Exercises for Chapter 1

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1 Solution Manual for A First Course in Abstract Algebra, with Applications Third Edition by Joseph J. Rotman Exercises for Chapter. True or false with reasons. (i There is a largest integer in every nonempty set of negative integers. Solution. True. If C is a nonempty set of negative integers, then C ={ n : n C} is a nonempty set of positive integers. If a is the smallest element of C, which exists by the Least Integer Axiom, then a c for all c C, so that a c for all c C. (ii There is a sequence of 3 consecutive natural numbers containing exactly 2 primes. Solution. True. The integers 48 through 60 form such a sequence; only 3 and 9 are primes. (iii There are at least two primes in any sequence of 7 consecutive natural numbers. Solution. False. The integers 48 through 4 are 7 consecutive natural numbers, and only 3 is prime. (iv Of all the sequences of consecutive natural numbers not containing 2 primes, there is a sequence of shortest length. Solution. True. The set C consisting of the lengths of such (finite sequences is a nonempty subset of the natural numbers. (v 79 is a prime. Solution. True. 79 < 8 = 9, and 79 is not divisible by 2, 3,, or 7. (vi There exists a sequence of statements S(, S(2,... with S(2n true for all n and with S(2n false for every n. Solution. True. Define S(2n to be the statement n = n, and define S(2n to be the statement n = n. (vii For all n 0, we have n F n, where F n is the nth Fibonacci number.

2 2 Solution. True. We have 0 = F 0,= F,= F 2, and 2 = F 3. Use the second form of induction with base steps n = 2 and n = 3 (verifying the inductive step will show why we choose these numbers. By the inductive hypothesis, n 2 F n 2 and n F n. Hence, 2n 3 F n.butn 2n 3 for all n 3, as desired. (viii If m and n are natural numbers, then (mn! =m!n!. Solution. False. If m = 2 = n, then (mn! = 24 and m!n! = 4..2 (i For any n 0 and any r =, prove that + r + r 2 + r 3 + +r n = ( r n+ /( r. Solution. We use induction on n. When n =, both sides equal + r. For the inductive step, note that [ + r + r 2 + r 3 + +r n ]+r n+ = ( r n+ /( r + r n+ = r n+ + ( rr n+ r = r n+2. r (ii Prove that n = 2 n+. Solution. This is the special case of the geometric series when r = 2; hence, the sum is ( 2 n+ /( 2 = 2 n+. One can also prove this directly, by induction on n 0..3 Show, for all n, that 0 n leaves remainder after dividing by 9. Solution. This may be rephrased to say that there is an integer q n with 0 n = 9q n +. If we define q =, then 0 = q +, and so the base step is true. For the inductive step, there is an integer q n with 0 n+ = 0 0 n = 0(9q n + = 90q n + 0 = 9(0q n + +. Define q n+ = 0q n +, which is an integer..4 Prove that if 0 a b, then a n b n for all n 0. Solution. Base step. a 0 = = b 0, and so a 0 b 0. Inductive step. The inductive hypothesis is a n b n.

3 3 Since a is positive, Theorem.4(i gives a n+ = aa n ab n ; since b is positive, Theorem.4(i now gives ab n bb n = b n+.. Prove that n 2 = 6 n(n + (2n + = 3 n3 + 2 n2 + 6 n. Solution. The proof is by induction on n. When n =, the left side is and the right side is =. For the inductive step, [ n 2 ]+(n + 2 = 3 n3 + 2 n2 + 6n + (n + 2 = 3 (n (n (n +, after some elementary algebraic manipulation..6 Prove that n 3 = 4 n4 + 2 n3 + 4 n2. Solution. Base step: When n =, both sides equal. Inductive step: [ n 3 ]+(n + 3 = 4 n4 + 2 n3 + 4 n2 + (n + 3. Expanding gives 4 n n n2 + 3n +, which is 4 (n (n (n Prove that n 4 = n + 2 n4 + 3 n3 30 n. Solution. The proof is by induction on n. If n, then the left side is, while the right side is = as well. For the inductive step, [ n 4] + (n + 4 = n + 2 n4 + 3 n3 30 n + (n + 4. It is now straightforward to chec that this last expression is equal to (n (n (n (n +..8 Find a formula for (2n, and use mathematical induction to prove that your formula is correct. Solution. We prove by induction on n that the sum is n 2. Base Step. When n =, we interpret the left side to mean. Of course, 2 =, and so the base step is true. Inductive Step (2n + (2n + = (2n ]+(2n + = n 2 + 2n + = (n + 2.

