Evaluate (√3 - i) ^ (3/4) using DeMoivre's theorem
I got
2^(3/4) (cos (-π/8) i sin (-π/8))
2^(3/4) (cos (-7π/8) i sin (-7π/8))
2^(3/4) (cos (-13π/8) i sin (-13π/8))
2^(3/4) (cos (-19π/8) i sin (-19π/8))
But according to my book this is wrong. How do you get the right answer?
Here's a link to a lesson on DeMoivre's theorem (in case anyone requires a refresher)
http://www.letu.edu/people/stevearmstrong/Math1252... 5.3.ppt
Update:On second calculation I got what Wyom got, but this is still wrong according to the book. I think this is probably a mistake on the book's part (I've seen a number of other mistakes by the book)
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I think it should go as:
(√3 - i) ^ (3/4)
2^(3/4) (cos (-π/6) + i sin (-π/6))^(3/4)
2^(3/4) (cos (2*m*π - π/6) + i sin (2*m*π - π/6))^(3/4)
2^(3/4) (cos (3/2*m*π - π/8) + i sin (3/2*m*π - π/8)) (applying theorem)
for m = 0
2^(3/4) (cos (π/8) - i sin (π/8))
for m = 1
2^(3/4) (cos (11π/8) + i sin (11π/8))
for m = 2
2^(3/4) (cos (23π/8) + i sin (23π/8))
for m = 3
2^(3/4) (cos (35π/8) + i sin (35π/8))
does that match?
2^(3/4)*[cis(-pi/6)]^(3/4)
=2^(3/4)*[cis(-pi/2)]^(1/4)
=2^(3/4)*[cis(2n*pi-pi/2)]^(1/4)
=2^(3/4)*[cis(2n*pi-pi/2)/4] where n=0,1,2,3
Hence, The answers are
2^(3/4)*[cis(-pi/8)]
2^(3/4)*[cis(3pi/8)]
2^(3/4)*[cis(7pi/8)]
2^(3/4)*[cis(11pi/8)]