Let the complex numbers α and 1 α ¯ lie on the circles z - z 0 2 = 4 and z - z 0 2 = 16 respectively, where…

Given: ${\left|\mathrm{z}-{\mathrm{z}}_0\right|}^2=4$

$\Rightarrow \left(\mathrm{z}-{\mathrm{z}}_0\right)\left(\overline{\mathrm{z}}-{\overline{\mathrm{z}}}_0\right)=4$

$\Rightarrow \mathrm{z}\overline{\mathrm{z}}-{\overline{\mathrm{z}}}_0\mathrm{z}-{\mathrm{z}}_0\overline{\mathrm{z}}+{\left|{\mathrm{z}}_0\right|}^2=4$

$\Rightarrow {\left|\mathrm{z}\right|}^2-{\overline{\mathrm{z}}}_0\mathrm{z}-{\mathrm{z}}_0\overline{\mathrm{z}}=2 \left\{\text{as} {\left|{\mathrm{z}}_0\right|}^2=2\right\}$

It is given that, $\mathrm{\alpha }$ lies on ${\left|z-z_0\right|}^2=4$.

$\Rightarrow {\left|\mathrm{\alpha }\right|}^2-\mathrm{\alpha }{\overline{\mathrm{z}}}_0-{\mathrm{z}}_0\overline{\mathrm{\alpha }}-2=0 ...\left(\mathrm{i}\right)$

We know that, $\overline{\mathrm{\alpha }}·\mathrm{\alpha }={\left|\mathrm{\alpha }\right|}^2={\left|\overline{\mathrm{\alpha }}\right|}^2$

${\left|z-z_0\right|}^2=16$

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

$\Rightarrow \left(1-\overline{\mathrm{\alpha }}{\mathrm{z}}_0\right)\left(1-\mathrm{\alpha }{\overline{\mathrm{z}}}_0\right)=16|\mathrm{\alpha }{|}^2$

$\Rightarrow 1-\overline{\mathrm{\alpha }}{\mathrm{z}}_0-\mathrm{\alpha }{\overline{\mathrm{z}}}_0+|\mathrm{\alpha }{|}^2{\left|{\mathrm{z}}_0\right|}^2=16|\mathrm{\alpha }{|}^2$

$\Rightarrow 1-\overline{\mathrm{\alpha }}{\mathrm{z}}_0-\mathrm{\alpha }{\overline{\mathrm{z}}}_0=14|\mathrm{\alpha }{|}^2 ...\left(\mathrm{ii}\right)$

Using, $\left(\mathrm{i}\right) \text{and} \left(\mathrm{ii}\right)$

$\Rightarrow 5|\mathrm{\alpha }{|}^2=1$

$\Rightarrow 100|\mathrm{\alpha }{|}^2=20$