jmc

Question

A function f is defined on the complex numbers by f(z)=(a+bi)z, where a and b are positive numbers. This function has the property that for each complex number z, f(z) is equidistant from both z and the origin. Given that |a+bi|=8, find b^2.

Accepted Answer

From the given property, |f(z) - z| = |f(z)|.Then |(a + bi) z - z| = |(a + bi)z|,so |a + bi - 1||z| = |a + bi||z|. Since this holds for all complex numbers z, |a + bi - 1| = |a + bi| = 8.Then (a - 1)^2 + b^2 = 64 and a^2 + b^2 = 64. Subtracting these equations, we get 2a - 1 = 0, so a = 12. Hence, b^2 = 64 - a^2 = 64 - 14 = 2554.