Favorite Topics from Complex Arithmetic, Analysis and Related Algebra

Favorite Topics from Complex Arithmetic, Analysis and Related Algebra construction at 09FALL/complex.tex Franz Rothe Department of Mathematics University of North Carolina at Charlotte Charlotte, NC 3 frothe@uncc.edu September 7, 011 Contents 1 Some Number Theory 1.1 Prime numbers................................ 1. Rational and irrational............................ 5 Algebra of the Complex Numbers 5.1 Arctan identities............................... 6. Gaussian primes................................ 7.3 Sums of two squares............................. 11. Roots..................................... 16.5 The solution of the reduced cubic equation................. 19.6 The square root of a complex number.................... 3.7 Pythagorean triples............................... Roots of unity................................. 7 3 Polynomials 35 3.1 Rational zeros of polynomials........................ 35 3. Gauss Lemma................................ 36 3.3 Eisenstein s irreducibility criterium..................... 0 3. Descartes Rule of Signs........................... 0

Polynomials over the Complex Numbers 5.1 A proof of the Fundamental Theorem of Algebra.............. 5. Meditation on Descartes rule of signs.................... 7.3 Estimation of zeros with Rouché s Theorem................ 51. Perron s irreducibility criterium....................... 5.5 The limiting case............................... 53 5 The Residue Theorem and its Consequences 56 5.1 The residue theorem and the logarithmic residue theorem......... 56 5. Comparing two functions........................... 5 1

1 Some Number Theory 1.1 Prime numbers Proposition 1.1 (Euclidean Property). If a number c divides the product ab and gcd (c, a) = 1, then c divides b. Standard proof. By the extended Euclidean algorithm, there exist integers s, t such that 1 = sa + tc b c = sab c + tb The second line results by multiplication of both sides with b. Because c divides ab, the c right hand side is an integer. Hence c divides b, as to be shown. Second proof. Both numbers a and c are divisors of both products ab and ac. Hence both a and c are divisors of the greatest common divisor Hence the integer G := gcd (ab, ac) (1.1) q := ac G is a divisor of both a and c. Hence q is a divisor of gcd (a, c) = 1, which was assumed to be one. Hence q = 1, and ac = G is a divisor of ab. This implies that c is a divisor of b, as to be shown. Definition 1.1 (prime number). A prime number is an integer p, which is divisible only by 1 and itself. Euclid and many other mathematicians have shown that there exist infinitely many prime numbers. We put them into the increasing sequence p 1 =, p = 3, p 3 = 5, p = 7,... It is rather easy to see that for every positive integer, there exists a decomposition into prime factors. Let a and b be any positive integers. There exist sequence α i 0 and β i 0, with index i = 1,,... and only finitely many terms nonzero such that (1.) a = p α i i, b = p β i i i 1 i 1 The uniqueness of the prime decomposition turns out harder to prove. Astonishingly, the proof depends on Euclid s lemma, the proof of which in turn relies on the extended Euclidean algorithm.

Proposition 1. (Euclid s Lemma). If a prime number divides the product of two integers, the prime number divides at least one of the two integers. Reason. Let p be the prime number, and the integers be a and b. We assume that p divides the product ab, but p does not divide a. We need to show that p divides b. Because p does not divide a, the definition of a prime number implies gcd (a, p) = 1. By the extended Euclidean algorithm, there exist integers s, t such that 1 = sa + tp Hence b p = sab p + tb Because p divides ab, the right hand side is an integer. Hence p divides b, as to be shown. Proposition 1.3 (Monotonicity). Let a and b have the prime decompositions (1.3) a = p α i i, b = p β i i i 1 i 1 The number b is a divisor of a if and only if β i α i for all i 1. Reason. If β i α i for all i 1, then a = qb with q = i 1 p α i β i i and hence b is a divisor of a. Conversely, assume that b is a divisor of a. We need to show that β i α i for all i 1. Proceed by induction on b. If b = 1, then β i = 0 for all i 1, and the assertion is true. Here is the induction step b < n b = n : Let p i be any prime factor of b, which means that β i 1. Because p i divides b and b divides a, the prime p i divides a. By Euclid s Lemma Proposition 1., p i is a divisor of one of the primes p j occurring in the prime decomposition of a. Hence α j 1. Because different primes cannot divide each other, this implies i = j and p i = p j. Hence b p j < n is a divisor of a p j. By the induction assumption, this implies β i α i for all i j, as well as β j 1 α j 1 and hence β i α i for all i 1. Proposition 1. (Uniqueness of prime decomposition). The prime decomposition of any positive integer is unique. Reason. Assume a = p α i i and a = p β i i i 1 i 1 Because a divides a, the fact given above both tells that β i α i and α i β i for all i 1. Hence β i = α i for all i 1. 3

