328 Answers to Exercises and Solutions to ProblemsOn the other hand, suppose that Ibd - cl < 1 - d2 and lb + dl < II+ cl.
AsinCasel,itcanbearguedthatl-b+c-dandl+b+c+dhave
the same sign. Suppose 1 - b + c - d < 0 and 1 + b + c + d < 0. Then
lb + dl < -(l + c). S ince^0 <^1 - d2 and bd - c <^1 - d2, it follows that
d(b + d) < 1 + c < 0, so that II+ cl < IdI lb + dJ < lb + dl, a contradiction.
Hencel-b+c-d>Oandl+b+c+d>O.Notingthat-l+b-c+d
and 1 + b + c + d are the respective values of the polynomial at -1 and 1,
we see that one of the following possibilities obtains:(i) -l<p<l<qlr(ii) p 5 q < -1< r < 1(iii) -1 < p 5 q 5 P < 1.Ad (i), 1 - qr < 0 and (1 - pq)(l - pr)(l - qr) > 0 implies that p > 0 and
pq < 1 < pr < pqr = -d, which contradicts 1 - d2 > 0. Ad (ii), 1 - pq < 0
implies that r < 0 and d = lpqrl > lprl > 1 > lqrl, which contradicts
1 - d2 > 0. Hence (iii) must hold.
3.9. (d) If the zeros of a manic cube are r, u + iv, u - iv, with r, U, v real,
then the coefficients are positive w r + 2u < 0, 2ru + u2 + v2 > 0 and
r(u2 + v2) < 0. The last inequality forces r to be negative. However, we
can choose u to be a small positive value and v a large value to achieve the
other two inequalities. For example, taking r = -3, u = 1, v = 3 leads to
the polynomial t3 + t2 + 4t + 30.
3.10. (d) Let p(t) = 8a2t3 + 8abt2 + 2(b2 + ac)t + (bc - ad). We are given
that p(t) has a real zero u. If bc > ad, then p(t) > 0 for t 2 0, so that
u < 0. If bc 5 ad, then p(t) = 2t(2at + b)2 + 2act + (bc - ad) shows that
p(t) has no negative zero. If bc = ad, then u must be zero and it can be
checked directly that f(&@) = 0. If bc < ad, then u > 0.
3.11. (a) The reasoning, analogous to that of Section 4.5, can be outlined
as follows. If the Nyquist diagram does make at least one circuit of the
origin, then by shrinking the semi-circle in the right side of the z-plane
down to a point, we find that its image curve must at some stage pass
through the origin. Suppose the Nyquist diagram does not make a circuit
of the origin. Let D be the image of the circle of radius M. Since the disc
of radius M contains all the zeros of the polynomial, D must wind around
the origin n times (n being the degree of the polynomial). Consider three
curves:
DI: the image of the semi-circular arc {z : IzI = M, Rez 2 0},
D2: the image of the semi-circular arc {Z : 1.~1 = M, Re,r 2 0},
Da: the image of the line segment {Z : z = yi, y real, IyI < M}.
D = D1 U D2. A point tracking around D traces around D1 and D2 in
succession. D1 U 03 does not make any circuit of the origin, so that all the
winding around the origin is done by the D2 part of the curve. By making