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306 Answers to Exercises and Solutions to Problems

6.1. Let mi be the multiplicity of the zero ti. We have that p(t) =

(t - tl)mlpl(t), where m(h) # 0. S ince t2 - tl # 0, t2 must be a zero

of pi(t) of multiplicity m2 so that p(t) = (t - tl)“‘(t - t2)mzp2(t), where

pz(t) vanishes at neither t i nor t2. Continuing in this way, we find that

p(t) = q(t)(t - tl)ml... (t - tk)m* ) where q(t) is a polynomial over C which

does not vanish at any ti. If degq(t) 1 1, then q(t) must have a zero, by

the Fundamental Theorem. But such a zero would be an additional zero of

p(t), yielding a contradiction. Hence q(t) must be a constant.

6.3. (a) If si is a zero of p(t), so is Zi. Therefore, p(t) is divisible by

(t - Si)(t - Si).

(b) p(t) can be written as the product of linear polynomials over C. Lin-

ear factors corresponding to nonreal zeros can be paired off and combined

into real quadratic factors, as in (a).

6.4. For any complex number k, p(t)-k is a polynomial over C with exactly

n zeros counting multiplicity. The only values of k for which the zeros are

not distinct are those for which p(t) - H and its derivative p’(t) have a zero

in common. Thus, p(t) - k has a multiple zero only when k = p(r) for some

zero T of p’(t). S ince p’(t) has only finitely many zeros, the result follows.

6.6. We can write

a# + u,-~t”-’ + * * * + a0 = u,(t - q).. * (t - rn).

The result follows from expanding the right side and comparing coefficients.
6.7. Since 0 = ~(a,) = II(a, - bj), it follows for some j, a, = bj. Let
j = n, say. Then
772-l n-1
n (t - u;) = n(t - bj).
i=l j=l
Continue the argument, pairing off and cancelling factors t - ui and t - bj.
6.8. Suppose p(t) and q(t) have degree not exceeding n and p(oi) = q(oi).

Then (p- a>(t) h as d e g ree not exceeding n and at least n +^1 zeros, whence
it is the zero polynomial. The result follows.

6.9.

p(t) = C(t2 - 2t + l)m fi(t2 - (ri + l/Yi)t + 1)“’

i=l

for complex zeros ri with multiplicity mi within the closed unit disc. Each
quadratic factor is a reciprocal polynomial and it is straightforward to prove
that their product is also a reciprocal polynomial.

6.10. Let f(ui) be p rime for 0 5 i 5 2n, and suppose, if possible that
f(t) = g(t)h(t) where degg(t) = m 2 1, degh(t) = n - m 2 1. Then
either Ig(ui)l = 1 f or more than 2m of the oi or Ih( = 1 for more than