Question

If $|z|\le 1,|w|\le 1,$ then

• A. $|z-w{|}^2=(|z|-|w|{)}^2+(\arg z-\arg w{)}^2$
• B. $|z-w{|}^2\le (|z|-|w|{)}^2+(\arg z-\arg w{)}^2$
• C. $|z-w{|}^2\ge (|z|-|w|{)}^2+(\arg z-\arg w{)}^2$
• D. None of the above

Answer 1

The correct option is B $|z-w{|}^2\le (|z|-|w|{)}^2+(\arg z-\arg w{)}^2$

We know that, $|z-w{|}^2=|z{|}^2+|w{|}^2-2Re\left(z\overline{w}\right)$
$=|z{|}^2+|w{|}^2-2|z||w|\cos (\alpha -\beta ),$
$($where $\alpha =\arg z,\beta =\arg w)$
So, $|z-w{|}^2=(|z|-|w|{)}^2+2|z||w|(1-\cos (\alpha -\beta ))$
$\le (|z|-|w|{)}^2+4{\sin }^2{\left(\frac{\alpha -\beta }{2}\right)}^2(\because |z|\le 1,|w|\le 1)$
$\le (|z|-|w|{)}^2+4{\left(\frac{\alpha -\beta }{2}\right)}^2(\because \sin \theta <\theta \text{for}\theta \in (0,\frac{\pi }{2}))$
$\le (|z|-|w|{)}^2+(\alpha -\beta {)}^2$
$\le (|z|-|w|{)}^2+(\arg z-\arg w{)}^2$