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Solutions to Problems; Chapter 5 339

4.22. Denote the left side by p(t). S ince p(t) is positive for large It 1, p(t)
assumes its minimum value on R at some point c; we must have that
p’(c) = 0. But then p(c) = p’(c) + c2”/(2n)! > 0, so that p(t) has a positive
minimum and is therefore everywhere positive.
4.23. Suppose p(t) is real iff t is real. Then for all real values of k, p(t) + k
has the same property. Hence all the zeros of p(t) + k are real for each value
of k. By Exercise 5.2.16, this can occur only if degp(t) = 1.
4.24. Let q(t) = set + alt2/2 + ... + a,t”+‘/(n + 1). Then q(0) = 0,
q’(t) = p(t) and q’(0) = a0 > 0, so that q(t) is increasing at t = 0.
However, q(l)-q(-1) = 2(ae+a2/3+a4/5+...) < 0, so that q(1) < q(-1).
Hence q(t) cannot be increasing on the whole of the interval [-1, 11, so that
q’(t) = p(t) must assume both positive and negative values there. Hence,
for some r E [-l,l], p(r) = 0.
4.25. Let the zeros of p(z) be ri (1 5 i 5 n). Then, for z # ri,

n - kp’(z)/p(z) = n - kC(z - ri)-‘.

Suppose that ]z] > R+ lkl, then

It. - ril > IzI - lril > (R+ lkl) - R = lkj.

Hence IkC(z - ri)-‘1 5 IklC(lz - ril)-’ < n, SO that np(z) - kp’(z) # 0.
The result follows.
4.26. Let d be the greatest common divisor of p and q, and let zd = 1.
Then zP+‘J = zp = 1 and so z is a zero of the polynomial. On the other
hand, if ]z] = 1 and bzf’ = azP+Q + b - a, then

b = lbzpl = ]azP+q + b - al 5 a + (b - a) = b,

so that lazP+q + b - al = a + (b - a). Hence zP+q = 1 and so zp = 1. Hence
Zd = 1.
4.27. Suppose there exists a root z with ]z] # 1. Since the reciprocal of any
zero is also a zero, we may suppose that z = r(cos 0 + i sin 0) with r > 1.
Since z”(z - u) = (1 - uz), we have that (z]~“]z - u12 = ]I- uz]‘. Thus

r2”(r2-2rucos0+u2)=(l-2rucosB+u2r2).

Since r > 1 and r2 - 2rucose + u2 1 (r - 1~1)~ > 0,


r2 -2rucosB+u2 < 1 - 2ru case + u2r2^3 (r” - l)(l - u”) < 0,

which contradicts ]u] 5 1 and r > 1. Hence ]z] = 1 for each zero z.
A number of solutions to this problem is provided in Amer. Math. Monthly
72 (1965), 1143-1144. One of them yields the result as a special case of the

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