Solutions to Problems; Chapter (^5 341)
4.31. Let Rez > a. Then
Iz + a,-11 > Re(z + a,+l) > a + Re(a,-1)
= la,-21aP1 + la,-31a-2 + f.. + lacla-(n-l)
Ian-21 Iz(-l + Ian-31 lzlm2 +. .. + laOI lzl-(“-l)
- If(z)1 2 Izn + a,-lz”-‘I - Ian-21 lzlnm2 - ... - Ia01 > 0
Let Rez < -b. Then
-1% + a,-11 < Re(z + a,-1) < -b + Re(a,-1)
= -[la,-2(b-’ + Ia,-31bm2 + ... + (a01 lbl-(“-l)]
< -Ian-21 (z(-l - lan-3) )zlm2 -.. f - Ia01 lzlsCnel)
3 If(z)1 2 Izn + an-lznmll - Ian-21 lzl”-2 - ... > 0.
4.32. q(t) = f(t)g(t) where f(t) = tp(t) + p’(t) and g(t) = tp’(t) + p(t).
The function h(t) = tp(t) h as at least n + 1 distinct zeros (including 0),
so that h’(t) = g(t) h as at least n distinct real zeros, all distinct from the
zeros of p(t) exceeding 1.
Let {ul,... , u,} be the set of simple zeros of p(t) exceeding 1, and
{Vl,... ,v,} be the set of multiple zeros. Note that r + s = n. Each vi
is a zero of both p(t) and p’(t) and so is a zero of f(t). Prom a sketch of a
graph, it can be seen that the sign of f(ui) = p’(ui) alternates with i and
that f(t) has at least r - 1 distinct zeros apart from the vi. Hence f(t) has
at least r + s - 1 = n - 1 zeros exceeding 1.
It remains to check that these zeros are distinct from the zeros of g(t)
identified above. Suppose f(w) = g(w) = 0, 1~1 # 1. Then (w - l)@(w) -
p’(w)) = 0, so that p(w) = p’(w) = (1 + w)-‘g(w) = 0 so w is not one of
the zeros of g(t) distinct from those of p(t).
4.33. The result is clear if degp(z) = 1. If n = 2 and p(x) = ax2 + ba: + c,
then the left side is the discriminant b2 - 4ac which is nonnegative. Suppose
the result has been established for polynomials of degree up to n - 1. Let
p(x) be as specified and write it as (z - r)q(z), where r is a real zero of
p(z) and degq(c) = n - 1. Then
(n - 1)2(p’(~))2 - n(n - l)p(z)p”(2) = n(z - r)2[(n - 2)q’(x)2
- (n - 1)7(~)d’(~:)l+ ki(z)(z - r> - (n - lM~)l” 1 0.
4.34. Let (u, v) be a given open interval and let ai/bi (i = 1,2,,.. , n - 1)
be any n - 1 rationals in the interval. The polynomial g(t) = II(bit - a;) is
a polynomial of degree n - 1 all of whose zeros lie within the interval (u, v).
The derivative g’(t) h as n - 2 zeros rj with u < r1 < r2 <... < r-,-z < V.
The n numbers g(u), g(ri), g(ra),... ,g(rnT2), g(v) are all nonzero and
alternate in sign.