Solution
$\alpha$ satisfies equation $\left|z-z_{0}\right|^{2}=4 \Rightarrow\left|\alpha-z_{0}\right|^{2}=4\left(\because|z|^{2}-z \bar{z}\right)$
\[
\begin{aligned}
& \Rightarrow\left(\alpha-z_{0}\right)\left(\bar{\alpha}-\bar{z}_{0}\right)=4 \Rightarrow \alpha \bar{\alpha}-\alpha \bar{z}_{0}-z_{0} \bar{\alpha}+\left|z_{0}\right|^{2}=4\left(\because\left|z_{0}\right|=\sqrt{2}\right) \\
& \Rightarrow|\alpha|^{2}-\alpha \bar{z}_{0}-z_{0} \bar{\alpha}=2
\end{aligned}
\]
$\left|\mathrm{z}-\mathrm{z}_{0}\right|^{2}=16$ is satisfied by $\frac{1}{\bar{\alpha}} \Rightarrow\left(\mathrm{z}-\mathrm{z}_{0}\right)\left(\overline{\mathrm{z}}-\overline{\mathrm{z}}_{0}\right)=\left(\frac{1}{\bar{\alpha}}-\mathrm{z}_{0}\right)\left(\frac{1}{\alpha}-\overline{\mathrm{z}}_{0}\right)=16$
\[
\begin{aligned}
& \Rightarrow\left(1-\bar{\alpha} z_{0}\right)\left(1-\alpha \bar{z}_{0}\right)=16 \alpha \bar{\alpha} \Rightarrow 1-\bar{\alpha} z_{0}-\alpha \bar{z}_{0}+\alpha \bar{\alpha} z_{0} \bar{z}_{0}=16 \alpha \bar{\alpha} \text { using (1) } \\
& \Rightarrow 1+2-|\alpha|^{2}+|\alpha|^{2}\left|z_{0}\right|^{2}=16|\alpha|^{2} \Rightarrow 3-|\alpha|^{2}+2|\alpha|^{2}=16|\alpha|^{2} \Rightarrow 15|\alpha|^{2}=3 \Rightarrow 5|\alpha|^{2}=1 \Rightarrow 100|\alpha|^{2}=20
\end{aligned}
\]
.