4 4.9 Find a formula for + n j= j! j, and use induction to prove that your formula is correct. Solution. A list of the sums for n =, 2, 3, 4, is2, 6, 24, 20, 720. These are factorials; better, they are 2!, 3!, 4!,!, 6!. We have been led to the guess S(n : + j! j = (n +!. j= We now use induction to prove that the guess is always true. The base step S( has already been checed; it is on the list. For the inductive step, we must prove n+ S(n + : + j! j = (n + 2!. Rewrite the left side as [ + j= j= ] j! j + (n +!(n +. By the inductive hypothesis, the braceted term is (n +!, and so the left side equals (n +!+(n +!(n + = (n +![ + (n + ] = (n +!(n + 2 = (n + 2!. By induction, S(n is true for all n..0 (M. Barr There is a famous anecdote describing a hospital visit of G. H. Hardy to Ramanujan. Hardy mentioned that the number 729 of the taxi he had taen to the hospital was not an interesting number. Ramanujan disagreed, saying that it is the smallest positive integer that can be written as the sum of two cubes in two different ways. (i Prove that Ramanujan s statement is true. Solution. First, 729 is the sum of two cubes in two different ways: 729 = ; 927 = Second, no smaller number n has this property. If n = a 3 + b 3, then a, b 2. It is now a matter of checing all pairs a 3 + b 3 for such a and b.

5 (ii Prove that Ramanujan s statement is false. Solution. One must pay attention to hypotheses. Consider a 3 +b 3 if b is negative: 728 = ( 0 3 = ( 3.. Derive the formula for n i= i by computing the area (n + 2 of a square with sides of length n + using Figure.. Solution. Compute the area A of the square in two ways. On the one hand, A = (n + 2. On the other hand, A = D +2 S, where D is the diagonal and S is the staircase. Therefore, S = 2 [ (n + 2 (n + But S is the sum we are seeing. ] = 2n(n Figure n = 2 (n2 + n Figure n = 2 n(n +.2 (i Derive the formula for n i= i by computing the area n(n + of a rectangle with height n + and base n, as pictured in Figure.2. Solution. Compute the area R of the rectangle in two ways. On the one hand, R = n(n +. On the other hand, R = 2 S, where S is the shaded region (whose area is what we see. (ii (Alhazen Forfixed, use Figure.3 to prove (n + i = i= i + + i= ( i p. i= p=

6 6 Solution. As indicated in Figure.3, a rectangle with height n + and base n i= i can be subdivided so that the shaded staircase has area n i= i +, while the area above it is + ( ( ( n. One can prove this, for fixed, by induction on n Figure.3 Alhazan s Dissection (iii Given the formula n i= i = 2n(n +, use part (ii to derive the formula for n i= i 2. Solution. ( i (n + i = i 2 + p Therefore, and so = = i 2 + i p=0 2 i(i + i (n + 2 i = 3 2 i 2, i 2 = 2 3 (n + 2 2n(n + = 3 2 (2n + n(n + = 6 (2n + n(n +. i.

7 7.3 (i Prove that 2 n > n 3 for all n 0. Solution. Base step. 2 0 = 024 > 0 3 = 000. (Note that 2 9 = 2 < 9 3 = 729. Inductive step Note that n 0 implies n 4. The inductive hypothesis is 2 n > n 3 ; multiplying both sides by 2 gives 2 n+ = 2 2 n > 2n 3 = n 3 + n 3 n 3 + 4n 2 = n 3 + 3n 2 + n 2 > n 3 + 3n 2 + 4n = n 3 + 3n 2 + 3n + n n 3 + 3n 2 + 3n + = (n + 3. (ii Prove that 2 n > n 4 for all n 7. Solution. Base step. 2 7 = 3, 072 > 7 4 = 83, 2. (Note that 6 4 = (2 4 4 = 2 6. Inductive step. Note that n 7 implies n 7. The inductive hypothesis is 2 n > n 4 ; multiplying both sides by 2 gives 2 n+ = 2 2 n > 2n 4 = n 4 + n 4 n 4 + n 3 n 4 + 4n 3 + n 3 n 4 + 4n 3 + 7n 2 n 4 + 4n 3 + 6n 2 + n 2 n 4 + 4n 3 + 6n 2 + n n 4 + 4n 3 + 6n 2 + 4n + = (n Around 30, N. Oresme was able to sum the series n= n/2 n by dissecting the region in Figure.4 in two ways. Let A n be the vertical rectangle with base 2 n and height n, so that area(a n = n/2 n, and let B n be horizontal rectangle with base 2 n + + and height. Prove that 2 n+ n= n/2 n = 2. Solution. You may assume that n=0 ar n = a/( r if 0 r <. Now compute the area using A n = B n.

