Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review - Page 470: 139



Work Step by Step

Multiplying the numerator and the denominator by $i$, the expression $ \dfrac{2+3i}{2i} $ simplifies to \begin{array}{l} \dfrac{2+3i}{2i} \cdot \dfrac{i}{i} \\\\= \dfrac{2i+3i^2}{2i^2} \\\\= \dfrac{2i+3(-1)}{2(-1)} \\\\= \dfrac{-3+2i}{-2} \\\\= -\dfrac{-3+2i}{2} \\\\= \dfrac{3-2i}{2} .\end{array}
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