Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Test - Page 471: 30

Answer

$\dfrac{-3+5i}{2}$

Work Step by Step

Multiplying by the conjugate of the denominator, the given expression, $ \dfrac{1+4i}{1-i} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{1+4i}{1-i}\cdot\dfrac{1+i}{1+i} \\\\= \dfrac{1(1)+1(i)+4i(1)+4i(i)}{1^2-i^2} \\\\= \dfrac{1+i+4i+4i^2}{1-(-1)} \\\\= \dfrac{1+i+4i+4(-1)}{1+1} \\\\= \dfrac{1+i+4i-4}{2} \\\\= \dfrac{-3+5i}{2} .\end{array}
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