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: 86

Answer

$-2+i$

Work Step by Step

Step 1: Multiplying both the numerator and the denominator of the expression by the complex conjugate of the denominator: $\frac{-3+4i}{2-i}\times\frac{2+i}{2+i}$ Step 2: $\frac{-3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{(-3+4i)(2+i)}{(2)^{2}-(i)^{2}}$ Step 3: $\frac{(-3+4i)(2+i)}{(2)^{2}-(i)^{2}}=\frac{-6-3i+8i+4i^{2}}{4-(-1)}=\frac{-6+5i+4(-1)}{5}$ Step 4: $\frac{-6+5i+4(-1)}{5}=\frac{-10+5i}{5}=\frac{5(-2+i)}{2}=-2+i$
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