Prove that if \(c>0\), \[ \abs{w_1} = c\abs{w_2} \implies \abs{w_1 - c^2 w_2} = c\abs{w_1 - w_2} .\]
Prove that if \(c>0\) and \(c\neq 1\), with \(z_1\neq z_2\), then the following equation represents a circle: \[ \abs{z-z_1 \over z-z_2} = c .\] Find its center and radius.
Hint: use part (a)
Solution: part 1
\[ \abs{w_1 - c^2 w_2}^2 &= (w_1 - c^2 w_2) ( \bar{w_1} - c^2 \bar{w_2} ) \\ &= \abs{w_1}^2 + c^4 \abs{w_2}^2 - 2c^2 \Re(w_1 \bar{w_2}) \\ &= {\color{green} c^2 \abs{w_2}^2 } + c^4 \abs{w_2}^2 - 2c^2 \Re(w_1 \bar{w_2}) \\ &= c^2 \abs{w_2}^2 + {\color{green} c^2 \abs{w_1}^2 } - 2c^2 \Re(w_1 \bar{w_2}) \\ &= c^2 \abs{w_1 - w_2} ,\] where we’ve applied the assumption \(\abs{w_1} = c\abs{w_2}\) twice.
Solution: part 2
Using part 1: \[ w_1\da z-z_1, w_2 \da z-z_2 \implies \abs{w_1} &= c\abs{w_2} \\ \implies \abs{w_1 - c^2 w_2} &= c \abs{w_1 - w_2} \\ \implies \abs{ z-z_1 - c^2 (z-z_2) } &= \abs{(z-z_1) - (z-z_2)} \\ \implies \abs{(1-c^2) z - z_3} &= \abs{ z_2 - z_1 } \\ \implies \abs{z-z_4} &= r ,\] where the \(z_i\) and \(r\) are all constant, so this is the equation of a circle.