Answer
$g(3i)=-9-27i$
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 $
3i
$ in the function above results to
\begin{array}{l}\require{cancel}
g(z)=z^2(z+1)
\\\\
g(3i)=(3i)^2(3i+1)
\\\\
g(3i)=9i^2(3i+1)
\\\\
g(3i)=27i^3+9i^2
.\end{array}
Using $i^2=-1,$ the function above is equivalent to
\begin{array}{l}\require{cancel}
g(3i)=27i^3+9i^2
\\\\
g(3i)=27i^2\cdot i+9i^2
\\\\
g(3i)=27(-1)\cdot i+9(-1)
\\\\
g(3i)=-27i-9
\\\\
g(3i)=-9-27i
.\end{array}
