Let the complex numbers alpha and frac{1}{bar | vedclass

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(B) Given $|z-z_0|^2=4$ and $\alpha$ lies on it,so $|\alpha-z_0|^2=4$.$(\alpha-z_0)(\bar{\alpha}-\bar{z}_0)=4 \Rightarrow |\alpha|^2 - \alpha\bar{z}_0

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$20$

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(B) Given $|z-z_0|^2=4$ and $\alpha$ lies on it,so $|\alpha-z_0|^2=4$.$(\alpha-z_0)(\bar{\alpha}-\bar{z}_0)=4 \Rightarrow |\alpha|^2 - \alpha\bar{z}_0 - \bar{\alpha}z_0 + |z_0|^2 = 4$.Since $z_0 = 1+i$,$|z_0|^2 = 1^2+1^2 = 2$.So,$|\alpha|^2 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) = 4 - 2 = 2$ ... $(i)$.Given $\frac{1}{\bar{\alpha}}$ lies on $|z-z_0|^2=16$,so $|\frac{1}{\bar{\alpha}}-z_0|^2=16$.$|\frac{1-\bar{\alpha}z_0}{\bar{\alpha}}|^2 = 16 \Rightarrow \frac{|1-\bar{\alpha}z_0|^2}{|\alpha|^2} = 16$.$|1-\bar{\alpha}z_0|^2 = 16|\alpha|^2 \Rightarrow (1-\bar{\alpha}z_0)(1-\alpha\bar{z}_0) = 16|\alpha|^2$.$1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) + |\alpha|^2|z_0|^2 = 16|\alpha|^2$.$1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) + 2|\alpha|^2 = 16|\alpha|^2$.$1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) = 14|\alpha|^2$ ... $(ii)$.Subtracting $(ii)$ from $(i)$:$(|\alpha|^2 - (\alpha\bar{z}_0 + \bar{\alpha}z_0)) - (1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0)) = 2 - 14|\alpha|^2$.$|\alpha|^2 - 1 = 2 - 14|\alpha|^2$.$15|\alpha|^2 = 3 \Rightarrow |\alpha|^2 = \frac{3}{15} = \frac{1}{5}$.Therefore,$100|\alpha|^2 = 100 	imes \frac{1}{5} = 20$.

Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $|z-z_0|^2=4$ and $|z-z_0|^2=16$ respectively,where $z_0=1+i$. Then,the value of $100|\alpha|^2$ is.

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