# Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set: 113

$g(3i)=-9-27i$

#### 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}

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