A Fifth Root
Suppose z is a complex number such that z 5 = − 3 + i and z = 5 2 ( cos n ∘ + i sin n ∘ ) , where 2 7 0 < n < 3 6 0 . Find the exact value of n .
Details and assumptions
i is the imaginary unit, where i 2 = − 1 .
The answer is 318.
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2 solutions
how did u get 5n=150+360 x ???
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Suppose cis ( a ) ∘ = − 2 3 + 2 i . We plot the latter on the unit circle, and find that a = 1 5 0 works. However, notice that if we go around the unit circle once, we will land on the same point again. Therefore we can add or subtract 3 6 0 ∘ from the answer and still get a valid answer. Therefore a = 1 5 0 + 3 6 0 x . Simply replacing a = 5 n gives the desired expression.
By De Moivre's theorem, z 5 = ( 5 2 ( cos ( n ) + i sin ( n ) ) ) 5 = 2 ( cos ( 5 n ) + i sin ( 5 n ) ) . But z 5 is also − 3 + i = 2 ( − 2 3 + 2 1 i ) = 2 ( cos ( 1 5 0 + 3 6 0 k ) + i sin ( 1 5 0 + 3 6 0 k ) ) for integer k . Therefore 5 n = 1 5 0 + 3 6 0 k ⟺ n = 3 0 + 7 2 k , the value of n such that 2 7 0 < n < 3 6 0 is 3 1 8 .
By De Moivre's Theorem, z 5 = 2 cis ( 5 n ) ∘ . We set this equal to − 3 + i and divide by 2 to get cis ( 5 n ) ∘ = − 2 3 + 2 i . Therefore 5 n = 1 5 0 + 3 6 0 x , where x is an integer. Solving for n gives n = 3 0 + 7 2 x , and finding that x = 4 puts us in the desired range gives that n = 3 1 8 .