Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.1 - Complex Numbers - Exercise Set: 34

Answer

$16+30i$

Work Step by Step

RECALL: (1) $(a-b)^2 = a^2-2ab+b^2$ (2) $\sqrt{-1}=i$ (3) $i^2=-1$ (4) For any real number $a \gt 0$, $\sqrt{-a} = i\sqrt{a}$. Use rule (4) above to obtain: $=(-5-i\sqrt{9})^2 \\=(-5-i\sqrt{3^2})^2 \\=(-5-3i)^2$ Use rule (1) above with $a=-5$ and $b=3i$ to obtain: $=(-5)^2-2(-5)(3i) + (3i)^2 \\=25+30i+9i^2$ Use rule (3) above to obtain: $=25+30i+9(-1) \\=25+30i-9 \\=16+30i$
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