Proposition 1.5. Let a and b have the prime decompositions (1.3). The prime decompositions of the greatest common divisor and least common multiple are (1.) (1.5) gcd (a, b) = i 1 lcm (a, b) = i 1 p min[α i,β i ] i p max[α i,β i ] i 10 Problem 1.1. Check these formulas for a = 1001, b = 1. Proof of Proposition 1.5. Let g be the righthand side of equation (1.5). check properties (i) and (ii) defining the greatest common divisor. We need to (i) g divides both a and b. Check. This is clear, because both min[α i, β i ] α i and min[α i, β i ] β i for all i 1. (ii) If any positive integer h divides both a and b, then h divides the greatest common divisor g. Check. Let (1.6) h = i 1 p γ i i be the prime decomposition of h. Because h divides both a and b, monotonicity implies that both γ i α i and γ i β i for all i 1. Hence γ i min[α i, β i ] for all i 1, which easily implies that h is a divisor of g. Let l be the righthand side of equation (1.5). We need to check properties (i) and (ii) defining the least common multiple. (i) The number l is a multiple of both a and b. Check. This is clear, because both max[α i, β i ] α i and max[α i, β i ] β i for all i 1. (ii) If any positive integer k is a multiple of both a and b, the integer k is a multiple of the least common multiple l.

Check. Let (1.7) k = i 1 p γ i i be the prime decomposition of k. Because k is a multiple of both a and b, monotonicity implies that both γ i α i and γ i β i for all i 1. Hence γ i max[α i, β i ] for all i 1, which easily implies that k is a multiple of l. 1. Rational and irrational Proposition 1.6. For any natural numbers r 1 and a 1, the r-th root r a is only a rational number if it is even an integer. Proof. The assertion is clear for r = 1 or a = 1, hence we may assume r and a. Now assume that r a = m n n r a = m r with natural m, n. We show that any prime number p dividing n has to divide m, too. Hence after cancelling common factors, we get n = 1, and hence the root is an integer. Now assume that the prime p divides n. Hence p r divides n r, which in turn divides an r = m r. Hence p r divides m r. Hence, by Euclid s Lemma p divides m, as claimed. As already explained, the assertion follows. Proposition 1.7. For any natural numbers r and any a, which is not the r-th power of an integer, the root r a is irrational. Especially, the roots of the primes are all irrational. Proof. If the root r a is rational, it is even an integer m, and hence a = m r is the r-th power of m. Take the contrapositive: If a is not the r-th power of an integer, the root r a is irrational. Algebra of the Complex Numbers Definition.1 (Gaussian integer ). A complex number with integer real- and imaginary part is called a Gaussian integer. 5

