A complex polynomial has degree 1992 and distinct zeros. Show that we can find complex numbers zn, such that if p1(z) = z - z1 and pn(z) = pn-1(z)2 - zn, then the polynomial divides p1992(z).
Solution
Let the polynomial of degree 1992 be q(z). Suppose its roots are w1, w2, ... , w1992. Let S1 = {w1, ... , w1992}. We now define S2 as follows. Let z1 = (w1 + w2)/2 and take S2 to be the set of all numbers (w - z1)2 with w in S1. Note that w = w1 and w = w2 give the same number. It is possible that other pairs may also give the same number. But certainly |S2| <= |S1| - 1. We now repeat this process until we get a set with only one member. Thus if Si has more than one member, then we take we take zi to be an average of any two distinct members. Then we take Si+1 to be the set of all (w - zi)2 with w in Si. So after picking at most 1991 elements zi we have a set SN with only one member. Take zN to be that one member, so that SN+1 = {0}. Now if N < 1991, take the remaining zi to be 0 until we reach z1992.
Now as we allow z to take the values in S:
p1(z) = (z - z1) takes 1992 possible values;
p2(z) = (z - z1)2 - z2 takes at most 1991 possible values;
p3(z) = ( (z - z12 - z2)2 - z3 takes at most 1990 possible values;
p4(z) = ( ( (z - z1)2 - z2)2 - z3)2 - z4 takes at most 1989 possible values;
...
p1992(z) takes only the value 0.
So every root of q(z) is also a root of p1992(z). Hence q(z) must divide p1992(z).
© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002