# Roots of elementary monomials

Let $m_\lambda(X_1,X_2,...X_N)$ be a monomial symmetric function with partition $\lambda$.

For example:

$$m_{(3,1,1)}(X_1,X_2,X_3) =X_1^3X_2X_3 + X_1X_2^3X_3 + X_1X_2X_3^3$$

Is there a general formula for roots of $m_\lambda$ if $X_j$ is restricted to elements of $\mathbb C$ with $\| X_j \|=1$?

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Could you define what is a symmetric monomial with partition $\lambda$ ? –  user18119 Nov 12 '11 at 22:39
Ah sorry, that's not very good explained. See en.wikipedia.org/wiki/Symmetric_polynomial –  draks ... Nov 12 '11 at 23:03
"monomial" means one term; what you have are polynomials. –  Gerry Myerson Nov 13 '11 at 3:25
The usual term is not a "symmetrical monomial", it is a "monomial symmetric function". –  Phira Nov 13 '11 at 12:48

I doubt it. Let's look at $m_{(1,0,0,0,0)}$. You want the solutions of $$a+b+c+d+e=0$$ with all variables on the unit circle. It will be hard enough to find a formula for that special case, much less for the general case.
Note that $m_{(1,0)}$ is $a+b=0$ which is solved by $a=e^{it}$, $b=e^{i(t+\pi)}$. Then $m_{(1,0,0)}$ is $a+b+c=0$, which forces $a,b,c$ to be vertices of an equilateral triangle. Next, $m_{(1,0,0,0)}$ is $a+b+c+d=0$, and with a bit of work you can show that $a,b,c,d$ must be vertices of a rectangle. But once you get up to 5 unknowns the geometric argument doesn't give you anything that simple.
Not sure I follow you. $m_{(2,0)}$ gets us $a^2+b^2=0$ which is still a one-dimensional problem. –  Gerry Myerson Nov 19 '11 at 11:05