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Hello everyone I have to find the sum of all the square roots answers of the equation $(z + 1)^n = (z - 1)^n$.

I tried to use a = $z +1 , b = z -1$ and place $a^n - b^n = 0 = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-1})$

And invert this to a polynomial and use Vieta's formula for finding the sum of all the square roots answers but I didn't success

someone can help me please?

Bernard
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xx01
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  • https://math.stackexchange.com/questions/607487/the-roots-of-the-equation-zn-1zn and https://math.stackexchange.com/questions/1562603/solve-the-equation-z13iz-13-0 – lab bhattacharjee Apr 19 '20 at 15:03
  • What is the exact wording of the question you are asked to solve? The way you are asking isn't that clear. Do you want the sum of the squares of the roots of the polynomial or sum of the "square roots" of the roots of the polynomial? – achille hui Apr 19 '20 at 15:08
  • The sum of the squares of the polynomial. I meant if $x_1 , x_2 , x_3 , ... , x_n$ are the roots of the polynomial so I need to find $x_1^2 + x_2^2 + x_3^n + ... + x_n^2$ – xx01 Apr 19 '20 at 15:12
  • In that case, you just need to expand the difference $(z+1)^n - (z-1)^n$ to the form $2n (z^{n-1} - A z^{n-2} + B z^{n-3} + \cdots)$ using binomial theorem. By Vieta's formula, the sum you want equals to $A^2 - 2B$ – achille hui Apr 19 '20 at 15:33
  • What is A and what is B? – xx01 Apr 19 '20 at 15:43
  • Someone can post an answer or hint please? – xx01 Apr 19 '20 at 18:47

1 Answers1

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Hint

Since $z\ne 1$ and $z\ne -1$, you can write $$({z+1\over z-1})^n=1$$hence all the roots of the equation can be obtained from $${z+1\over z-1}=e^{j{2\pi\over n}k}\quad,\quad k=0,1,\cdots n-1$$

Mostafa Ayaz
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  • Can you solve all the question please? – xx01 Apr 20 '20 at 09:37
  • Did you make some try after the hint? – Mostafa Ayaz Apr 20 '20 at 09:39
  • I didn't understand what you do? – xx01 Apr 20 '20 at 10:28
  • Because $1$ and $-1$ are not among the roots of the equation, you can divide both sides of the equation to $(z-1)^n$. The equation is reduced to $u^n=1$ which is a famous equation with roots being equal to $e^{jk{2\pi\over n}}$. You can conclude this result by using the polar notation of $u$ as $u=re^{j\phi}$ – Mostafa Ayaz Apr 20 '20 at 11:16