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Further complex numbers — Question 184

Original QCE Vault practice · 10 marks

Q184 · Practice questionTechnology-freeComplex unfamiliar10 marks

QUESTION 184 (10 marks)

Let ω=cis⁡(2π/3)\omega=\operatorname{cis}(2\pi/3), so the cube roots of unity are 1,ω,ω21,\omega,\omega^2.
a)
Show that ω+ω2=−1\omega+\omega^2=-1 and ωω2=1\omega\omega^2=1.
[2 marks]
b)
Determine 12−1+12−ω+12−ω2\frac1{2-1}+\frac1{2-\omega}+\frac1{2-\omega^2} exactly.
[4 marks]
c)
For real a>1a>1, let R(a)=1a−1+1a−ω+1a−ω2R(a)=\frac1{a-1}+\frac1{a-\omega}+\frac1{a-\omega^2}. Show R(a)=3a2a3−1R(a)=\frac{3a^2}{a^3-1}, then evaluate ∫23R(a) da\int_2^3R(a)\,da exactly.
[4 marks]
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