Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 54


$\sqrt \frac{-81}{16}=\frac{9}{4}i$

Work Step by Step

$i=\sqrt -1$ hence we can write $\sqrt \frac{-81}{16}$ as $\sqrt -1\times\sqrt \frac{81}{16}=\sqrt -1\times\frac{9}{4}=i\times\frac{9}{4}=\frac{9}{4}i$. Hence the answer written in the form $a+bi$ where $a$ is the real part and $b$ is the imaginary part is $\frac{9}{4}i$ as $a=0$ and $b=\frac{9}{4}$
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