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How did it became -2pi/27 from the equation (-2pi/3+2pi(0))/9? Please explain.

How did it became -2pi/27 from the equation (-2pi/3+2pi(0))/9? Please explain.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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How did it became -2pi/27 from the equation (-2pi/3+2pi(0))/9? Please explain.
The polar form of –256 – 256/3i is 512 (cos (-) + i sin (-)) (for steps, see polar form calculator). According to the De Moivre's Formula, all n-th roots of a complex number r (cos (0) + i sin (0)) 0..n – 1. zk), k = are given by ri (cos (0+2rk) + i sin (0+2nk n We have that r = 512, 0 : and n = 3 9. k = 0: 512 ( cos + i sin - +2-7-0 9. 9. 2 (cos (-) i sin (-)) = 2 cos () – 2i sin ()
Transcribed Image Text:The polar form of –256 – 256/3i is 512 (cos (-) + i sin (-)) (for steps, see polar form calculator). According to the De Moivre's Formula, all n-th roots of a complex number r (cos (0) + i sin (0)) 0..n – 1. zk), k = are given by ri (cos (0+2rk) + i sin (0+2nk n We have that r = 512, 0 : and n = 3 9. k = 0: 512 ( cos + i sin - +2-7-0 9. 9. 2 (cos (-) i sin (-)) = 2 cos () – 2i sin ()
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