Question

Function f(z)=2z-(3+i)

Answer 1

F(z) = 2z - (3+i)

1) second interate for z0 = i

first iterate = f(z0) = f(i) = 2i - 3 - i = i - 3 = z1
second iterate = f(z1) = f(i-3) = 2(i-3) - 3 - i = 2i - 6 - 3 - i = i - 9

2) third iterate for z0= 3 - i

first iterate = f(z0) = f(3 - i) = 2(3-i) - (3+i) = 6 -2i -3 - i =3 - 3i = z1
second iterate = f(z1) = f(3 -3i) = 2(3-3i) - (3+i) = 6 - 6i -3 -i = 3 - 7i = z2
third iterate = f(z2) = f(3 - 7i) = 2 (3-7i) - (3+i) = 6 - 14i - 3 - i = 3 -15i = z3.

3) first iterate for z0=0.5+i

first iterate = f(z0) = f(0.5 +i) = 2(0.5+i) - (3+i) = 1 +2i - 3 - i = - 2 + i = z1

4) third iterate for z0=-2-5i

first iterate = f(z0) = 2(- 2 - 5i) - (3 + i) = - 4 - 10i - 3 - i = -7 - 11i = z1
second iterate = f(z1) = 2( - 7 - 11i) - (3+i) = -14 - 22i - 3 - i = -17 -23i = z2
third iterate = f(z2) = 2(-17-23i) -(3+i) = -34 - 46i -3 - i = -37 - 47i = z3