## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises: 54

#### Answer

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

#### 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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