Answer
$-e^{2}$
Work Step by Step
Given: $e^{2+i\pi}$
$e^{2+i\pi}=e^{2}e^{i\pi}$
$=e^{2}(cos(\pi)+isin(\pi))$
$=e^{2}(-1+i0)$
$=-e^{2}$
Multivariable Calculus, 7th Edition
by Stewart, James
Published by
Brooks Cole
ISBN 10:
0-53849-787-4
ISBN 13:
978-0-53849-787-9
Appendix G - Exercises - Page A12: 45
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