Find the first three iterates of the function f(z) = 2z + (3 - 2i) with an initial value of z0 = 4.

First Iterate: 11-2i

Second Iterate: 25-6i

Third Iterate: 53-14i

Step-by-step explanation:

Given:

f(z)=2z+(3-2i)

And  

z_0=4

For the first iterate we have to put the value of z0 into the function in place of z.

First Iterate:

z_1=f(z_0 )=2z_0+(3-2i)

=2(4)+(3-2i)

=8+3-2i

=11-2i

Second Iterate:

For the second iterate we have to put the value of z1 into the function in place of z.

z_2=f(z_1 )=2z_1+(3-2i)

=2(11-2i)+(3-2i)

=22-4i+3-2i

=25-6i

Third Iterate:

For the third iterate we have to put the value of z2 into the function in place of z.

z_3=f(z_2 )=2z_2+(3-2i)

=2(25-6i)+(3-2i)

=50-12i+3-2i

=53-14i

answer: the answer is 35%

range is infinite on both x and y axis positive and negative

step-by-step explanation: