Question

prove that sum of the 99th powers of the roots of the equation z^7-1 = 0 is zero​

Answer 1

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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Answer 2

Concept Introduction:-

It might resemble a word or a number representation of the quantity's arithmetic value.

Given Information:-

We have been given that $z^7-1 = 0$.

To Find:-

We have to find that prove the sum of the $99$th powers of the roots of the equation $z^7-1 = 0$ is zero​.

Solution:-

According to the problem

The roots of $z^7-1 = 0$ are,

$\frac{2\pi}{7}+sin \frac{2\pir}{7},$$r=0,1,2,3,.....,6$

Let $\alpha=cos\frac{2\pi}{7}+sin \frac{2\pi}{7}$

Then the roots are $1,\alpha ^2,........,\alpha ^6$

$1+\alpha ^{99}+....+(\alpha^6) ^{99}=\frac{1-\alpha ^{99} \times 7}{1-\alpha ^{99}}=0$

since $\alpha ^{99}^{\times 7}=1\\\alpha ^{99}\neq 1$

Final Answer:-

The correct answer is the prove that the the sum of the $99$th powers of the roots of the equation $z^7-1 = 0$ is zero​.

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