Moivre 2

Find the cube roots of 125(cos 288° + i sin 288°).


Result

r =  5
α1 =  96 °
α2 =  216 °
α3 =  336 °

Solution:

r=1253=5r=\sqrt[3]{ 125 } = 5
α1=288/3+0⋅ 360/3=96∘\alpha _1=288/3+0 \cdot \ 360/3 = 96 ^\circ
α2=288/3+1⋅ 360/3=216∘\alpha _2=288/3+1 \cdot \ 360/3 = 216 ^\circ
α3=288/3+2⋅360/3=336∘ z1=5(cos⁡96+isin⁡96) z2=5(cos⁡216∘+isin⁡216∘) z3=5(cos⁡336∘+isin⁡336∘)\alpha _3 = 288/3+2\cdot 360/3 = 336 ^\circ \ \\ z_1 = 5 ( \cos 96 + i \sin 96 ) \ \\ z_2 = 5 ( \cos 216 ^\circ + i \sin 216 ^\circ ) \ \\ z_3 = 5 ( \cos 336 ^\circ + i \sin 336 ^\circ )

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