8 8 B 3 B 2 B A A 2 A 3 B Figure.4 Oresme s Dissections. Let g (x,...,g n (x be differentiable functions, and let f (x be their product: f (x = g (x g n (x. Prove, for all integers n 2, that the derivative f (x = g (x g i (xg i (xg i+(x g n (x. i= Solution. Base step. Ifn = 2, this is the usual product rule for derivatives. Inductive step. Define h(x = g (x g n (x = f (x/g n+ (x. Rewrite what has to be shown: Now f (x = f (x = (h(xg n+ (x n+ j= g j (x f (x. g j (x = h (xg n+ (x + h(xg n+ (x [ g = = i (xh(x i= g i (x = n+ j= g j (x f (x. g j (x ] g n+ (x + [ ] f (x g n g n+ (x.6 Prove, for every n N, that ( + x n + nx whenever x R and + x > 0. Solution. We prove the inequality by induction on n. The base step n = says + x + x, which is obviously true. For the inductive step,

9 9 we record the inductive hypothesis: ( + x n + nx. Multiplying both sides of this inequality by the positive number + x preserves the inequality: ( + x n+ = ( + x( + x n ( + x( + nx = + (n + x + nx 2 + (n + x, because nx Prove that every positive integer a has a unique factorization a = 3 m, where 0 and m is not a multiple of 3. Solution. Model your solution on the proof of Proposition.4. Replace even by multiple of 3 and odd by not a multiple of 3. We prove this by the second form of induction on a. The base step n = holds, for = 3 0 is a factorization of the desired ind. For the inductive step, let a. If a is not a multiple of 3, then a = 3 0 a is a good factorization. If a = 3b, then b < a, and so the inductive hypothesis gives 0 and an integer c not divisible by 3 such that b = 3 c. It follows that a = 3b = 3 + c, which is a factorization of the desired ind. We have proved the existence of a factorization. To prove uniqueness, suppose that n = 3 m = 3 t m, where both and t are nonnegative and both m and m are not multiples of 3; it must be shown that = t and m = m. We may assume that t. If > t, then canceling 3 t from both sides gives 3 t m = m. Since t > 0, the left side is a multiple of 3 while the right side is not; this contradiction shows that = t. We may thus cancel 3 from both sides, leaving m = m..8 Prove that F n < 2 n for all n 0, where F 0, F, F 2,... is the Fibonacci sequence. Solution. The proof is by the second form of induction. Base step: F 0 = 0 < = 2 0 and F = < 2 = 2. (There are two base steps because we will have to use two predecessors for the inductive step. Inductive step:

10 0 If n 2, then F n = F n + F n 2 < 2 n + 2 n 2 (by inductive hypothesis < 2 n + 2 n = 2 2 n = 2 n. By induction, F n < 2 n for all n 0. Notice that the second form is the appropriate induction here, for we are using two predecessors, S(n 2 and S(n, to prove S(n..9 If F n denotes the nth term of the Fibonacci sequence, prove that m F n = F m+2. n= Solution. By Theorem., we have F n = (α n β n for all n. Hence, m F n = n= m n= m = n= (α n β n (α n β n ([ α = m+ α ] [ β m+ ]. β Now α(α =, so that /( α = α; similarly, /( β = β. Therefore, m n= F n = ([ α m+ α ] [ β m+ ] β = (( α[( α m+ ] [( β( β m+ ] = [ (α β + (α m+2 β m+2 ]. This is the desired formula, for (α β = and (α m+2 β m+2 = F m Prove that 4 n+ + 2n is divisible by 2 for all n. Solution. We use the second form of induction.

