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(A) Let $ z = \frac{\beta-\alpha}{1-\bar{\alpha} \beta} $. We want to find $ |z| $.Consider $ |z|^2 = z \cdot \bar{z} = \left( \frac{\beta-\alpha}{1-\

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(A) Let $ z = \frac{\beta-\alpha}{1-\bar{\alpha} \beta} $. We want to find $ |z| $.Consider $ |z|^2 = z \cdot \bar{z} = \left( \frac{\beta-\alpha}{1-\bar{\alpha} \beta} \right) \left( \frac{\bar{\beta}-\bar{\alpha}}{1-\alpha \bar{\beta}} \right) $.Expanding the numerator: $ (\beta-\alpha)(\bar{\beta}-\bar{\alpha}) = \beta \bar{\beta} - \beta \bar{\alpha} - \alpha \bar{\beta} + \alpha \bar{\alpha} = |\beta|^2 - \beta \bar{\alpha} - \alpha \bar{\beta} + |\alpha|^2 $.Expanding the denominator: $ (1-\bar{\alpha} \beta)(1-\alpha \bar{\beta}) = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + \bar{\alpha} \alpha \beta \bar{\beta} = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2 |\beta|^2 $.Since $ |\beta|=1 $,we have $ |\beta|^2 = 1 $.Substituting $ |\beta|^2 = 1 $ into the expressions:Numerator: $ 1 - \beta \bar{\alpha} - \alpha \bar{\beta} + |\alpha|^2 $.Denominator: $ 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2 (1) = 1 - \alpha \bar{\beta} - \bar{\alpha} \beta + |\alpha|^2 $.Since the numerator and denominator are equal,$ |z|^2 = 1 $,which implies $ |z| = 1 $.

Column-$I$	Column-$II$
$A. Z_1 Z_2$	$1. \text{imaginary axis}$
$B. Z_1 + Z_2 = 0$	$2. \text{Im}(-Z_2)$
$C. \text{Im}(Z_1)$	$3. \|Z_1\|^2$
$D. \text{Re}(Z_1)$	$4. \text{Re}(Z_2)$

If $ \alpha $ and $ \beta $ are two different complex numbers with $ |\beta|=1 $,then $ \left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right| $ is equal to

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