The question is if $w$ is any real number , the equation $z^k = w$ has $k$ roots,
$k ∈ n$.
Show that the sum of the k roots is zero.
Usually with these root questions I convert the right hand side to polar form so in this case
$z^k = w cis 0$
$z = w^{1/k} cis 1/k$ $(0 + 2πn)$ (n is any integer)
$z = w^{1/k} cis 2πn/k$
My thought was writing $k$ as $1,2,3,4,5$ and so on but that would be too long , is there an easier method of solving this?
If $r_1, r_2, \dots, r_k$ are the roots of the polynomial $z^k-w$, then $$z^k - w = (z-r_1)(z-r_2)\cdots(z-r_k)$$
Assuming $k>1$, the coefficient of $z^{k-1}$ on the left side is $0$, while on the right side, it is $-r_1-r_2-...-r_k$, so:
$$0=-r_1-r_2-...-r_k $$
and hence
$$0=r_1+r_2+...+r_k$$
Not entirely certain but I would look at this.
$$z=w^{\frac{1}{k}}*(cos(2\pi\frac{n}{k})+isin(2\pi\frac{n}{k}))$$ then $$z^k=w*(cos(2\pi n) +isin(2\pi n)=w$$ Hence $z$ is a root for every $n=0,1,2...k$
This isn't true for all $k.$ (Consider the $k=1$ case when $w\ne0$. Also, consider the $k>1$ case when $w=0.$) However, for $k\ge 2$ and $w\ne0,$ it is true. Geometrically, the points $|w|^{1/k}\operatorname{cis}\frac{2\pi n}{k}$ (with $n=0,1,...,k$) are evenly-spaced points on the circle of radius $|w|^{1/k}$ about the origin. Their "center of gravity" (so to speak) is the origin, and this is obtained by adding them all together and dividing the result by $k.$ Of course, the division by $k$ isn't the reason that the center of gravity is the origin: it's because their sum is $0.$
Another approach you might take is to show that if $p(x)$ is any polynomial of degree $2$ or more, then the coefficient of its linear term is the opposite of the sum of the roots. Hence, if the polynomial has no linear term, as is the case for $z^k-w$ for $k\ge 2,$ then the sum of its roots is $0.$
