Chapter 4 Quadratic Functions and Factoring - 4.6 Perform Operations with Complex Numbers - 4.6 Exercises - Problem Solving - Page 281: 73

$c=-0.5i$ does belong to the Mandelbrot set.

Let $f (z)= z ^2 +(-0.5i)$ $z_0=0$ $|z_0|=0$ $z_1=f(0)=0^2+(-0.5i)=-0.5i$ $|z_1|=\frac{1}{2}$ $z_2=f(-0.5i)=(0.5i)^2+(-0.5i)=-0.25-0.5i$ $|z_2|=\frac{\sqrt 5}{4}$ $z_3=f(-0.25-0.5i)=(-0.25-0.5i)^2+(-0.5i)=-\frac{3}{16}-0.25i$ $|z_3|=\frac{5}{16}$ $c=-0.5i$ does belong to the Mandelbrot set.