College Trigonometry, 2011a

1002 Applications of Trigonometry
(w k ) n = ( √ n
rcis ( θ
n + 2π n k)) n
= ( n√ r) n cis ( n · [ θ
n + 2π n k])
= rcis (θ +2πk)
DeMoivre’s Theorem
Since k is a whole number, cos(θ +2πk) = cos(θ) and sin(θ +2πk) =sin(θ). Hence, it follows that
cis(θ +2πk) =cis(θ), so (w k ) n = rcis(θ) =z, as required. To show that the formula in Theorem
11.17 generates n distinct numbers, we assume n ≥ 2 (or else there is nothing to prove) and note
that the modulus of each of the w k is the same, namely n√ r. Therefore, the only way any two of
these polar forms correspond to the same number is if their arguments are coterminal – that is, if
the arguments differ by an integer multiple of 2π. Suppose k and j are whole numbers between 0
and (n − 1), inclusive, with k ≠ j. Sincek and j are different, let’s assume for the sake of argument
(
k−j
n
)
. For this to be an integer multiple of 2π,
that k>j. Then ( θ
n + 2π n k) − ( θ
n + 2π n j) =2π
(k − j) must be a multiple of n. But because of the restrictions on k and j, 0