Answer
$g(1+i)=-2+4i$
Work Step by Step
Factoring the $GCF=z^2$ and using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the given function, $
g(z)=\dfrac{z^4-z^2}{z-1}
,$ simplifies to
\begin{array}{l}\require{cancel}
g(z)=\dfrac{z^4-z^2}{z-1}
\\\\
g(z)=\dfrac{z^2(z^2-1)}{z-1}
\\\\
g(z)=\dfrac{z^2(z+1)(z-1)}{z-1}
\\\\
g(z)=\dfrac{z^2(z+1)(\cancel{z-1})}{\cancel{z-1}}
\\\\
g(z)=z^2(z+1)
.\end{array}
Substituting $z$ with $
1+i
$ in the function above results to
\begin{array}{l}\require{cancel}
g(z)=z^2(z+1)
\\\\
g(1+i)=(1+i)^2(1+i+1)
\\\\
g(1+i)=[(1)^2+2(1)(i)+(i)^2](2+i)
\\\\
g(1+i)=(1+2i+i^2)(2+i)
\\\\
g(1+i)=1(2)+1(i)+2i(2)+2i(i)+i^2(2)+i^2(i)
\\\\
g(1+i)=2+i+4i+2i^2+2i^2+i^3
\\\\
g(1+i)=2+5i+4i^2+i^3
.\end{array}
Using $i^2=-1,$ the function above is equivalent to
\begin{array}{l}\require{cancel}
g(1+i)=2+5i+4i^2+i^3
\\\\
g(1+i)=2+5i+4i^2+i^2\cdot i
\\\\
g(1+i)=2+5i+4(-1)+(-1)\cdot i
\\\\
g(1+i)=2+5i-4-i
\\\\
g(1+i)=-2+4i
.\end{array}
