Answer: choice D) 4 + 3i
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Plug in the initial value z = 2 + 3i. Simplify.
f(z) = z^2 + c
f(2+3i) = (2+3i)^2 + c
f(2+3i) = (2+3i)*(2+3i) + c
f(2+3i) = 2(2+3i) + 3i(2+3i) + c
f(2+3i) = 4 + 6i + 6i + 9i^2 + c
f(2+3i) = 4 + 12i + 9i^2 + c
f(2+3i) = 4 + 12i + 9(-1) + c
f(2+3i) = 4 + 12i - 9 + c
f(2+3i) = -5 + 12i + c
The initial z value is z = 2+3i, which when plugged into f(z) leads to the term z1 = -1+15i. The input is 2+3i and the output is -1+15i
Set f(2+3i) equal to the output z1 = -1+15i and solve for c
f(2+3i) = z1
-5 + 12i + c = -1 + 15i
c = -1 + 15i + 5 - 12i
c = 4 + 3i
So if c = 4 + 3i, then f(z) = z^2 + c will have f(2+3i) = -1 + 15i
