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Solutions to Problems; Chapter 8 385

so that

(u + bi)2 = (c + di) + 2(u - bi)(cosO + isine)

=+ I(u + bi)2 - (c + di)l = 21~ - bil

* (u2 - b2 - c)~ + (2ub - d)2 = 4(u2 + b2).

[Solution by Georges Gonthier.]

Third solution. We have

(Q2 - b2 - c)/2 = cos(v + w) + cos(u + w) + cos(u + v)... (1)

(2ub - d)/2 = sin(v + w) + sin(u + w) + sin(u + v)... (2)

u2 -I- b2 - 3 = cos(v - w) + cos(w - u) + cos(u - v).... (3)

The process that yields (3) f rom the equations for a and b yields the fol-

lowing equation from (1) and (2):

[(u2 + b2 - c)/212 + [(2ub - d)/212 - 3

= cos((u + w) - (u + v)) + cos((u + v) - (v + 20)) + cos((v + w) - (u + w))

= u2+b2-3.

[Solution by Alexander Pruss.]

38.

X-^2

Y2 = Cot2 e + tan2 e - sec2 e - ~08~ e +^4

= --I + cos2 eccsc2 e - 1) + 4 = 3 + ~05~ e Cot2 e

while

xy = csc e - cotecose+tanesece-sine

= csce(i - ~0s~ e) - sine + tan e set e = tan e set e.

Hence x2y2(z2 - y2 - 3) = 1.

39. The polynomial equals

((3/4) - x) + (x2 + x4)( 1 - x) + Lr6 = (x - 1)(X5 + X3 + x) + (3/4)

is clearly positive for x 5 3/4 and x > 1. The derivative of the polynomial
is

6x5 - 5z4 + 4x3 - 3x2 + 2x - 1 = ( 1/2)[(3x4 + 2x2 + 1)(4x - 3) + (1 - x4)]

which is positive for 314 5 x < 1. Hence the given function increases on

the closed interval [3/4, l] and so is positive there.