Answer:
C
Step-by-step explanation:
We have: (I rewrote the function)
[tex]f(z_n)={z_{n-1}} ^2+c[/tex]
Given that:
[tex]\displaystyle c=2-3i \text{ and } z_0 = 1 + 2 i[/tex]
The first iterate will be:
[tex]\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}[/tex]
The second iterate will be:
[tex]\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}[/tex]
And the third iterate will be:
[tex]\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}[/tex]
Hence, our answer is C.
