146 4. Equations(b) Use (a) to argue that, when ]%I = r, then p(z) is in the annulus
with center 0, inner radius (1/2)r” and outer radius (3/2)r”.
(c) Use (a) to argue further that, as t makes one counterclockwise
circuit of the circle with center 0 and radius r, p(z) must make
n circuits of the origin.- Sketch a proof of the fundamental theorem along the following lines.
(a) It suffices to prove the theorem for manic polynomials p(z).
(b) If p(0) = 0, the theorem holds. It suffices therefore to establish
the result when p(0) # 0.
(c) If r is sufficiently small, then as z traces around C,, its image
p(z) will not make a circuit of the origin.
(d) If r is sufficiently large, then as z traces around C,, its image
p(z) will make at least one circuit of the origin.
(e) Let r grow slowly from small to large values. How will the image
D, of C, vary? Why is it plausible to infer that for some value
of r, D, must contain the origin?
4.6 Consequences of the Fundamental Theorem
The fundamental theorem has a number of consequences about the fac-
torization of a polynomial and the extent to which a polynomial can be
determined by certain of its values.
Exercises
- Let p(t) be a polynomial over C with positive degree n. By the Fun-
damental Theorem it has at least one complex zero, but, it is known
that there are no more than n distinct zeros (Exercise 2.2.4). Let
these be tl, t2,... ,tk where 1 < k 5 n. Show that there are positive
integers rni and a constant c for which
(i) ml + m2 +. a. + mk = n;
(ii) p(t) = c(t - tl)“l(t - t2)m2 f -. (t - tk)m*.- Prove that a polynomial over C is irreducible if and only if its degree
is 1. [A field with the property that only linear polynomials are irre-
ducible is said to be algebraically closed; thus, this result states that
C is an algebraically closed field. Over such a field, every nonconstant
polynomial has a zero.] - Let p(t) be a polynomial over R. p(t) can have zeros which are either
real or nonreal; however, the complex conjugate of any zero is also a
zero (Exercise 1.3.14).