उत्तर (20)

समाधान

$|z - z_0|^2 = 4$

$\Rightarrow (\alpha - z_0)(\bar{\alpha} - \bar{z}_0) = 4$

$\Rightarrow \alpha \bar{\alpha} - \alpha \bar{z}_0 - z_0 \bar{\alpha} + |z_0|^2 = 4$

$\Rightarrow |\alpha|^2 - \alpha \bar{z}_0 - z_0 \bar{\alpha} = 2$

$|z - z_0|^2 = 16$

$\Rightarrow \left(\frac{1}{\bar{\alpha}} - z_0\right)\left(\frac{1}{\alpha} - \bar{z}_0\right) = 16$

$\Rightarrow (1 - \bar{\alpha} z_0)(1 - \alpha \bar{z}_0) = 16|\alpha|^2$

$\Rightarrow 1 - \bar{\alpha} z_0 - \alpha \bar{z}_0 + |\alpha|^2 |z_0|^2 = 16|\alpha|^2$

$\Rightarrow 1 - \bar{\alpha} z_0 - \alpha \bar{z}_0 = 14|\alpha|^2$

समीकरण (1) और (2) से

$\Rightarrow 5|\alpha|^2 = 1$

$\Rightarrow 100|\alpha|^2 = 20$