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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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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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