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Lilit [14]
2 years ago
13

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.

Mathematics
1 answer:
Artist 52 [7]2 years ago
3 0

Answer:

C

Step-by-step 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.

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