11 Base Step. Ifn =, then 4 n+ + 2n = 6 + = 2, which is obviously divisible by 2. Since our inductive step will involve two predecessors, we are obliged to chec the case n = 2. But = = 89 = 2 9. Inductive Step. 4 n+2 + 2n+ = 4 4 n n = 4 4 n+ + (4 2n 4 2n + 2 2n = 4(4 n+ + 2n + 2n ( 2 4. Now the last term is divisible by 2; the first term, by the inductive hypothesis, and the second because 2 4 = 2..2 For any integer n 2, prove that there are n consecutive composite numbers. Conclude that the gap between consecutive primes can be arbitrarily large. Solution. The proof has nothing to do with induction. If 2 a n +, then a is a divisor of (n +!; say, (n +! =da for some integer d. It follows that (n +!+a = (d + a, and so (n +!+a is composite for all a between 2 and n Prove that the first and second forms of mathematical induction are equivalent; that is, prove that Theorem.4 is true if and only if Theorem.2 is true. Solution. Absent..23 (Double Induction Let S(m, n be a doubly indexed family of statements, one for each m 0 and n 0. Suppose that (i S(0, 0 is true; (ii if S(m, 0 is true, then S(m +, 0 is true; (iii if S(m, n is true for all m 0, then S(m, n + is true for all m 0. Prove that S(m, n is true for all m 0 and n 0. Solution. Conditions (i and (ii are the hypotheses needed to prove, by (ordinary induction that the statements S(m, 0 are true for all m 0. Now consider the statements T (n : S(m, n is true for all m 0. We prove that all the statements T (n are true by induction on n 0. The base step has been proved above, and condition (iii is precisely what is needed for the inductive step.

12 2.24 Use double induction to prove that (m + n > mn for all m, n 0. Solution. According to Exercise.23, there are three things to verify. (i S(0, 0: ( = 0. (ii S(m, 0 S(m+, 0: if(m+ 0 > m, then (m+2 0 >(m+ 0 = 0? (iii S(m, n S(m, n + : does (m + n > mn imply (m + n+ > m(n +? Yes, because (m + n+ = (m + (m + n >(m + mn = m 2 n + mn > mn + m, for m 2 n mn and mn m. Notice that 2 n > n is the special case S(0, n..2 For every acute angle θ, i.e., 0 <θ<90, prove that sin θ + cot θ + sec θ 3. Solution. That θ is an acute angle implies that the numbers sin θ, cot θ, and sec θ are all positive. The inequality of the means gives [ 3 (sin θ + cot θ + sec θ ] 3 sin θ cot θ sec θ. Now sin θ cot θ sec θ = sin θ cos θ sin θ so that [ 3 (sin θ + cot θ + sec θ ] 3 and cos θ =, 3 (sin θ + cot θ + sec θ. Therefore, sin θ + cot θ + sec θ Isoperimetric Inequality. (i Let p be a positive number. If is an equilateral triangle with perimeter p = 2s, prove that area( = s 2 / 27. Solution. This is an elementary fact of high school geometry.

13 3 (ii Of all the triangles in the plane having perimeter p, prove that the equilateral triangle has the largest area. Solution. Use Heron s formula: if a triangle T has area A and sides of lengths a, b, c, then A 2 = s(s a(s b(s c, where s = 2 (a + b + c. The inequality of the means gives [ ] (s a + (s b + (s c 3 (s a(s b(s c = A2 3 s, with equality holding if and only if s a = s b = s c. Thus, equality holds if and only if a = b = c, which is to say T is equilateral..27 Prove that if a, a 2,...,a n are positive numbers, then (a + a 2 + +a n (/a + /a 2 + +/a n n 2. Solution. By the inequality of the means, [(a + a 2 + +a n /n] n a a n and [(/a + /a 2 + +/a n /n] n /a /a n. Now use the general fact that if p q > 0 and p q > 0, then pp qq to obtain (a + a 2 + +a n /n] n [(/a + /a 2 + +/a n /n] n (a a n (/a /a n. But the right side is a a n (/a (/a n =, so that (a + a 2 + +a n /n] n [(/a + /a 2 + +/a n /n] n. Taing nth roots gives (a +a 2 + +a n (/a +/a 2 + +/a n n True or false with reasons. (i For all integers r with 0 < r < 7, the binomial coefficient ( 7 r is a multiple of 7. Solution. True. (ii For any positive integer n and any r with 0 < r < n, the binomial coefficient r is a multiple of n. Solution. False. (iii Let D be a collection of 0 different dogs, and let C be a collection of 0 different cats. There are the same number of quartets of dogs as there are sextets of cats. Solution. True. (iv If q is a rational number, then e 2πiq is a root of unity. Solution. True.

