Question

Ind the first three iterates of the function f(z) = z2 + c with a value of c = 1 + 2i and an initial value of z0 = 0.

Answer 1

Answer:  1+2i, -2+6i, -31-22i

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Work Shown:

Plug in z = 0 and c = 1+2i. Simplify.

f(z) = z^2 + c

f(0) = 0^2 + 1+2i

f(0) = 1+2i

The output 1+2i will be the input for the next iteration

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Plug in z = 1+2i. Keep c = 1+2i the same.

f(z) = z^2 + c

f(1+2i) = (1+2i)^2 + 1+2i

f(1+2i) = 1+4i+4i^2 + 1+2i

f(1+2i) = 1+4i+4(-1) + 1+2i

f(1+2i) = 1+4i-4 + 1+2i

f(1+2i) = -2+6i

The output -2+6i will be the input for the next iteration.

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Plug in z = -2+6i. Keep c = 1+2i the same.

f(z) = z^2 + c

f(-2+6i) = (-2+6i)^2 + 1+2i

f(-2+6i) = 4-24i+36i^2 + 1+2i

f(-2+6i) = 4-24i+36(-1) + 1+2i

f(-2+6i) = 4-24i-36 + 1+2i

f(-2+6i) = -31-22i