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338 Answers to Exercises and Solutions to Problems

where g(z) is a sum of the form C f zk with /Z 2 2. By an estimate similar

to that above,

I(1 - z)(l + z + .zfla +.. *)I > (1 - ]z( - ]z]2)(1 - ]z])-’ > 0.

Thus, the result follows.

4.21. Suppose that all the zeros lie on a straight line. Since the sum of

the zeros is 0, the line must pass through 0, so that the zeros are -UJ, uw,

(1 - U)UI for some complex w and real u with 0 5 u 5 1. We have that

[-u - (1 - U) + U(1 - u)]U? = 12(1+ ifi)

+ (u2 - u + 1)2u2 = -24(cos r/3 + isin 7r/3)

= 24(cos 4x/3 + i sin 4~13).

Hence w = fr(cos2?r/3 + isin2rr/3), where (u2 - u + l)r2 = 24.

By considering the product of the zeros, we find that

a = w3(u - u2) = fr3(u - u”) = ~t24~/~(u - u”)(u” - u + 1)-3/2

= &243/2(1/4 - v)(3/4 + v)-~/~

where 0 5 v = (l/2 - u)” 5 l/4. Hence a must be real.

(Remark. One has the suspicion that Ial is minimized when the equation
has a double root. This will occur when u = 1 -u = l/2. In fact, by finding
the zeros j&( 1 - i&) of the derivative 3[z2 + 4( 1 + id)], we can identify
the roots of the equation in the critical situation as &fi(l - ifi) with
multiplicity 2 and ~2fi(l - ifi). This will occur when a = ~32fi. In
view of this, we would expect 243/2[1/4 - v][3/4 + vlT3j2 to assume its
maximum value of 32& when v = 0. Thus, we look at the difference of the
squares of these two quantities.)
Now

Hence

2l’ - 243[1/4 - v12[3/4 + VI-”

= 2’[3/4 + ~]-~[4(3/4 + v)~ - 27(1/4 - v)“]

= 27[3/4 + v]-~v(~v - 9)2 2 0.

JaJ = 243/2[1/4 - v][3/4 -I- v13j2 5 fi” = 324.
On the other hand, suppose a is real and satisfies ]a] 5 32fi. Then we
can find u such that ~t24~/~u(l - u)(l - u + u2)-‘j2 assumes the value
a. Then, if r and w are determined by the equations above, we find that
the symmetric functions in -w, uw, (1 - u)w yields the coefficients of the
equation and therefore must be its roots.
This result is generalized in El. Math. 12 (1957), 12.