Wilson Stothers' Inversive Geometry and CabriJava Pages

Definitions

For a complex polynomial f(X)

n(f) = the number of distinct zeros of f
m(f) = the number of simple zeros of f

The polynomial f is lacunary of index k if it can be written as f(X) = (a(X)X^k+i+1+b(X))X^j, where

i,j,k ≥ 0,
a(0), b(0) ≠ 0,
deg(b) = i.

Theorem

If f is lacunary of index k, then

n(f) ≥ k+1
m(f) ≥ k+1-deg(b)-deg(a).

The proof depends on the simple

Lemma

If f is a non-constant complex polynomial, and d=gcd(f,f').
Then f = f*d, and

n(f) = deg(f*) = deg(f) - deg(d),

m(f) = deg(f*) - n(d) ≥ deg(f) - 2deg(d).

Proof of Lemma
Observe that, if α is a root of f of multiplicity m+1, then it is a root of f' (and hence of d) of multiplicity m.
Thus the roots of f* are the roots of f, and each occurs exactly once.
Part (1) is immediate.
For part (2), we note that the simple roots of f are the roots of f* which are not roots of d.

Proof of Theorem
Let f be lacunary of index k.
We may as well assume that j = 0, so that f(0) = b(0) ≠ 0.
Then f(X) = a(X)X^k+i+1+b(X)
and f'(X) = A(X)X^k+i+b'(X), where A(X) = (k+i+1)a(X)+a'(X)X.
As d(X) = gcd(f(X),f'(X)), it divides b'(X)f(X)-b(X)f'(X) = (a(X)b'(X)X-A(X)b(X))X^k+i.
But f(0) ≠ 0, so X does not divide f(X), and hence does not divide d(X).
Thus d(X) divides e(X) =(a(X)b'(X)X-A(X)b(X)).
The latter is non-zero (the value at 0 is A(0)b(0) ≠ 0),
so deg(d) ≤ deg(e) = deg(a)+i.
By the Lemma, n(f) ≥ deg(f) - (deg(a)+i) = (deg(a)+k+i+1) - (deg(a)+i) = k+1.
The second part follows at once as i = deg(b).

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