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Answers to Exercises; Chapter 5 327

]w/u] > 1, so that ]w] > u. A similar argument using (b) and-the fact that
anvn 1 an-lvnS1 2... 2 alv 2 a0 yields lw] < v.
3.8. (a) Case 1: Both zeros are nonreal. If the zeros are u + vi and u - vi,
then (1 + c) + b = (1 - u)” + v2 and (1 + c) - b = (1 + u)~ + v2 are both
positive, so that Ibl < 1 + c always holds. Hence c < 1 u2 + v2 < 1
both zeros lie in the interior of the complex unit disc.
Case 2: Both zeros are real. If the zeros are r and s, then (1 + c) - b =
(l+r)(l+s) and (l+c)+b = (1-r)(l-s). Suppose that IrI < 1, Is] < 1.
Then -1 < c = rs < 1 j 0 < 1 + c < 2. Also, 1 f r and 1 f s are positive,
so that (1 + c) f b > 0 + b < 1 + c. On the other hand, if lb1 < 1 + c < 2,
then (1 f r)(l f s) > 0, so that 1 + r and 1 + s have the same sign, as do
1 - r and 1 - s. Since lrsl = ICI < 1, this forces -1 < r, s < 1, as required.
For other treatments of this criterion, consult problems El313 in Amer.
Math. Monthly 65 (1958), 284, 776 and 1029 in Crux Math. 11 (1985), 83;
12 (1986), 191-193.
(b) Case 1: The teros are nonreal. Let the zeros be r, u+vi, u-vi. Then

1 - b + c - d = (1 + r)[(l + u)” + v”]

l+b+c+d=(l-r)[(l-~)~+v~]

1 - c + bd - d2 = [l - (u” + v2)][(1 - ru)2 + r2v2].

Hence-1<r<1,u2+v2<1~]b+d]<1+candc-bd<l-d2.

Let all the zeros lie within the open disc defined by ]z] < 1. Then the

product -d of the zeros satisfies IdI < 1. Hence 1 - d2 > 0 and d(b + d) 5

Id(b+d)l<Ib+dl<1+csothatbd-c<1-d2.HenceIc-bdl<1-d2

and lb + dl < 1 + c = ]I+ cl.

On the other hand, suppose lbd - cl < 1 - d2 and Jb + dl < 11 + cl.

Clearly c - bd < 1 - d2. We show that^1 - b + c - d and^1 + b + c + d

must both be positive. Suppose, if possible, that 1 - b + c - d 5 0 and

1 + b + c + d 2 0. Then 1 + c and -(l + c) do not exceed b + d so that

II+ 4 5 lb + 4, a contradiction. On the other hand, if^1 - b + c - d^2 0

and 1 + b + c + d 5 0, then 1 + c and -(l + c) do not exceed -(b + d), so

that 11 + cl 2 -(b + d) = lb + dl, a contradiction. Since^1 - b + c - d and

1 + b + c + d also cannot both be negative, we must have lb + dl < 1 + c

and all zeros lie inside the open unit disc.

Case 2: All zeros are real. Let the zeros be p, q, r with p < q < r. Then

1 - b + c - d = (1 +p)(l + q)(l + r)

l+b+c+d=(l-p)(l-q)(l-r)

1 -c+bd-d2 = (l-pq)(l-pr)(l -qr).

Let all the zeros lie within the open unit disc. From the above equations,

Ib+dl<l+c=Il+~landc-bd<l-d~.AsinCasel,itcanbeshown

thatbd-c<l-d2,sothat]bd-c]<l-d2.