Answer
$g(2-3i)=-51-21i$
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 $
2-3i
$ in the function above results to
\begin{array}{l}\require{cancel}
g(z)=z^2(z+1)
\\\\
g(2-3i)=(2-3i)^2(2-3i+1)
\\\\
g(2-3i)=[(2)^2-2(2)(3i)+(3i)^2](3-3i)
\\\\
g(2-3i)=(4-12i+9i^2)(3-3i)
\\\\
g(2-3i)=4(3)+4(-3i)-12i(3)-12i(-3i)+9i^2(3)+9i^2(-3i)
\\\\
g(2-3i)=12-12i-36i+36i^2+27i^2-27i^3
\\\\
g(2-3i)=12-48i+63i^2-27i^3
.\end{array}
Using $i^2=-1,$ the function above is equivalent to
\begin{array}{l}\require{cancel}
g(2-3i)=12-48i+63i^2-27i^3
\\\\
g(2-3i)=12-48i+63i^2-27i^2\cdot i
\\\\
g(2-3i)=12-48i+63(-1)-27(-1)\cdot i
\\\\
g(2-3i)=12-48i-63+27i
\\\\
g(2-3i)=-51-21i
.\end{array}
