Weve seen this before its the angle of elevation

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Unformatted text preview: x number in polar form z = |z |cis(θ) and solve wn = z in the same manner as above, we arrive at the following theorem. Theorem 11.17. The nth roots of a Complex Number: Let z = 0 be a complex number with polar form z = rcis(θ). For each natural number n, z has n distinct nth roots, which we denote by w0 , w1 , . . . , wn − 1 , and they are given by the formula wk = √ n rcis θ 2π + k n n The proof of Theorem 11.17 breaks into to two parts: first, showing that each wk is an nth root, and second, showing that the set {wk : k = 0, 1, . . . , (n − 1)} consists of n different complex numbers. To show wk is an nth root of z , we use DeMoivre’s Theorem to show (wk )n = z . 16 The reader is challenged to find all of the complex solutions to w5 = 32 using the techniques in Chapter 3. 11.7 Polar Form of Complex Numbers (wk )n = √ n rcis θ n + √n = ( n r) cis n n 2π nk θ π · n + 2n k 853 DeMoivre’s Theorem = rcis (θ + 2πk ) 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 (wk )n = rcis(θ) = z , as required....
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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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