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 $|z| < 1$
$\left|\frac{w+a}{1+w}\right| < 1$
$\Rightarrow \quad|w+a|<|1+w \bar{a}|$
$\Rightarrow(w+a)(\bar{w}+\bar{a}) < (1+w \bar{a})(1+\bar{w} a)$
$\Rightarrow \quad w \bar{w}+w \bar{a}+a \bar{w}+a \bar{a} < 1+\bar{w} a$
$+w \bar{a}+w \bar{w} a \bar{a}$
$\Rightarrow \quad|w|^2+|a|^2 < 1+|u|^2|\alpha|^2$
$\Rightarrow|w|^2|a|^2-|w|^2-|a|^2+1 > 0$
$\Rightarrow \quad\left(|w|^2-1\right)\left(|a|^2-1\right) > 0$
$\Rightarrow \quad|u|^2-1 < 0 \quad[\because|a| < 1]$
$\therefore \quad|w| < 1,|z| < 1$
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Let $\mathrm{A}_{\mathrm{k}}=\mathrm{a}_1{ }^2-\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2-\mathrm{a}_4{ }^2+\ldots+\mathrm{a}_{2 \mathrm{k}-1}{ }^2-\mathrm{a}_{2 \mathrm{k}}{ }^2$.
If $\mathrm{A}_3=-153, \mathrm{~A}_5=-435$ and $\mathrm{a}_1{ }^2+\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2=66$, then $\mathrm{a}_{17}-\mathrm{A}_7$ is equal to....................
$A:\frac{1}{{\log_2 \left| z \right|}} - \frac{1}{{\log_2 \left| z \right| - 1}} - 1 < 0$ and $\left( {B:\operatorname{Im} \left( z \right) = 0} \right).$ The range of values of $Re(z)$
lying in the region $A \cap B$ is