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Unformatted text preview: (β ) − sin(α) sin(β ))
Since i2 = −1
+ i (sin(α) cos(β ) + cos(α) sin(β )) Factor out i
= cos(α + β ) + i sin(α + β ) Sum identities = cis(α + β ) Deﬁnition of ‘cis’ Putting this together with our earlier work, we get zw = |z ||w|cis(α + β ), as required.
Moving right along, we next take aim at the Power Rule, better known as DeMoivre’s Theorem.11
We proceed by induction on n. Let P (n) be the sentence z n = |z |n cis(nθ). Then P (1) is true, since
z 1 = z = |z |cis(θ) = |z |1 cis(1 · θ). We now assume P (k ) is true, that is, we assume z k = |z |k cis(kθ)
for some k ≥ 1. Our goal is to show that P (k + 1) is true, or that z k+1 = |z |k+1 cis((k + 1)θ). We
z k+1 = z k z Properties of Exponents = |z |k cis(kθ) = |z |k |z | =
11 (|z |cis(θ)) Induction Hypothesis cis(kθ + θ) |z |k+1 cis((k Product Rule + 1)θ) Compare this proof with the proof of the Power Rule in Theorem 11.14. 11.7 Polar Form of Complex Numbers 849 Hence, assuming P (k ) is true, we have that P (k + 1) is true, so by the...
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