.1 Arctan identities What is remarkable about the following products? (a) ( + i) (3 + i) = 5 + 5i (b) ( + i) (7 i) = 5 + 5i (c) (3 + i) (7 + i) = 50 + 50i (d) (5 + i) (39 i) = 11 + 11i Each time, the real- and imaginary parts turn out to be equal. Do there exists more examples with this special property? For any positive x > 0, the principal argument of the complex number x + i is Arg (x + i) = arctan 1. Too, Arg (1 + i) = arctan 1 = π. Taking the arguments of the x formulas (a) through (d), one gets: (a) (b) arctan 1 + arctan 1 3 = π arctan 1 arctan 1 7 = π (c) (d) arctan 1 3 + arctan 1 7 = π arctan 1 5 + arctan 1 39 = π Of course, left- and right-hand side of these formulas could still differ by an integer multiple of π. But it is clear that this cannot happen in the given four examples. Do there exist more similar formulas? Open Problem (May be difficult). How many solutions has the equation (.1) c + 1 = N with natural numbers N and c? Open Problem (May be difficult). Are there more than four solutions to the equation (.) M (c k + 1) = N k=1 with natural numbers N and M, and c k for k = 1,..., M? 6

. Gaussian primes It is straightforward to see that the Gaussian integers are a ring. The units in any ring are defined as its invertible elements. Clearly there are just four units 1, i, 1, i in the ring of Gaussian integers. Furthermore, division with remainder is possible. Hence the Euclidean algorithm works in the Gaussian integers. Consequently, any Gaussian integer can be decomposed uniquely into a product of irreducible elements unique up multiplication by the four units. Hence the Gaussian integers are a unique factorization domain, abbreviated UFD. The irreducible elements are called Theorem.1 (Gaussian primes). The Gaussian primes are: (a) the number 1 + i. (b) the usual primes p = 3, 7, 11, 19,... which are p 3 mod. (c) all pairs p + iq, p iq where p + q = r is a prime r 1 mod. Indeed, for every usual prime r = 5, 13, 17, 9,... which is r 1 mod there exists a unique decomposition into a sum of two integer square. The decomposition r = p + q becomes unique by the additional requirement that the (negative or positive) integer p 1 mod and q is even and positive. Here is a table with the smallest examples. Too, I list the squares a + ib = (p + iq). r p q a b 5 1 3 13 3 5 1 17 1 15 9 5 1 0 37 1 6 35 1 1 5 9 0 53 7 5 61 5 6 11 60 73 3 53 9 5 39 0 97 9 65 7 101 1 10 99 0 109 3 10 91 60 113 7 15 5 137 11 105 At that point, my heating system had been repaired, and I stopped. 10 Problem.1. Prove that there are infinitely many primes p 3 mod. 7

Answer. Let p 1,... p N be any primes congruent to 3 modulo. The number has the following properties: (a) P is odd. (b) P 3 mod. P := 1 + (c) P is not divisible by p ν for any ν = 1... N. By (a) and (b), P cannot be the product of primes which are all congruent to 1 modulo. Hence there exists a prime p congruent to 3 modulo dividing P. By item (c), this prime p is different from p 1,... p N. Hence there exist at least N + 1 different primes equivalent to 3 modulo. Since we know that such primes exist, there do exist infinitely many. 10 Problem.. Prove that there are infinitely many primes r 1 mod. Answer. Let r 1,... r N be any primes congruent to 1 modulo. Define the number P := 1 + Indeed P has in the Gaussian integers the factoring [ ] [ ] N N P = 1 + i r ν 1 i r ν ν=1 Neither of the two factor is divisible by 1 + i nor any prime p 1 mod. Hence we get the following properties: (a) P is odd. (b) P 1 mod. (c) P is not divisible by r ν for any ν = 1... N. N ν=1 N ν=1 (d) P is not divisible by any prime p 3 mod. By (a) and (d), P is the product of primes congruent to 1 modulo only. By item (c), all these factors are different from r 1,... r N. Hence there exist at least N + 1 different primes equivalent to 1 modulo. Since we know that such primes exist, there do exist infinitely many. p ν r ν ν=1

