Answer:
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
