If $w_1,w_2, ...,w_m$ are distinct root of the equation $x^m-1=0$ then show that $(a+bw_1)^m+(a+bw_2)^m+(a+bw_3)^m+\cdots+(a+bw_m)^m=m(a^m+b^m)$

Question

Taken $x^m-1=(x-w_1)(x-w_2)\cdots (x-w_m)$. But all are in summation in proof. How to handle it.

Use the binomial theorem to write it as $$\sum_{j=1}^m (a+b\omega_j)^m = \sum_{k=0}^m{m\choose k} a^{m-k}b^k \sum_{j=1}^m\omega_j^k$$ Related: Sum over the powers of the roots of unity $\sum \omega_j^k$ — Winther, Dec 14 '16 at 18:37

score 0 · Accepted Answer · answered Dec 14 '16 at 18:47

We have $\sum\limits_{i=0}^m(a+bw_1)^m$, using newton this is equal to:

$\sum\limits_{i=1}^m\sum\limits_{j=1}^m \binom{n}{j}a^jb^{m-j}w_i^{m-j}$.

Changing the order of the sums we get: $\sum\limits_{j=1}^m\sum\limits_{i=1}^m \binom{n}{j}a^jb^{m-j}w_i^{m-j}=\sum\limits_{j=1}^m\binom{n}{j}a^jb^{m-j}\sum\limits_{i=1}^m w_i^{m-j}$.

now, it is easy to prove looking at the unit circle that $\sum\limits_{i=1}^m w_i^{m-j}$ consists of the sum of all of the $\frac{m}{(m-j,m)}$ roots of unity, taken $(m-j,m)$ times each. So this is equal to $0$ unless $(m-j,m)=m$. In other words if $j=0$ or $j=m$.

So the sum is just equal to:

$\binom{m}{0}b^m(m)+\binom{m}{m}b^n(m)=m(a^m+b^m)$

If $w_1,w_2, ...,w_m$ are distinct root of the equation $x^m-1=0$ then show that $(a+bw_1)^m+(a+bw_2)^m+(a+bw_3)^m+\cdots+(a+bw_m)^m=m(a^m+b^m)$

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