Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises - Page 113: 101


$(\displaystyle \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i )^{2}=i$ So, $\displaystyle \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ is a square root of $i$.

Work Step by Step

$(\displaystyle \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i )^{2}\qquad$...Combine the fractions $=(\displaystyle \frac{\sqrt{2}+\sqrt{2}i}{2})^{2}\qquad$...Apply exponent rule: $(\displaystyle \frac{a}{b})^{n}=\frac{a^{n}}{b^{n}}$ $=\displaystyle \frac{(\sqrt{2}+\sqrt{2}i)^{2}}{2^{2}}\qquad$...Apply Perfect Square Formula: $(a+b)^{2}=a^{2}+2ab+b^{2},a=\sqrt{2},b=\sqrt{2}i$ $=\displaystyle \frac{(\sqrt{2})^{2}+2\sqrt{2}\sqrt{2}i+(\sqrt{2}i)^{2}}{4}$ $=\displaystyle \frac{2+2\cdot 2i+2i^{2}}{4}\qquad$...Apply imaginary number rule: $\quad i^{2}=-1$ $=\displaystyle \frac{2+4i+2\cdot(-1)}{4}\qquad$...Simplify. $=\displaystyle \frac{4i}{4}$ $=i$ Therefore, $(\displaystyle \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i )^{2}=i$
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