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put this solution on YOUR website!I am studying DeMoivre's Theorum in Aplications of Trigonometry, I was asked to find the real and complex solution to the equation
Possible answers
a. -2,2,
b. 2,
,
c. -2,
,
d.
,
,
2cis(4PI/3)}}
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The polar form for (0,8) is z = 8cis(0+2(pi)n) where n=0,1,2,3,...
The cube root is (cube rt 8)cis((2/3)(pi)n) where n=0,1,2,...
For n=0 the cube root is 2cis(0) = 2*1 = 2
For n=1 the cube root is 2cis((2/3)(pi))
For n=2 the cube root is 2cis((4/3)(pi))
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These are in answer "b".
Cheers,
Stan H.
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