Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.1 - Complex Numbers - Exercise Set - Page 314: 23



Work Step by Step

Rationalize the denominator by multiplying the conjugate of the denominator, which is $1-i$, to both the numerator and the denominator to obtain: $=\dfrac{2i(1-i)}{(1+i)(1-i)} \\=\dfrac{2i-2i^2}{(1+i)(1-i)}$ Use the rule $(a+b)(a-b) = a^2-b^2$ to obtain: $=\dfrac{2i-2i^2}{1^2-i^2} \\=\dfrac{2i-2i^2}{1-i^2}$ Use the fact that $i^2=-1$ to obtain: $=\dfrac{2i-2(-1)}{1-(-1)} \\=\dfrac{2i+2}{1+1} \\=\dfrac{2+2i}{2} \\=\dfrac{2}{2} + \dfrac{2}{2}i \\=1+i$
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