Find the first three iterates of the function f(z) = 2z + (3 - 2i) with an initial value of z0 =1 + 2i

The three iterates are:

z_1=8i+3z

1

=8i+3

z_2=14i+9z

2

=14i+9

z_3=26i+21z

3

=26i+21

Step-by-step explanation:

We are asked to find the first three iterates of the function:

f(z) =2z+(3-2i)f(z)=2z+(3−2i) with an initial value:

z_0=5iz

0

=5i

The first iterate is:

\begin{gathered}z_1=f(z_0)=2z_0+(3-2i)z_1=2\times 5i+3-2iz_1=10i-2i+3z_1=8i+3\end{gathered}

z

1

=f(z

0

)=2z

0

+(3−2i)

z

1

=2×5i+3−2i

z

1

=10i−2i+3

z

1

=8i+3

Now, the second iterate is:

\begin{gathered}z_2=f(z_1)=2z_1+(3-2i)z_2=2\times (8i+3)+3-2iz_2=16i+6+3-2iz_2=14i+9\end{gathered}

z

2

=f(z

1

)=2z

1

+(3−2i)

z

2

=2×(8i+3)+3−2i

z

2

=16i+6+3−2i

z

2

=14i+9

Third iterate is:

\begin{gathered}z_3=f(z_2)=2z_2+(3-2i)z_3=2\times (14i+9)+3-2iz_3=28i+18+3-2iz_3=26i+21\end{gathered}

z

3

=f(z

2

)=2z

2

+(3−2i)

z

3

=2×(14i+9)+3−2i

z

3

=28i+18+3−2i

z

3

=26i+21