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Question 984794: Find the fifth roots of 32(cos 280° + i sin 280°).
Answer by srinivas.g(540)   (Show Source):
You can put this solution on YOUR website! i) Let z = 32(cos 280° + i sin 280°) = 32{cos(k*360° + 280°) + isin(k*360° + 280°)},
where k is an integer.
ii) ==> z^(1/5) = [32{cos(k*360° + 280°) + isin(k*360° + 280°)}]^(1/5)
==> z^(1/5) = 2[cos{(k*360° + 280°)/5} + isin{(k*360° + 280°)/5}]
[Application of De-Moievere's theorem]
==> z^(1/5) = 2[cos(72k + 56) + isin(72k + 56)]
We can now get the 5 roots, by assigning k = 0, 1, 2, 3 & 4
When k = 0, 1st root = 2(cos 56° + isin 56°)
When k = 1, 2nd root = 2(cos 128° + isin 128°)
When k = 2, 3rd root = 2(cos 200° + isin200°)
When k = 3, 4th root = 2(cos 272° + isin 272°)
When k = 4, 5th root = 2(cos 344° + isin344°)
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