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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set: 116

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}
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