Question

Show that the sum of the 99th powers of the roots of the equation x^5=1 is zero
​

Answer 1

Step-by-step explanation:

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.

Answer 2

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