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: 36

Answer

$-7+4i\sqrt{11}$

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: $=(-2-i\sqrt{11})^2$ Use rule (1) above with $a=-2$ and $b=i\sqrt{11}$ to obtain: $=(-2)^2-2(-2)(i\sqrt{11}) + (i\sqrt{11})^2 \\=4+4i\sqrt{11}+i^2(11) \\=4+4i\sqrt{11} + 11i^2$ Use rule (3) above to obtain: $=4+4i\sqrt{11} + 11(-1) \\=4+4i\sqrt{11} -11 \\=-7+4i\sqrt{11}$
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