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Let $\alpha$ be a fixed non-zero complex number with $| a | < 1$ and $w=\left(\frac{z-\alpha}{1-\bar{a} z}\right)$, where $z$ is a complex number. Then,
there exists a complex number $z$ with $|z| < 1$ such that $|w| > 1$
$|u| > 1$ for all $z$ such that $|z| < 1$
$|w| < 1$ for all $z$ such that $|z| < 1$
there exists $z$ such that $|z|<1$ and $|w|=1$
Solution
(c)
We have,
$w=\left(\frac{z-a}{1-\bar{a} z}\right),|a| < 1$
$\Rightarrow w(1-\bar{a} z)=z-a$
$\Rightarrow \quad(w-w \bar{a} z)=z-a$
$\Rightarrow \quad w+a=z(1+w \bar{a})$
$\Rightarrow \quad z=\frac{w+a}{1+w \bar{a}}$
When $|z| < 1$
$\left|\frac{w+a}{1+w}\right| < 1$
$\Rightarrow \quad|w+a|<|1+w \bar{a}|$
$\Rightarrow(w+a)(\bar{w}+\bar{a}) < (1+w \bar{a})(1+\bar{w} a)$
$\Rightarrow \quad w \bar{w}+w \bar{a}+a \bar{w}+a \bar{a} < 1+\bar{w} a$
$+w \bar{a}+w \bar{w} a \bar{a}$
$\Rightarrow \quad|w|^2+|a|^2 < 1+|u|^2|\alpha|^2$
$\Rightarrow|w|^2|a|^2-|w|^2-|a|^2+1 > 0$
$\Rightarrow \quad\left(|w|^2-1\right)\left(|a|^2-1\right) > 0$
$\Rightarrow \quad|u|^2-1 < 0 \quad[\because|a| < 1]$
$\therefore \quad|w| < 1,|z| < 1$
