College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Mid-Chapter Check Point - Page 166: 43

Answer

$\dfrac{1+i}{1-i}=i$

Work Step by Step

$\dfrac{1+i}{1-i}$ Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator: $\dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\cdot\dfrac{1+i}{1+i}=\dfrac{(1+i)^{2}}{1^{2}-i^{2}}=\dfrac{1+2i+i^{2}}{1-i^{2}}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=\dfrac{1+2i-1}{1-(-1)}=\dfrac{2i}{1+1}=\dfrac{2i}{2}=i$
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