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设z1=cosα+isinα,|z1|=1
z2=cosβ+isinβ,z2=1,
z1+z2=(cosα+cosβ)+i(sinα+sinβ)
z1+z2=1/2+√3i/2,
cosα+cosβ=1/2,(1)
(sinα+sinβ)=√3/2,(2)
(1)两边平方+(2)两边平方,
2+2(cosαcosβ+sinαsinβ)=1/4+3/4=1,
cos(α-β)=-1/2,
cos(α-β)=cos2π/3,
α-β=2π/3,
α=β+2π/3,
代入(1)式,cos(β+2π/3)+cosβ=1/2,
sin(π/6-β)=sinπ/6,
β=0,α=2π/3,
z1=-1/2+√3i/2,
z2=1.
或z2=-1/2+√3i/2,
z1=1.
1年前
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