Subject: Mathematics — Complex Numbers
Question: Let complex numbers $\alpha$ and
$\frac{1}{\bar{\alpha}}$ lie on circles
$\left( x - x_{0} \right)^{2} + \left( y - y_{0} \right)^{2} = r^{2}$
and
$\left( x - x_{0} \right)^{2} + \left( y - y_{0} \right)^{2} = 4r^{2}$.
respectively. If $z_{0} = x_{0} + iy_{0}$ satisfies the equation
$2\left| z_{0} \right|^{2} = r^{2} + 2$, then $|\alpha|$
Answer: $\frac{1}{\sqrt{7}}$
Solution: Since, $\alpha$ lies on the circle
$\left( x - x_{0} \right)^{2} + \left( y - y_{0} \right)^{2} = r^{2}$\
$${\therefore\left| \alpha - z_{0} \right|^{2} = r^{2}
}{\Rightarrow \ \left( \alpha - z_{0} \right)\left( \bar{\alpha} - {\bar{z}}_{0} \right) = r^{2}
}{\Rightarrow \alpha\bar{\alpha} - \alpha{\bar{z}}_{0} - \bar{\alpha}z_{0} + z_{0}{\bar{z}}_{0} = r^{2}
}{\Rightarrow |\alpha|^{2} + \left| z_{0} \right|^{2} - \alpha{\bar{z}}_{0} - \bar{\alpha}z_{0} = r^{2}
}$$Also $\frac{1}{\bar{\alpha}}$ lies on the circle
$\left( x - x_{0} \right)^{2} + \left( y - y_{0} \right)^{2} = 4r^{2}$\
$${\therefore\left| \frac{1}{\bar{\alpha}} - z_{0} \right|^{2} = 4r^{2} \Rightarrow \left( \frac{1}{\bar{\alpha}} - z_{0} \right)\left( \frac{1}{\alpha} - {\bar{z}}_{0} \right) = 4r^{2}
}{\Rightarrow \frac{1}{\alpha\bar{\alpha}} - \frac{z_{0}}{\alpha} - \frac{{\bar{z}}_{0}}{\bar{\alpha}} + z_{0}{\bar{z}}_{0} = 4r^{2}
}{\Rightarrow \frac{1}{|\alpha|^{2}} - \frac{z_{0}\bar{\alpha}}{|\alpha|^{2}} - \frac{{\bar{z}}_{0}\alpha}{|\alpha|^{2}} + \left| z_{0} \right|^{2} = 4r^{2}
}{\Rightarrow \ 1 + |\alpha|^{2}\left| z_{0} \right|^{2} - z_{0}\bar{\alpha} - {\bar{z}}_{0}\alpha = 4r^{2}|\alpha|^{2}}$$
On subtracting equation (i) from (ii), we get\
$$1 - |\alpha|^{2} + \left| z_{0} \right|^{2}\left( |\alpha|^{2} - 1 \right) = r^{2}\left( 4|\alpha|^{2} - 1 \right)
$$or
$\ \left( |\alpha|^{2} - 1 \right)\left( \left| z_{0} \right|^{2} - 1 \right) = r^{2}\left( 4|\alpha|^{2} - 1 \right)$\
Using $\left| z_{0} \right|^{2} = \frac{r^{2} + 2}{2}$, we get\
$${\left( |\alpha|^{2} - 1 \right)\frac{r^{2}}{2} = r^{2}\left( 4|\alpha|^{2} - 1 \right)
}{\Rightarrow |\alpha|^{2} - 1 = 8|\alpha|^{2} - 2 \Rightarrow |\alpha| = \frac{1}{\sqrt{7}}}$$