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: 33

Answer

$-8i$

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: $=(-2+i\sqrt{4})^2 \\=(-2+i\sqrt{2^2})^2 \\=(-2+2i)^2$ Use rule (1) above with $a=-2$ and $b=2i$ to obtain: $=(-2)^2+2(-2)(2i) + (2i)^2 \\=4-8i+4i^2$ Use rule (3) above to obtain: $=4-8i+4(-1) \\=4-8i-4 \\=-8i$
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