Answer
$e^{\pi i}+1=0$
Work Step by Step
We are given that $e^{a+ib}=e^a (\cos b+i \sin b)$
Plug $a=0; b=\pi$
$e^{0+\pi i}=e^{0} (\cos \pi+i \sin \pi)$
This gives: $e^{\pi i}=-1+0$
and $e^{\pi i}+1=0$
Hence, the result has been proved.
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 13, Trigonometric Ratios and Functions - 13.3 Evaluate Trigonometric Functions of Any Angle - 13.3 Exercises - Skill Practice - Page 871: 34
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