Complex Analysis Problem?
Hello,
I have to do this problem without using any major theorems or tools because the class hasn't covered anything yet and I am stuck.
let P (z) = a0 + a1z + · · · + ak−1zk−1 + zk be a polynomial of degree k ∈ N
with complex coefficients a0, . . . , ak−1. Suppose that P (z) = the product of (z − zj )., j=1....k
Find an expression for the sum (zj)^2, j=1....k in terms of the coefficients of P .
b) Let n ∈ N and ζk = exp(2πik/n) for k=1....n-1. Find a simple expression
for the sum 1/(ζk − 1)^2 k=1,...., n-1 in terms of n.
Any help will be appreciated.
Comments
It helps to see a concrete case. Suppose P has degree 3 and we call the roots p, q, and r. Then P(z) = (z - p)(z - q)(z - r) and expanding the product on the right hand side and collecting like terms shows you that
P(z) = z^3 + (-(p + q + r)) z^2 + (pq + pr + qr)z - pqr.
If you square the coefficient of z^2 you get (p^2 + q^2 + r^2) + (2pq + 2pr + 2qr), which is almost what you want, but for this second term which I put in parentheses. But notice this second term is exactly twice the coefficient of z in P(z). This shows how to obtain the answer in the degree 3 case.
In general it is very similar. The coefficient of z^(k-1) is -(z_1 + z_2 + ... + z_k). If you square this and collect like terms you get
[the sum of the squares of the z_j] + 2 [the sum, over all pairs (i,j) with i < j, of z_i z_j].
This second sum in brackets is precisely the coefficient of z^(k-2). So the answer in the general case is
a_{k-1}^2 - 2 a_{k-2}.
It is more generally true that *any* polynomial in z_1, z_2, ..., z_k that is "symmetric" in these quantities (ie, it's the same if you permute the indices 1, 2, ..., k--- like z_1^2 + ... + z_k^2, for example) can be given in terms of the coefficients a_0, a_1, ..., a_{k-1} of the polynomial P having z_1, ..., z_k as roots. This is sometimes called the "fundamental theorem of symmetric polynomials" and there are general algorithms for finding the formula that would produce the above formula that we found by hand in this special case.
In (b) note that zeta_1, ..., zeta_{n-1}, together with the number 1, are all of the roots of the polynomial f(z) = z^n - 1. The polynomial (z + 1)^n - 1 therefore has as roots zeta_1 - 1, zeta_2 - 1, ..., zeta_n - 1, and also 1 - 1 = 0. If you expand this out this is the polynomial
z^n + C(n,n-1) z^(n-1) + ... + C(n,1) z = sum_{j=1}^n C(n,j) z^j
where C(n,j) = n!/(j! (n-j!) is the binomial coefficient (there is no j = 0 term because we subtracted the 1 off of (z + 1)^n). If you factor a z out of this you get a polynomial of degree n-1, namely
q(z) = z^(n-1) + C(n,n-1) z^(n-2) + ... + C(n,1) = sum_{j=1}^n C(n,j) z^{j-1}
which has exactly the numbers zeta_1 - 1, ..., zeta_{n-1} - 1 as its roots (1 - 1 = 0 is not root).
If you look at z^(n-1) q(1/z) it turns out to be a polynomial also (if you define it in the obvious way at 0): it is the polynomial
1 + C(n,n-1) z + ... + C(n,1) z^(n-1) = sum_{j=1}^n C(n,j) z^(n-j)
and by construction its roots are precisely 1/(zeta_1 - 1), 1/(zeta_2 - 1), ...., 1/(zeta_{n-1} - 1). In order to apply (a) we need a polynomial with these roots and leading term 1, so we will have to divide this polynomial by C(n,1). We get the polynomial
h(z) = [1/C(n,1)] + [C(n,n-1)/C(n,1)] z + ... + z^(n-1) = sum_{j=1}^n [C(n,j)/C(n,1)] z^(n-j)
Applying the formula from (a) we see that the sum of the squares of the numbers 1/(zeta_j - 1) is
[the coefficient of z^(n-2) in h(z)]^2 - 2 [the coefficient of z^(n-3) in h(z)] = [C(n,2)/C(n,1)]^2 - 2 [C(n,3)/C(n,1)]
And plugging in the explicit formulas C(n,1) = n and C(n,2) = n(n-1)/2 and C(n,3) = n(n-1)(n-2)/6 and simplifying you get (-n^2 + 6n - 5)/12.
The key insight in (b) I think is the realization that if you know a polynomial Q(z) having the nonzero numbers w_1, w_2, ..., w_k as its roots, then you can immediately write down a polynomial T(z) having the numbers 1/w_1, 1/w_2, ..., 1/w_k as its roots. The way to do it is to note that if you divide both sides of the equation Q(z) = 0 by z^k (here k is the degree of Q) you immediately get an equation expressing that a certain polynomial in 1/z is also equal to 0. The general formula for this polynomial T(z) in terms of Q is z^k Q(1/z), as we saw here.
Having this realization allows you to do (b) by (1) finding a polynomial with the zeta_j as roots (this is easy), (2) using polynomial from (1) to find a polynomial with only the numbers zeta_j - 1 as roots (also easy), (3) using the polynomial from (2) to find a polynomial with only the numbers 1/(zeta_j - 1) as roots (this requires the realization we mentioned above), and finally (4) applying (a) to the polynomial you got out of (3) to get the formula for the sum of the 1/(zeta_j - 1)^2.