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

Apply the function repeatedly, using each result as the next input. Starting with z₀ = −1 − 2i, we get z₁ = 2(−1 − 2i) + (3 − 2i) = 1 − 6i. Then z₂ = 2(1 − 6i) + (3 − 2i) = 5 − 14i, and z₃ = 2(5 − 14i) + (3 − 2i) = 13 − 30i. This matches choice a: 1 − 6i, 5 − 14i, 13 − 30i.

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