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: 79

Answer

$$(2+i)(2-i)(4+3i)=20+15i$$

Work Step by Step

$$A=(2+i)(2-i)(4+3i)$$ For $(2+i)(2-i)$, we can still use the formula $$(a+b)(a-b)=a^2-b^2$$ which means $$(2+i)(2-i)$$ $$=4-i^2$$ $$=4-(-1)$$ $$=5$$ Now we can apply the result just found to $A$ $$A=5(4+3i)$$ $$A=20+15i$$
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