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352 Answers to Exercises and Solutions to Problems

there must be zeros in the other, in order for the imaginary parts to cancel
in the sum. Any zero of p(z) satisfies the equation, so that if p(z) has any
nonreal zeros, the result follows. Suppose p(z) has only real zeros. Then
P(Z) = CH(Z - ri) w h ere all the ri are real and c is nonreal. The polynomial
p(-Z) - 1 = CH(-Z - ri) + 1 assumes a nonreal value when z is real and
not equal to -ri for any i and assumes the value 1 for z = -ri. Hence all
the zeros of p(-z) - 1, which also satisfy the equation, are nonreal and the
result follows.
If p(z) = iz + l/2, then the equation becomes (iz + 1/2)2 = 0, both of
whose roots are in the upper half plane.

4.14. The polynomials zk(t f 1) satisfy the requirements, so we may assume
that the degree n exceeds 1. Suppose that the leading coefficient is 1, so
that the polynomial p(t) has the form

t" + cn-ltn-l + c,-2t+-2 +.. * + qt + co

with zeros ri. Note that all the ri fail to vanish. Then

0 5 Crf = czml - 2c,-z =s- cn-z = -13 Crf = 3.

Similarly,
0 2 Criy2 = (c~/co)~ - 2(cz/cc) * Crf2 = 3.
Hence
e(rf + ri”) = 6.
i=l

Since rf + rr2 1 2 for each i, it follows that n 2 3. If n = 3, then lril = 1
for each i and the only possibilities are t3 + t2 - t - 1 = (2 + 1)2(t - 1) and
t3 - t2 - t + 1 = (t - 1)2(t + 1). If n = 2, then t2 - t - 1 and t2 + t - 1
are the only possibilities and both satisfy the requirements. If the leading
coefficient is to be -1, then the only possibilities are the negatives of these
polynomials.

4.15. Let the zeros of j(x) be ri and those of f’(x) be sj. Then, since
f’(x) = f(x)C(x - ri)-l. F or each j, 0 = f’(sj) = f(sj)E(sj - ri)-l. Since
f(Sj) # 0, C(Sj - ri)-’ = 0.

4.16. The purported roots are of the form w = v3(4v3 - 3~)~’ =
(4 - 3~~)~’ where 2v = u + u-l and u is a primitive 7th root of unity.
Since 4v2 = u2 + us2 + 2 and 8v3 = u3 + um3 + 6v, it follows that

8v3 + 4v2 -4v-l=O.

Thus, the three zeros of 8t3+4t2-4t- 1 are vi = cos(2~/7), vz = cos(4%/7)
and so the three zeros of t3 + 4t2 - 4t - 8 are v;i,
the cubic whose zeros are w,T’ where wi =