## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$g(1+i)=-2+4i$
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}