Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.1 Complex Numbers - 8.1 Exercises - Page 358: 85

Answer

$-1-2i$

Work Step by Step

Step 1: Multiplying both the numerator and the denominator of the expression by the complex conjugate of the denominator: $\frac{1-3i}{1+i}\times\frac{1-i}{1-i}$ Step 2: $\frac{1-3i}{1+i}\times\frac{1-i}{1-i}=\frac{(1-3i)(1-i)}{(1)^{2}-(i)^{2}}$ Step 3: $\frac{(1-3i)(1-i)}{(1)^{2}-(i)^{2}}=\frac{1-i-3i+3i^{2}}{1+1}=\frac{1-4i+3(-1)}{2}$ Step 4: $\frac{1-4i+3(-1)}{2}=\frac{1-4i-3}{2}=\frac{-2-4i}{2}=-1-2i$
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