Question

If $|z|\le 1$ and $|\omega |\le 1$

Then show that $|z-\omega {|}^2\le (|z|-|\omega |{)}^2+(argz-arg\omega {)}^2$.

Answer 1

From the figure
$\alpha =(argz-arg\omega )$ ....... (1)
and $si{n}^2\alpha /2\le (\alpha /2{)}^2$ ...... (2)
In $\Delta$OAB, from cosine rule
$(AB{)}^2=(Oa{)}^2+(OB{)}^2-2OA.OBcos\alpha$
$\Rightarrow |z-\omega {|}^2=|z{|}^2+|\omega {|}^2-2|z||\omega |cos\alpha$
$\Rightarrow |z-\omega {|}^2=(|z|-|\omega |{)}^2+2|z||\omega |(1-cos\alpha )$
$\Rightarrow |z-\omega {|}^2=(|z|-|\omega |{)}^2+4|z||\omega |si{n}^2\alpha /2$
$\Rightarrow |z-\omega {|}^2\le (|z|-|\omega |{)}^2+|z||\omega |{\alpha }^2$ [from (2)]
$\Rightarrow |z-\omega {|}^2\le (|z|-|\omega |{)}^2+{\alpha }^2$ $\{\because |z|\le 1,|\omega |\le 1\}$
$|z-\omega {|}^2\le (|z|-|\omega |{)}^2+(argz-arg\omega {)}^2$ {from (1)}

Ans: 1