Find the first three iterates of the function f(z) = 2z + (3 - 2i) with an initial value of
z0 = -1 - 2i.
a.
1 - 6i, 5 - 14i, 13 - 30i
c.
5 - 2i, 13 - 2i, 29 - 2i
b.
2 - 5i, 9 - 17i, 21 - 37i
d.
3 - 2i, 17 - 2i, 23 - 2i

select the best answer from the choices provided

a
b
c
d

a.  1 - 6i, 5 - 14i, 13 - 30i

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 = 1 - 6i

z1 =  1 - 6i  ==>  f(z) = 2( 1 - 6i ) + 3 - 2i = 2 - 12i + 3 -2i = 5 - 14i

z2 = 5 - 14i ==>  f(z) = 2(5 - 14i) = 10 - 28i + 3 - 2i = 13 - 30i

z3 =13 - 30i ==>  f(z) = 2(13 - 30i) + 3 - 2i = 26 - 60i + 3 - 2i = 29 -62i

£56

step-by-step explanation:

u divide 24 by 3 and u get 8 so if u do 8 time 7 or 8+8+8+8+8+8+8 u get 56 and 56 divided by 7 is 8.

let's fu(!

step-by-step explanation: