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Solutions to Problems; Chapter (^5 335)
The distance between the outer zeros of x3+px+q is equal to the distance
between the outer zeros of the given polynomial, and it lies between
x:4 - tg = 3(-p/3)‘12 = (a2 - 3b)l12
and
x2 - Xl = 2(-p) ‘I2 = (2/d)(a2 - 3b)‘i2.
4.16. ‘It is clear that .r6 + 6% + 10 = 0 has no pure imaginary roots. Since
6% + 10 2 0 when z > -513 and z6 + 6% 1 0 when t < -513, there are no
real roots. Since the coefficients are real, the number of zeros in the first
and fourth quadrants are equal as are the number of zeros in the second
and third quadrants. This it suffices to consider zeros z = r(cos 0 + isin 0)
for which r > 0 and 0 < 0 < 7r, i.e. sin0 > 0.
For such a root
r6cos60+6rcosB+10=0... (1)
r6sin68+6rsin0=0.... (2)
From (2), r5 = -6 sin e/sin 68 + sin 60 < 0 + n/6 < i9 < n/3, ?r/2 < 0 <
2~/3 or 5~/6 < 6’ < r. Substituting for r5 into (1) yields
0 = r(r5 cos 68 + 6 cos 0) + 10 = (6r sin 50)/(sin 60) + 10
so that r = -(5 sin 68)/(3 sin se).
Hence e must satisfy the equation f (8) = 0 where
f (0) = 6.35 sin 0 sin5 58 - 55 sin6 68.
Note that f(O) is positive for 0 = 7r/6, ?r/2 and 5?r/6, and negative for
0 = 7r/3 and 2rr/3. Since f(r) = 0, we look at f(r - 4) = f(4), where 4 is
very small. In this case, sin+ & 4, sin54 L ~I#J and sin64 - 64, so
f (4) 2 6.35 - 55@- - 55 .6’@ < 0.
Thus f(4) must vanish in each of the open intervals
(0, r/6), (r/6, r/3), (43~/2), (~/2,27r/3), (2a/3,5~/6), (5~/6,4.
Note that the vanishing of f(d) entails that sin50 must be positive. Since
sin60 < 0 for e in the second, fourth and sixth intervals, we obtain a
positive value of r and a viable solution (r,d) to the equations (1) and (2).
(The zeros off (6) in the other three intervals lead to inadmissible solutions
with r < 0.)
Thus there are six zeros of z6 + 6% + 10, one in the first quadrant whose
argument is between 916 and n/3, two others in each of the second and
third quadrants, and one in the fourth quadrant.