Question:
prove the sum of 99th powers of the roots of equation x>7-=0 is zero and hence deduce the roots of x>6+x>5+x>4+x>3+x>2+x+1=0
Answer:
Given, x7 - 1 = 0
=> x7 = 1
=> x = (1)1/7
=> x = (cos 0 + i*sin 0)1/7
=> x = cos(2π/7) + i*sin (2π/7)
So, 7 root of unity of 1, a, a2 ,a3 ,a4 ,a5 ,a6
Where a = cos(2π/7) + i*sin (2π/7)
Now,
199 + a99 + (a2 )99 + (a3 )99 + (a4 )99 + (a5 )99 + (a6 )99
= 1 + a99 + a2(99) + a3(99) + a4(99) + a5(99) + a6(99)
= 1*{1 -(a99 )7 }/(1 - a99 )
= {1 -(cos(2π/7) + i*sin (2π/7)99(7) }/{1 - cos(2π/7) + i*sin (2π/7)99 }
= {1 - cos 2π*99 - i*sin 2π*99}/{1 - cos(198π/7) + i*sin (198π/7) }
= (1 - 1 - i*0)/{1 - cos(198π/7) + i*sin (198π/7) }
= 0
So, the sum of 99th power of the roots of the equation x7 - 1 = 0 is zero.
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