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Solutions to Problems; Chapter 1 255

from which the result follows.

8.14. (a) The quadratic equation can be rewritten (z - u)~ = a2 - b. Let u
be such that u2 = a2 - b. Then the roots of the equation are -a + u and
--a - u. The roots of the quadratic and linear equations are collinear if and
only if, for some real t,

w = (1 - t)(-u + u) + t(-u - u) = --a + (1 - 2t)u

if and only if (20 + u)/ u is real, if and only if (w + u)-il is real.
(b) Suppose the roots of z2 + 2cz + d = 0 are -c + v and -c - v. Then
the four roots are collinear if and only if u and v are real multiples of a - c.
(Make a sketch.)

8.15. Let cos48 = (b2 - 8uc)/b2. Then cos2 28 = (b2 - 4uc)/b2, from which
it follows that 2cos2 0 - 1 = d-/b. It follows that -(b/u) cos2 0 is a
root of the given quadratic equation.

8.16. The equation z2 - 2(cos 4 + i sin $)z + 1 = 0 is equivalent to

[z - (cos qj + i sin +)I” = cos2$-sin2+- 1+2isin$cos4

= 2i sin ~(COS 4 + i sin f#~)

= 2 sin 4[cos(+ + 7r/2) + i sin(+ + r/2)].

Solving for z yields two roots

u=(cosf$+isin~)+~%Gj(cosB+isinf9)

v=(cos4+isin4)-&Gj(cosB+isinB)

where 0 = 412 + 7r/4. (Note that 7r/4 < e < 37r/4.)
Observe that

cos 4 = sin(a/2 + 4) = sin 28 = 2 sine cos e

sin~=-c0s(7r/2+~)=2sin2~-1=1-2Cos2e.

Then

u+i=(2sinB+&GKj)(cose+isin8)

v+i=(2sine-$GGj)(cosB+isin8).

Since 2sin 8 - ,/ZGGjJ = 2sin B - &%? B - 2 > 0, it follows that
arg(u + i) = arg(v + i) = 8.
Also,

u-i = 2 cos B(sin e - i cos e) + J~(COS e + i sin e)

= [-2ic0sB + ~iZEji~(COSe + isine)

v-i = [-2i cam e - &ZG~](COS e + i sin e)