4.7. Problems on Equations in One Variable 149The result we want is that all the zeros of the polynomial f’(t) also lie
in the polygonal region P.
Verify that this result is true for any polynomial of degree 2. What are
the possible shapes of P in this case?
To prove the result in general, we require some preliminary facts which
will help reduce the situation to the special case that two zeros are real
and the remainder lie in the upper half plane.
(1) If Sk is the set of zeros of fck)(t) (k = 0, 1,... , n - l), then the set of
zeros of the polynomial gck)(t), where g(t) = f(t + w),‘is Sk - w = {z-w :
% E Sk}.
(2) With Sk as in (1)) the set of zeros of htk)(t), where h(t) = f(wt), is
w--l& = {w-l% : % E Sk}.
(3) If rl, r2,... ,rk are the zeros of f(t) with respective multiplicities
ml, mz,... , mk, thenf’(t) ml
f(t)=t-rl+“‘+~*From (1) and (2), we see that the result holds for f(t) iff it holds for
either f(t + w) or f(wt) for fixed w # 0. By suitable choice of wr and wz,
we can arrange that any two zeros representing vertices of P correspond to
two real zeros u and w of the polynomial q(t) = f(wl(t + wg)) and that the
remaining zeros of q(t) 1 ie in the upper half plane, i.e. if q(w) = 0, then Im
w 1 0.
From (3), any zero of f’(t) w h ic h is not an ri must make the rational
function on the right side vanish. We show that the right side will not vanish
for t = z when Im z < 0. Look at each term individually, and observe that
Im(r - ri) < 0, so that Im(mi(z - pi)-‘) > 0. Hence Im(f’(f)/f(z)) is
positive.
Now go back to the function f(t) and argue that all of the zeros of f’(t)
lie in the same halfplane determined by each edge of P as contain the zeros
of f(t). Hence, all the zeros.of f’(t) must lie in P.
4.7 Problems on Equations in One Variable
- Solve x4 + a4 = 4az(x2 + a”).
- Solve the equation
(x2 - 2 - 2)4 + (2x + 1)” = (x2 + x - 1)4- Solve (x2 - 4)(x2 - 2x) = 2.
- Solve the equation (x + a)(x + 2a)(x + 3a)(x + 4a) = b4.