Find the first three iterates of the function f(z) = 2z + (3 - 2i) with an initial value of
z0 = 1 + 2i.
a.
5 + 2i, 13 + 2i, 29 + 2i
c.
5 - 2i, 13 - 2i, 29 - 2i
b.
3 + 2i, 17 + 2i, 23 + 2i
d.
3 - 2i, 17 - 2i, 23 - 2i

select the best answer from the choices provided

a
b
c
d

a.  5 + 2i, 13 + 2i, 29 + 2i

Step-by-step explanation:

We'll use the formula  f(z) = 2z + (3 - 2i)  for each iteration. The output of the first iteration will be come the input of the second iteration, and so on.

So, we start with z0 = 1 + 2i and we plug that into the base equation:

z0 = 1 + 2i ==> f(z) = 2(1 + 2i) + 3 - 2i = 2 + 4i + 3 - 2i = 5 + 2i

z1 = 5 + 2i ==> f(z) = 2(5 + 2i) + 3 - 2i = 10 + 4i + 3 - 2i = 13 + 2i

z2 = 13 + 2i ==> f(z) = 2(13 + 2i) + 3 - 2i = 26 + 4i  + 3 - 2i = 29 + 2i

z3 = 29 + 2i

-14 is the answer

hope this : ))

all but the second one

step-by-step explanation:

cuz that one goes over 180

mark brainliest