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**Unformatted text preview: **w1 = − 2+ i 2, w2 = − 2 − i 2,
to
√
and w3 = 2 − i 2. 854 Applications of Trigonometry √
√
3. For z = 2+ i 2, we have z = 2cis π . With r = 2, θ = π and n = 3 the usual computations
4
4
√
√
√
√
π
π
7π
yield w0 = 3 2cis 12 , w1 = 3 2cis 9π = 3 2cis 34 and w2 = 3 2cis 112 . If we were
12
to convert these to rectangular form, we would need to use either the Sum and Diﬀerence
Identities in Theorem 10.16 or the Half-Angle Identities in Theorem 10.19 to evaluate w0 and
w2 . Since we are not explicitly told to do so, we leave this as a good, but messy, exercise.
4. To ﬁnd the ﬁve ﬁfth roots of 1, we write 1 = 1cis(0). We have r = 1, θ = 0 and n = 5.
√
π
π
π
Since 5 1 = 1, the roots are w0 = cis(0) = 1, w1 = cis 25 , w2 = cis 45 , w3 = cis 65 and
π
w4 = cis 85 . The situation here is even graver than in the previous example, since we have
π
no identities developed to help us determine the cosine or sine of 25 . At this stage, we could
approximate our answers using a calculator, and we leave this to the Exercises.
Now that we have done some computations using Theorem 11.17, we take a step back to look
at things geometrically. Es...

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