14 4 (v Let f (x = ax 2 + bx + c, where a, b, c are real numbers. If z is a root of f (x, then z is also a root of f (x. Solution. True. (vi Let f (x = ax 2 + bx + c, where a, b, c are complex numbers. If z is a root of f (x, then z is also a root of f (x. Solution. False. (vii The primitive 4th roots of unity are i and i. Solution. True..29 Prove that the binomial theorem holds for complex numbers: if u and v are complex numbers, then (u + v n = u r n r v r. Solution. The proof of the binomial theorem for real numbers used only properties of the three operations: addition, multiplication, and division. These operations on complex numbers have exactly the same properties. (Division enters in only because we chose to expand (a + b n by using the formula for ( + x n ; had we not chosen this expository path, then division would not have been used. Thus, the binomial theorem really holds for commutative rings..30 Show that the binomial coefficients are symmetric : = r n r r=0 for all r with 0 r n. Solution. By Lemma.7, both ( r and n n r are equal to n! r!(n r!..3 Show, for every n, that the sum of the binomial coefficients is 2 n : = n n. Solution. By Corollary.9, if f (x = ( + x n, then there is the expansion ( f (x = + x + x n + + x n n. Evaluating at x = gives the answer, for f ( = ( + n = 2 n.

15 .32 (i Show, for every n, that the alternating sum of the binomial coefficients is zero: + +( 0 2 n = 0. n (ii Solution. If f (x = ( + x n, then f ( = ( n = 0; but the expansion is the alternating sum of the binomial coefficients. Use part (i to ( prove, for a given n, that the sum of all the binomial n coefficients r with r even is equal to the sum of all those r with r odd. Solution. By part (i, + ± = n Since the sign of r is ( r, the terms r with r even are positive while those with r odd are negative. Just put those coefficients with negative coefficient on the other side of the equation..33 Prove that if n 2, then ( ( r n r = 0. r r= Solution. Again, consider f (x = (+x n. There are two ways to describe its derivative f (x. On the one hand, f (x = n( + x n. On the other hand, we can do term-by-term differentiation: f (x = r= r x r r. Evaluating at x = gives f ( = n( n = 0, since n. On the other hand, we can use the expansion to see f ( = ( ( r n r. r r=.34 If r n, prove that = n r r. r Solution. Absent.

16 6.3 Let ε,...,ε n be complex numbers with ε j =for all j, where n 2. (i Prove that ε j ε j = n. (ii j= j= Solution. The triangle inequality gives u + v u + v for all complex numbers u and v, with no restriction on their norms. The inductive proof is routine. Prove that there is equality, ε j = n, j= if and only if all the ε j are equal. Solution. The proof is by induction on n 2. For the base step, suppose that ε + ε 2 =2. Therefore, 4 = ε + ε 2 2 = (ε + ε 2 (ε + ε 2 = ε 2 + 2ε ε 2 + ε 2 2 = 2 + 2ε ε 2. Therefore, 2 = + ε ε 2, so that = ε ε 2 = ε ε 2 cos θ = cos θ, where θ is the angle between ε and ε 2 (for ε = = ε 2. Therefore, θ = 0orθ = π, so that ε 2 =±ε. We cannot have ε 2 = ε, for this gives ε + ε 2 =0. For the inductive step, let n+ j= ε j =n +. If n j= ε j < n, then part (i gives ( n ε j + εn+ ε j + < n +, j= contrary to hypothesis. Therefore, n j= ε j = n, and so the inductive hypothesis gives ε,...,ε n all equal, say to ω. Hence, nj= ε j = nω, and so j= nω + ε n+ =n +.

17 7 The argument concludes as that of the base step. (n + 2 = (nω + ε n+ (nω + ε n+ = n 2 + 2nω ε n+ +, so that ω ε n+ =. By the base step, ω = ε n+, and the proof is complete..36 (Star of David Prove, for all n > r, that + r r + r r = +. r r r + r r r + r r + + r + Solution. Using Pascal s formula, one sees that both sides are equal to (n!n!(n +! (r!r!(r +!(n r!(n r!(n r +!..37 For all odd n, prove that there is a polynomial g n (x, all of whose coefficients are integers, such that sin(nx = g n (sin x. Solution. From De Moivre s theorem, we have cos nx + i sin nx = (cos x + i sin x n, sin nx = Im (cos x + i sin x n ( = Im i r r sin r x cos n r x. r=0

