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Solutions to Problems; Chapter 5 337

We know the real zero does not leave until a = 2. When will the nonreal

zero leave?)

Let a > 4127 and suppose that the three zeros are r, u + iv and u - iv.

Then

r+2u=l 2ru+u2+v2=0 r(u2+v2)=-a.

We eliminate u and P from these three equations. From the first and third

-a = (1 - 2u)(u2 + v2).

From the second and third

-2au + (u” + v2)2 = 0.

Hence

-a2 = [a - (u” + v2)“](u2 + v”).

Thus, u2 + v2 is the unique positive zero of the polynomial g(t) = t3 -

at - a2. This zero lies in the interval

(0, l] e 0 5 g(1) M 0 5 1 - a - a2 W a < (l/2)(& - 1).

[Note that 4/27 < (l/2)(& - 1) < 2.1

Hence, all three zeros of f(t) 1 ie in the closed unit disc iff 0 5 a <

(l/2)(&5 - 1).

4.19. We can make a change of variable t = s + (b + c)/2 to obtain a
polynomial g(s) = s3 + us2 - v2s - uv2 whose zeros are -u, -v, v where
u 2 v 2 0. The zeros of g(s) and g’(s), respectively, are equal to the zeros
of j(t) and f’(t) translated by -(b + c)/2. Hence it suffices to show the
result for g(s), i.e. that g’(s) = 3s2 + 2us - v2 has a zero in the closed
interval [0, v/3].
Clearly g’(0) 5 0, g’(v/3) = 2v(u - v)/3 2 0. If g’(0) = 0, then v = 0
and g(s) has 0 as a double zero, so that f(t) has b = c as a double zero.
If g’(v/3) = 0, then g(s) has -v = -U as a double zero, so that f(t) has
a = b as a double zero.

4.20. Observe that u = (l/2)(& - 1) is the unique positive zero of the
polynomial 1 - t - t2, so that, if ]z] < u, then 1 - ]z] - ]z12 > 0.
First, suppose that ni 12. Then, for 1.~1 < u < 1,

]l$ tnl + 2”’ +... + z-1 > 1 - ]z]“’ - ]Z]“2 -... - ]zI”k

1 1 - 1212 - ]%I3 -.. * > 1 - ]#(l - ]z])-i

= (1- ]z] - ]Z]“)(l - ]z])-i > 0.

On the other hand, let ni = 1. Observe that

(1 - ,Z)(1+ 2 + Z”Z +. *. + fn*) = 1 - z2 + g(z),