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 - Page 314: 31



Work Step by Step

RECALL: (1) $\sqrt{-1}=i$ (2) For any real number $a\gt0$, $\sqrt{-a} = i\sqrt{a}$. Use rule (2) above to obtain: $=5\cdot i\sqrt{16}+3 \cdot i\sqrt{81} \\=5i\sqrt{4^2} + 3i\sqrt{9^2} \\=5i(4) + 3i(9) \\=20i+27i \\=(20+27)i \\=47i$
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