Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises: 56

Answer

$(-2+3i)(4-2i)=-2+16i$

Work Step by Step

$(-2+3i)(4-2i)$ Evaluate the product by multiplying each term of the first factor by every term of the second factor: $(-2+3i)(4-2i)=-8+4i+12i-6i^{2}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=-8+4i+12i-6(-1)=-8+16i+6=-2+16i$
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