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**Unformatted text preview: **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 wk 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 . Since k and j are diﬀerent, let’s assume for the sake of argument
θ
π
π
θ
that k > j . Then n + 2n k − n + 2n j = 2π k−j . For this to be an integer multiple of 2π ,
n
(k − j ) must be a multiple of n. But because of the restrictions on k and j , 0 < k − j ≤ n − 1.
(Think this through.) Hence, (k − j ) is a positive number less than n, so it cannot be a multiple
of n. As a result, wk and wj are diﬀerent complex numbers, and we are done. By Theorem 3.14,
we know there at most n distinct solutions to wn = z , and we have just found all fo them. We
illustrate Theorem 11.17 in the next example.
Example 11.7.4. Use Theorem 11.17 to ﬁnd the following:
√
1. both square roots of z = −2 + 2i 3
2. the four fourth roots of z = −16
√
√
3. the th...

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