+489 A function \(f\) is defined on the complex numbers by
\(f(z) = (a + bi)z,\)
where \(a\) and \(b\) are positive real numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that
\(|a + bi| = 3\)
find \(a\) and \(b\)
+6250 I'll give you a hint.
\(\text{since }f(z) \text{ has the stated property }\forall z\\ f(1) \text{ is equidistant from }(0,0i), \text{ and }(1,0i)\\ \text{The locus of these points is }\left(\dfrac 1 2 , i y\right),~\forall y \in \mathbb{R}\\ \text{From this you can figure out }a\\ \text{Once you have }a \text{ finding }b \text{ is trivial}\)
