Remember that |a + bi| = a^2 + b^2.
So from equations, (z - 3)^2 + (z - 5)^2 = 4 and (z - 6)^2 + (z - 6)^2 = 4.
(x - 3)^2 = 4, (y - 5)^2 = 4, (x - 6)^2 = 4, (y - 6)^2 = 4
Two possible values of x, two possible value of y. Therefore, four solutions for z.