18 8 Write n = 2m +. Only odd powers of i are imaginary, so that, if r = 2 +, sin nx = ( n ( 2 + sin 2+ x cos 2(m x. 0 m But cos 2(m x = (cos 2 x m = ( sin 2 x m, and so we have expressed sin nx as a polynomial in sin x..38 (i What is the coefficient of x 6 in ( + x 20? Solution. Pascal s formula gives ( 20 6 = 484. (ii How many ways are there to choose 4 colors from a palette containing paints of 20 different colors? Solution. Pascal s formula gives ( 20 4 = 484. One could also have used part (i and Exercise Give at least two different proofs that a set X with n elements has exactly 2 n subsets. Solution. There are many proofs of this. We offer only three. Algebraic. Let X ={a, a 2,...,a n }. We may describe each subset S of X by a bitstring; that is, by an n-tuple where ɛ i = (ɛ,ɛ 2,...,ɛ n, { 0 ifa i is not in S ifa i is in S. (after all, a set is determined by the elements comprising it. But there are exactly 2 n such n-tuples, for there are two choices for each coordinate. Combinatorial. Induction on n (taing base step n = 0 is also fine; the only set with 0 elements is X =, which has exactly one subset, itself. If X has just one element, then there are two subsets: and X. For the inductive step, assume that X has n + elements, of which one is colored red, the other n being blue. There are two types of subsets S: those that are solid blue; those that contain the red. By induction, there are 2 n solid blue subsets; denote them by B. But, there are as many subsets R containing the red as there are solid blue subsets: each R arises by adjoining the red element to a solid blue subset, namely, B = R {red} (even the singleton subset consisting of the red element alone arises in this way, by adjoining the red element to. Hence, there are 2 n + 2 n = 2 n+ subsets.

19 9 Binomial Coefficients. If X has n elements, then the number of its subsets is the sum of the number of 0-subsets (there is only, the empty set, the number of - subsets, the number of 2-subsets, etc. But r is the number of r-subsets, as we have seen in the text, and so the total number of subsets is the sum of all the binomial coefficients, which is 2 n, by Exercise A weely lottery ass you to select different numbers between and 4. At the wee s end, such numbers are drawn at random, and you win the jacpot if all your numbers match the drawn numbers. What is your chance of winning? Solution. The answer is 4 choose, which is ( 4 =, 22, 79. The odds against your winning are more than a million to one..4 Assume that term-by-term differentiation holds for power series: if f (x = c 0 +c x +c 2 x 2 + +c n x n +, then the power series for the derivative f (x is f (x = c + 2c 2 x + 3c 3 x 2 + +nc n x n +. (i Prove that f (0 = c 0. Solution. f (0 = c 0, for all the other terms are 0. (If one wants to be fussy this is the wrong course for analytic fussiness then the partial sums of the series form the constant sequence c 0, c 0, c 0,... (ii Prove, for all n 0, that f (n (x = n!c n + (n +!c n+ x + x 2 g n (x, where g n (x is some power series. Solution. This is a straightforward proof by induction on n 0. The base step is obvious; for the inductive step, just observe that f n+ (x = ( f (n (x.asf (n (x is a power series, by assumption, its derivative is computed term by term. (iii Prove that c n = f (n (x(0/n! for all n 0. (Of course, this is Taylor s formula. Solution. If n = 0, then our conventions that f (0 (x = f (x and 0! = give the result. For the inductive step, use parts (i and (ii..42 (Leibniz Prove that if f and g are C -functions, then ( fg (n (x = f ( (x g (n (x. =0 Solution. The proof is by induction on n.

20 20 If n =, then the equation is precisely the product rule of Calculus: ( fg = f g + fg. For the inductive step, we have ( fg n+ =[( fg n ] [ = f g n ] = = = =0 f [ g n ] =0 f [ + g n + f g n +] =0 f + g n + =0 f g n +. Rewrite this last expression without the sigma notation: ( f 0 0 g n+ n ( + f g n n + + f g n + + ( + f 0 g n n + + f g n +. The coefficient of f g n + is thus ( + n ( = n+, by Lemma.7, as desired..43 Find i. Solution. By De Moivre s theorem, since i = e iπ/2,wehave =0 i = e iπ/4 = cos π/4 + i sin π/4 = i (i If z = r[cos θ + i sin θ], show that w = n r [cos(θ/n + i sin(θ/n] is an nth root of z, where r 0. Solution. By De Moivre s theorem, w n = r n ([cos(θ/n + i sin(θ/n] n = r[cos(θ + i sin(θ]. (ii Show that every nth root of z has the form ζ w, where ζ is a primitive nth root of unity and = 0,, 2,...,n.