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August 30, 2014

August 30, 2014

Posted by **britt** on Tuesday, January 17, 2012 at 11:28am.

{1 - i(/3)}^4

- trigonometry -
**Reiny**, Tuesday, January 17, 2012 at 2:11pmI will assume you know the theorem and the common variables used in it

r = √(1^2 + (1/3)^2 )

= √(1 + 1/9)

= √(10/9)

√10/3

let the angle be Ø

cosØ = 1/(√10/3) = 3/√10

sinØ = (-1/3) / (√10/3) = -3/√10 , so Ø is in IV and Ø = 341.565°

so (1 - (1/3) i )^4 = (√10/3)^4 [ cos 4(341.565° + i sin 4(341.564°)

= (100/81) [ .28 + (-.96i) ]

= 28/81 - 96/81i

= 28/81 - 32/27 i

=

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