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
\([-2+2i,-2-6i,-34+26i]\)
