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 _____.

$$\alpha$$ lies on $$|z - z_0|^2 = 4$$ and $$\frac{1}{\bar{\alpha}}$$ lies on $$|z - z_0|^2 = 16$$, where $$z_0 = 1+i$$.

$$|\alpha - z_0|^2 = 4$$ ... (1)

$$\left|\frac{1}{\bar{\alpha}} - z_0\right|^2 = 16$$ ... (2)

From (2): $$\left|\frac{1 - \bar{\alpha}z_0}{\bar{\alpha}}\right|^2 = 16$$, so $$\frac{|1-\bar{\alpha}z_0|^2}{|\alpha|^2} = 16$$.

Note $$|1 - \bar{\alpha}z_0|^2 = |1 - \bar{\alpha}z_0|^2$$. Also $$|\alpha - z_0|^2 = (\alpha - z_0)\overline{(\alpha - z_0)} = |\alpha|^2 - \alpha\bar{z_0} - \bar{\alpha}z_0 + |z_0|^2 = 4$$.

$$|z_0|^2 = 1 + 1 = 2$$. So $$|\alpha|^2 - \alpha\bar{z_0} - \bar{\alpha}z_0 + 2 = 4$$, i.e., $$|\alpha|^2 - \alpha\bar{z_0} - \bar{\alpha}z_0 = 2$$ ... (i)

$$|1-\bar{\alpha}z_0|^2 = 1 - \alpha\bar{z_0} - \bar{\alpha}z_0 + |\alpha|^2|z_0|^2 = 1 - \alpha\bar{z_0} - \bar{\alpha}z_0 + 2|\alpha|^2$$

From (i): $$\alpha\bar{z_0} + \bar{\alpha}z_0 = |\alpha|^2 - 2$$.

$$|1-\bar{\alpha}z_0|^2 = 1 - (|\alpha|^2 - 2) + 2|\alpha|^2 = 3 + |\alpha|^2$$

From (2): $$\frac{3 + |\alpha|^2}{|\alpha|^2} = 16$$

$$3 + |\alpha|^2 = 16|\alpha|^2$$

$$15|\alpha|^2 = 3$$

$$|\alpha|^2 = \frac{1}{5}$$

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

The answer is $$\boxed{20}$$.