Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 8 - Review - Exercises - Page 622: 35

Answer

$16*(-1/2+\sqrt 3*i)$

Work Step by Step

Let us first write the expression in cis notation, We see that $1^2+(-\sqrt 3)^2=4$ Thus the magnitude of the polar number is 2. Also from the unit circle, we recognize these values as 300 degrees We can rewrite this as (degrees) $(2cis(300))^4 = 2^4*cis(1200)=16*cis(120) = 16*(-1/2+\sqrt 3*i)$
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