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Stitz-Zeager_College_Algebra_e-book

# B using a graphing utility as needed verify that the

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Unformatted text preview: ree cube roots of z = 2 + i 2 4. the ﬁve ﬁfth roots of z = 1. Solution. √ π 1. We start by writing z = −2 + 2i 3 = 4cis 23 . To use Theorem 11.17, we identify r = 4, π θ = 23 and n = 2. We know that z has two square roots, and in keeping with the notation √ π in Theorem 11.17, we’ll call them w0 and w1 . We get w0 = 4cis (2π/3) + 22 (0) = 2cis π 2 3 √ (2π/3) 2π 4π and w1 = 4cis + 2 (1) = 2cis 3 . In rectangular form, the two square roots of √2 √ z are w0 = 1 + i 3 and w1 = −√− i 3. We can check our answers by squaring them and 1 showing that we get z = −2 + 2i 3. 2. Proceeding as above, we get z = −16 = 16cis(π ). With r = 16, θ = π and n = 4, we get the √ √ π π four fourth roots of z to be w0 = 4 16cis π + 24 (0) = 2cis π , w1 = 4 16cis π + 24 (1) = 4 4 4 √ √ π π π π π 2cis 34 , w2 = 4 16cis π + 24 (2) = 2cis 54 and w3 = 4 16cis π + 24 (3) = 2cis 74 . 4 √ √ √4√ √ √ Converting these√ rectangular form gives w0 = 2+ i 2,...
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