Question

If $|u|<1,|v|<1$, and $z=\frac{u-v}{1+\overline{u}v}$, then least value of $|z|$ is

• A. $\frac{|u|-|v|}{1+|u||v|}$
• B. $\frac{|u|+|v|}{1+|u||v|}$
• C. $\frac{||u|-|v||}{1-|u||v|}$
• D. None of these

Answer 1

The correct option is A

$\frac{|u|-|v|}{1+|u||v|}$

Let $u=r_1{e}^{i\varphi },v=r_2{e}^{i\theta }$
where $r_1<1,r_2<11+\overline{u}v=1+r_1{r}_2{e}^{i(\theta -\varphi )}$
$|1+\overline{u}v{|}^2=(1+r_1{r}_2\cos (\theta -\varphi ){)}^2+r_1^2{r}_2^2{\sin }^2(\theta -\varphi )=1+2r_1{r}_{2}cos(\theta -\varphi )+r_1^2{r}_2^2\le (1+r_1{r}_2{)}^2=(1+|u||v|{)}^2\Rightarrow |1+\overline{u}v|\le 1+|u||v|\Rightarrow \frac{1}{1+|u||v|}\le \frac{1}{|1+\overline{u}v|}$
Also, $||u|-|v||<|u-v|$
Thus, $\frac{||u|-|v||}{1+|u||v|}\le \left|\frac{u-v}{1+\overline{u}v}\right|$