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Answers to Exercises; Chapter 1 249

(d) By de Moivre’s Theorem,

cosn0 + isin n0

n

n

=

cc >

cos”-* e sins Oi*

S

a=0

n

= Y( )

2:

cos n-2r esin2r e(i2)r

r=O

+ “n$21 ( 2t “+ 1 ) COStl-2t-l ,in2f+l @j2t+l

= ‘z( ~r)(-l)~~0s+2rO(sin2B)‘+i{.ee}.

t-=0

Equating real parts, setting 0 = arccos x, cos 0 = x, sin2 0 = 1 - x2 yields

the results.

4.1. (a) Let p(t) = at3 + bt2 + ct + d. Then for any number P, p(t) -p(r) =
(t -r)[u(t2+rt +r2)+b(t+r)+c] = (t-r)[ut2+(ur+b)t +(ur2+br+c)].
The result follows immediately.
(b) One root is 1. By (a), we can write the given cubic as a product of
t - 1 and a quadratic polynomial. Indeed, it is equal to (t - l)(t” + t - 3).
The solutions are: 1, [-1 + &l/2, [-1 - a]/2.

4.2. The solutions are E = 1,2,9.

4.4. (d) If ui is the single root of the quadratic, we have q = -2~: and
p = -3ui, and the factorization can be checked.
(e) We can choose real r and 0 such that

U: = r3(cos 38 + isin 38)

71: = r3(cos3e - isin38).

The condition that uevo be real requires that ue = r(cos 0 + i sin 0) and
ve = r(cos t9 - i sin 0). Note that w = cos 27~13 + i sin 2~13.

4.5. (a) -3, (3 + iy/T?)/S, (3 - ifi)/2.
(b) u3 = 3.569O[cos 147.30°+isin 147.30’1 and u = l.OOOO+l.l549i. The
solutions are 2, 1, -3.

4.6. x = t + 5 yields t3 - 108t + 432 = 0, x = -7, 11, 11.

4.7. x3 + px + q = 0 being the equation, set x = u2b + ab2. An argument
similar to that in Exercise 4(a) leads to 3u3b3 = -p and p(u3 + b3) = 3q.
Having found the determination a and b for the cube roots of a3 and b3
respectively, the pairs (a, b), ( uw, bw), (uw2, 6w2) all yield the same values
of u2b + ub2.