Question

If $|u|<1,|v|<1,$ and $z=\frac{u-v}{1-uv}$, then the 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 C $\frac{||u|-|v||}{1-|u||v|}$

$\left|z\right|=\left|\frac{u-v}{1-uv}\right|$
$\Rightarrow |u-v|>|\left|u\right|-\left|v\right||$ ------(i)

Let $u=r{e}^{i\theta },v=s{e}^{i\alpha };r,s<1$

$\Rightarrow {|1-uv|}^2={|1-{rs.e}^{i(\theta +\alpha )}|}^2=1+r^2{s}^2{cos}^2(\theta +\alpha )+r^2{s}^{2}si{n}^2(\theta +\alpha )-2rs.cos(\theta +\alpha )$

$\Rightarrow 1+r^2{s}^2-2rs.cos(\theta +\alpha )\le {(1-rs)}^2={(1-\left|u\right|\left|v\right|)}^2$

Thus, $|1-uv|\le (1-\left|u\right|\left|v\right|)$ ------(ii)
The result follows by dividing (i) and (ii).
Hence, (c) is correct.