Q.
If 1,α1,α2,α3,α4 be the roots of x5−1=0, then the value of ω2−α1ω−α1ω2−α2ω−α2ω2−α3ω−α3ω2−α4ω−α4… is . (where ω is imaginary cube root of unity.)
547
167
Complex Numbers and Quadratic Equations
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Solution:
(z−1)(z−α1)……..(z−α4)=z5−1
Put z=ω,z=ω2 and divide (ω2−1)(ω2−α1)(ω2−α2)(ω2−α3)(ω2−α4)(ω−1)(ω−α1)(ω−α2)(ω−α3)(ω−α4)=ω10−1ω5−1 (ω2−α1)(ω2−α2)(ω2−α3)(ω2−α4)(ω−α1)(ω−α2)(ω−α3)(ω−α4) =(ω−1)2(ω2−1)2=(ω+1)2=ω4=ω