10 Problem.3 (The little forgotten theorem). Assume that (.3) z = a + ib = a + b e iπk l with real integers a, b, k, l such that a and b are relatively prime, and k and l are relatively prime. How many different nonzero values can z 0 assume? Prove that there are non besides the obvious eight ones. Answer. The complex number z can only assume the eight values 1, i, 1, i and ±1±i, but no other values. Because of the assumption gcd (a, b) = 1, the extended Euclidean algorithm yields integers s, t such that sa tb = 1, and hence (a + ib)(s + it) = 1 + i(at + bs). Hence no real prime can divide a + ib. We now repeatedly use the fact that the Gaussian integers are a unique factorization domain. Assume the Gaussian prime p + iq divides a + ib. Ruling out all other cases, we show the only possibility is p + iq = 1 + i. Because of the remark above, the case of a real prime p 3 mod, and q = 0, is impossible. Hence we are left with the task to rule out the case of a Gaussian prime p + iq with 5 p + q 1 mod. In the l-th power of the given equation (.3), (.) (a + ib) l = (a + b ) l the two sides sides are both Gaussian integers as well as real, hence real positive integers. Equation (.) implies that p + iq divides the natural number (a + b ) l. Hence p iq divides this same number, too. Since the Gaussian prime p iq divides (a+ib) l, unique factorization implies that p iq divides a + ib, too. Since 5 p + q 1 mod. the two numbers p + iq and p iq are relatively prime. Hence we conclude there product (p + iq)(p iq) = p + q divides a + ib, too. But we have already shown that no real prime divides a + ib. Hence the only possibilities left are a + ib = i u (1 + i) s with u = 0, 1,, 3 and s = 0, 1. Theorem. (The forgotten little theorem). A Gaussian integer does only have an argument iπk with integer k, l an argument rational in degree measurement if it l lies on the x-axis or on the y-axis, or on one of the two lines of slope ±1 through the origin. Proof. Assume that (.5) z = a + ib = a + b e iπk l with real integers a, b, k, l such that k and l are relatively prime. Let d := gcd (a, b) and define a := a d and b := b d. By the last problem, a + ib = i u (1 + i) s with u = 0, 1,, 3 and s = 0, 1. Multiplying both sides by d yields a + ib = i u (1 + i) s d with u = 0, 1,, 3 and s = 0, 1, and d any natural number. These are just the Gaussian integers on the four lines Z, i Z, (1 + i) Z, and (1 i) Z, as to be shown. 9

Corollary. Let a + ib be a Gaussian integer with a 0, b 0. Both real- and imaginary parts of (a + ib) l are nonzero if either l odd or a b. If l > 0 is even, the following happens: l Re (a + ib) l Im (a + ib) l l mod Re (a + ib) l = 0 a = b Im (a + ib) l 0 l 0 mod Re (a + ib) l 0 Im (a + ib) l = 0 a = b 10 Problem.. Give the reason for Corollary.. Proof of Corollary.. For any nonzero Gaussian integer a + ib, and integer l > 0, the equation (.6) (a + ib) l = i u (a + b ) l is equivalent to Re (a + ib) l = 0 for u odd, but equivalent to Im (a + ib) l = 0 for u even. In all cases, the equation (.7) (a + ib) l = (a + b ) l follows. Hence, as explained in the solution of Problem.3, we conclude that either a = 0, or b = 0, or a = b. Since we have excluded the first two possibilities, we conclude that a = ±b and hence equation.6 implies a l (1 ± i) l = i u l/ a l. Hence l = m is even and (±i) m = i u. Hence either l mod, and m, u both odd, and Re (a + ib) l = 0 or l 0 mod, and m, u both even, and Im (a + ib) l = 0. 10 Problem.5. (a + ib)l+ R l (a, b) := Re a b (a + ib)l+ S l (a, b) := Im ab(a b ) Calculate R 0, S 0, R, S. Prove that for all l 0 mod, R l and S l are integer homogenous and symmetric polynomials in a and b. Answer. (a + ib) R 0 (a, b) = Re = a b a b a b = 1 (a + ib)6 R (a, b) = Re = a6 15a b + 15a b b 6 = a 1a b + b a b a b (a + ib) S 0 (a, b) = Im ab(a b ) = a3 b ab 3 ab(a b ) = (a + ib) S (a, b) = Im ab(a b ) = a7 b 56a 5 b 3 + 56a 3 b 5 ab 7 = a a b + b ab(a b ) 10

