Question

Find the first three output values of the fractal-generating function f(z)=z^2-2+2i

Answer 1

f ( z ) = z² - 2 + 2 i1 ) z = 0:f ( 0 ) = 0² - 2 + 2 i = - 2 + 2 i2 ) f ( - 2 + 2 i ) = ( - 2 + 2 i )² - 2 + 2 i = = 4 - 8 i + 4 i² - 2 + 2 i = 4 - 8 i - 4 - 2 + 2 i = - 2 - 6 i3 ) f ( - 2 - 6 i ) = ( - 2 - 6 i )² - 2 + 2 i = 4 + 24 i + 36 i² - 2 + 2 i == 4 + 24 i - 36 - 2 + 2 i = - 34 + 26 i.

Answer 2

we have

`f(z)=z^2-2+2i`

remember that

`i^(2)=-1`

Step 1

Find the first output value for
`z=0`

substitute in the fractal-generating function

`f(0)=0^2-2+2i`

`f(0)=-2+2i`

Step 2

Find the second output value for
`z=-2+2i`

substitute in the fractal-generating function

`f(-2+2i)=(-2+2i)^2-2+2i`

`= 4-8i+4i^2-2+2i`

`=2-6i+4*(-1)\\= 2-6i-4\\=-2-6i`

so

`f(-2+2i)=-2-6i`

Step 3

Find the third output value for
`z=-2-6i`

substitute in the fractal-generating function

`f(-2-6i)=(-2-6i)^2-2+2i`

`=4+24i+36i^2-2+2i`

`= 2+26i+36*(-1)\\= 2+26i-36\\=-34+26i`

so

`f(-2-6i)=-34+26i`

therefore

the answer is

the first three output values of the fractal-generating function are

`[-2+2i,-2-6i,-34+26i]`