varphi$
${|z-w|}^2=r^2+R^2-2rR\cos (\theta -\varphi )$
$r^2+R^2-2rR+2rR[1-\cos (\theta -\varphi \left)\right]$
$={(r-R)}^2+2rR.2\sin {}^2\left\{\frac{\theta -\varphi }{2}\right\}$
$\le {(r-R)}^2+\mathrm{4.1.1}{\left(\frac{\theta -\varphi }{2}\right)}^2$
$={(\left|z\right|-\left|w\right|)}^2+{(Argz-Argw)}^2$
$R<1$, $\sin \psi <\psi$ where $\psi$ is a $+$ive angle. Also ${\sin }^2\psi <{\psi }^2$.