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248 Answers to Exercises and Solutions to Problems

The treasure is located at (l/2)( l+ 1 ‘) , midway between P and Q. Since this

does not depend on z, the position of the treasure is the same regardless

of the position of T.

3.8. For positive integer n, use induction. For n = -1, note that

(cosx-isinx)(cosx+isinx) = 1. For n = -6 < 0, we have that (cosx+

isin~)-~ = (cos Lx + isin kc)-’ = cos(-k)x + isin(-k)c.

3.9. Each equation implies that IzI = 1, so that z = cos6’ + isin 0 for some

8.

(a) cos 30 + i sin 3~9 = 1 implies that 38 is a multiple of 27r, thus the solu-

tions are 1, cos(2?r/3)+isin(27r/3) = $(-l+ifi), cos(2r/3)-isin(27r/3) =

4(-l - ifi).

(b) 1, i, -1, -i.

(c) 1, (1 + ifi)/Z, (-1 + h&/2, -1, (-1 - id)/2, (1 - i&)/2.

(d) 1, (1+ i)/& i, (-1+ i)/fi, -1, (-1 - i)/fi, -i, (1 - i)/fi.

In each case, the numbers are equally spaced on the perimeter of the unit

circle with center at the origin.

3.10. (a) Since x2 - y2 = u, 22~ = b, it follows that x2 and -y2 are the

roots of

t2 - at - b2/4 = 0.

Let c2 = u2 + b2, c > 0. Then t2 = $(c + a), y2 = i(c - u). If b > 0,

(x:,Y> = kh/m, Ad-); if b < 0, (2, Y> = (=h/m,

r/m). In the case that b = 0, c = Ial, and z = 0 or y = 0

according as a is negative or positive.

(b) 3 - 4i and -3 + 4i.

3.11. a) We have 2t = -3 f da. Applying Exercise 10 yields

+ -3 + 4i = f(1 + 2i). The roots are -2 - i and -1 + i.

(b) Note that the discriminant is -7 - 24i = (3 - 4i)2. The roots are

2 - 3i and -1 + i.

3.12. We can apply the quadratic formula for the roots of the equation.

Exercise 10 shows how the square root of the discriminant can be found.

3.13. Il+izl = Il-izl (l+iz)(l-27) = (l-iz)(l+zY) U i(z-YF) =
-i(z -F) 2 -F = 0 _ z is real.

3.15. (a) Let e = arccos x. The equation is equivalent to

cos(n + i)e + cos(n - qe = 2 cos d cos 8.

(b) x, 2x2 - 1, 4x3 - 32, 8x4 - 8x2 + 1.
(c) Since Tn+l = 2xTn - Tn-l, by induction it can be established that
T, is a polynomial of degree n.