By Von Heinrich Suter
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Using de Moivre’s theorem to expand (cos t + i sin t)5 , then using the binomial theorem, we have cos 5t + i sin 5t = cos5 t + 5i cos4 t sin t + 10i2 cos3 t sin2 t + 10i3 cos2 t sin3 t + 5i4 cos t sin4 t + i5 sin5 t. 1 Abraham de Moivre (1667–1754), French mathematician, a pioneer in probability theory and trigonometry. 40 2 Complex Numbers in Trigonometric Form Hence cos 5t + i sin 5t = cos5 t − 10 cos3 t (1 − cos2 t) + 5 cos t (1 − cos2 t)2 + i(5(1 − sin2 t)2 sin t − 10(1 − sin2 t) sin3 t + sin5 t).
A − b)2 + (b − c)2 + 2(a − b)(b − c) + (c − a)2 = 2(a − b)(b − c). 1 Algebraic Representation of Complex Numbers 19 It follows that (a − c)2 = (a − b)(b − c). Taking absolute values, we obtain β 2 = γα, where α = |b − c|, β = |c − a|, γ = |a − b|. In an analogous way, we obtain α2 = βγ and γ 2 = αβ. , (α − β)2 + (β − γ)2 + (γ − α)2 = 0. Hence α = β = γ. 9 Problems 1. Consider the complex numbers z1 = (1, 2), z2 = (−2, 3), and z3 = (1, −1). Compute the following: (a) z1 + z2 + z3 ; (b) z1 z2 + z2 z3 + z3 z1 ; (c) z1 z2 z3 ; z1 z2 z3 z 2 + z22 + + ; (f) 12 .
For n = 2, the equation Z 2 − 1 = 0 has the roots −1 and 1, which are the square roots of unity. 2. , the roots of equation Z 3 − 1 = 0, are given by εk = cos Hence ε0 = 1, 2kπ 2kπ + i sin for k ∈ {0, 1, 2}. 3 3 √ 2π 1 3 2π + i sin =− +i =ε ε1 = cos 3 3 2 2 √ 4π 1 3 4π + i sin =− −i = ε2 . ε2 = cos 3 3 2 2 They form an equilateral triangle inscribed in the circle C(O; 1) as in Fig. 7. 7. 3. For n = 4, the fourth roots of unity are εk = cos 2kπ 2kπ + i sin for k = 0, 1, 2, 3. 4 4 In explicit form, we have π π + i sin = i; 2 2 3π 3π ε2 = cos π + i sin π = −1 and ε3 = cos + i sin = −i.