SOLUTION: Use demoivre's theorem to find the three cube roots of -64i. Write your answers in both polar form and a + bi form.
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Question 1129850: Use demoivre's theorem to find the three cube roots of -64i. Write your answers in both polar form and a + bi form. Answer by ikleyn(52812) (Show Source):
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Use de Moivre's theorem to find the three cube roots of -64i. Write your answers in both polar form and a + bi form.
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de Moivre is the name of the French mathematician (1667-1754).
You can read about him from this Wikipedia article https://en.wikipedia.org/wiki/Abraham_de_Moivre
As any/every person's name, it should be written using first CAPITAL letter.
The trigonometric form (= the polar form) of the complex number "-64i" is 64*(cos(270°) + i*sin(270°)).
The modulus of "-64i" is 64, the argument is 270° = .
According to the general theory, there are three complex cube roots of "-64i". They have the modulus of = 4.
The first cube root has the argument of 90° = , one third of the argument of "-64i".
Each next cube root has the argument by = 120° = more than the previous one.
Thus the tree complex roots are
1) 4*(cos(90°) + i*sin(90°)) = ;
2) 4*(cos(90°+120°) + i*sin(90° + 120°)) = 4*(cos(210°) + i*sin(210°) = = .
3) 4*(cos(90°+240°) + i*sin(90° + 240°)) = 4*(cos(330°) + i*sin(330°)) = = .
