Question

Find the first three iterates of the function f(z) = z2 + c with a value of c = 2 - 3i and an initial value of z0 = 1 + 2i.

a.
-1 + i, 2 - 3i, -17 - 29i
c.
1 + i, 2 + 3i, -17 - 29i
b.
-1 + i, 2 - 5i, -19 - 23i
d.
1 + i, 2 + 3i, -19 - 23i

Answer 1

C

Explanation:

We have: (I rewrote the function)

`f(z_n)={z_(n-1)} ^2+c`

Given that:

`\displaystyle c=2-3i \text{ and } z_0 = 1 + 2 i`

The first iterate will be:

`\displaystyle \begin{aligned} f(z_1)&=(z_0)^2+c \\ &=(1+2i)^2+(2-3i) \\ &= (1+4i+4i^2)+(2-3i) \\ &=1+4i-4+2-3i \\ &=-1+i \end{aligned}`

The second iterate will be:

`\begin{aligned}f(z_2) &=(z_1)^2+c\\ &=(-1+i)^2+(2-3i) \\&= (1-2i+i^2)+(2-3i) \\&=1-2i-1+2-3i \\&=2-5i \end{aligned}`

And the third iterate will be:

`\begin{aligned} f(z_3)&=(z_2)^2+c\\ &=(2-5i)^2+(2-3i) \\ &=(4-20i+25i^2)+(2-3i) \\ &=4-20i-25+2-3i \\ &=-19-23i \end{aligned}`

Hence, our answer is C.