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342 Answers to Exercises and Solutions to Problems

Let c = min{lg(u)l, Ig(v)l, Ig(ri)l} and let k be a large positive integer

such that (IuI + IvI)“/k < c. Define f(t) = P/k + g(t). Then the signs of

f(t) agree with those of g(t) at u, rj, v and so the signs of f(t) at these

n points alternate. Hence f(t) has n - 1 zeros in the interval (u, v). The

polynomial kf(t) fulfils the requirements.

4.35. Since (l+iX)m = f(z)+ z ‘g( z ), 1 ‘t f o 11 owsthat (l-iz)m = f(z)-is(x).

Hence

2[af(z) + bg(z)] = (a + bi)(l - ix)“’ + (a - bi)(l + iz),.

Suppose that z is a zero of af(z) + bg(x). Then, clearly z # fi, and

II+ izlm Ia + bil

11 - izlm

= - = 1 * II+ izl = 11 - izl

la - bil

j z is real (Exercise 1.3.13).

Answers to Exercises

Chapter 6

1.1. Let the roots of the given equation be u, v, w.
First solution. u2 + v2 + w2 = (u + v + w)~ - 2(uv + uw + VW) = -5;
u2v2+u2w2+v2w2 = (uv+uw+vw) 2 -2uvw(u+v+w) = -11; UVW2 =
(UVW)~ = 100. The equation x3 + 5x2 - 11x - 100 = 0 has the required
roots.
Second solution. The roots of z3 + x2 + 3x + 10 = 0 are -u, -v, -w.
Since

(X2 - u2)(x2 - v2)(x2 - w2)

= (x - u)(x- v)(x - W)(X + u)(z + v)(x + w)

= (x3-x2+3x- 10)(x3 + x2 + 3x + 10)

= (x3 + 3x)2 - (x2 + lo)2 = x6 + 5x4 - 11X2 - 100.

If y = x2, then

(y - u2)(y - v2)(y - w”) = y3 + 5y2 - lly - 100

is a polynomial with the required zeros.

1.2. Since uvw = -l/2, the polynomial sought has zeros 3u, 3v, 3w and
so is 4t3 - 21t2 - 27t + 54.

1.3. With si representing the symmetric functions of m, n, p, q, we find
that the required polynomial is

0 - mn>(t - v>(t - mq)(t - w)(t - nq)(t - pq)