After some pain with the binomial formulas, one gets the general case with l = m: m +m ( ) m + R m (a, b) = ( 1) s a m+t b m t s s=0 t= m m +m ( ) ( 1) s m + S m (a, b) = a m+t b m t s + 1 s=0 t= m Since the binomial coefficients ( ) even odd are even, these are integer polynomials the other assertions are easy to check, too. 10 Problem.6 (Calculation to the bones). Calculate R m (1, 1) and S m (1, 1), confirming they are nonzero. Answer. R l + (a + b )abs l = Re (a + ib)l+ + (a + b )Im (a + ib) l a b With a = b = 1 and l = m, we get = Re (a + ib)l+ Re i(a ib)(a + ib) l+1 a b [a + ib i(a ib)](a + ib)l+1 = Re a b (1 i)(a + ib)l+1 = Re a + b R m (1, 1) + S m (1, 1) = Re (1 i)(1 + i) m+1 / = ( ) m With a = 1 + ε = z and b = 1, we get in the limit z 1 using l Hôpital s rule (1 + ε i)m+1 R m (1, 1) S m (1, 1) = lim Re (1 i) ε 0 ε d (z i)m+1 = Re (1 i) d z = Re (m + 1)(1 i) m+1 = (m + 1)( ) m Both formulas together imply S m = m( ) m 1. Actually, we check that for all m 0 R m = (m + 1)( ) m, S m = (m + )( ) m.3 Sums of two squares From the decomposition of a Gaussian integer into Gaussian primes, one can quite easily see how many solutions the integer equation m = a +b has for any given natural number m. 11

10 Problem.7. We begin with some examples. For this illustration, I count only the solutions of m = a + b with integers 0 a b, but give these solutions completely. Give a list for m = 1 through 30. Answer. m list of a + b 1 0 + 1 1 + 1 3 no solution 0 + 5 1 + 6 no solution 7 no solution + 9 0 + 3 10 1 + 3 11 no solution 1 no solution 13 + 3 1 0 + 15 1 + m list of a + b 16 0 + 17 1 + 1 3 + 3 19 no solution 0 + 1 no solution no solution 3 no solution no solution 5 0 + 5 = 3 + 6 1 + 5 7 no solution no solution 9 + 5 30 no solution It is still hard to find a pattern. Since the square of the absolute value of a complex number is the sum of the squares of its real- and imaginary parts, we can use complex arithmetic. Here are some further examples for illustration. (a) 1 + i = 1 + = 5 (b) 3 + i = 3 + = 13 (c) The square of the absolute value of (1 + i) ( 3 + i) = 7 i, and (1 + i) ( 3 i) = 1 i implies 5 13 = 7 + = 1 + = 65. (d) Squaring part (c) yields (1 + i) ( 3 + i) = ( 3 + i)(5 1i) = 33 + 56i (1 + i) ( 3 i) = ( 3 + i)(5 + 1i) = 63 16i The square of the absolute value yields 5 13 = 33 + 56 = 63 + 16 = 65 Are there more ways to solve a + b = 65? Indeed, there are still two further solutions obtained as absolute squares of 13( 3+i) = 39+5i and 5(5 1i) = 5 60i. Last not least, 65 is a perfect square. 39 + 5 = 5 + 60 = 0 + 65 = 65 1

(e) The absolute square of (1 + i)(1 + i) = 1 + 3i yields 10 = 1 + 3. 10 Problem.. Prove that a natural number is a perfect square if and only if it has an odd number of divisors. Answer. If d [1, m) is a divisor of m N, then m ( m, m] is different one. In d that way, all divisors of m appear in pairs. The only exception is m in the case that m is a perfect square. Proposition.1. The integer equation m = a + b is solvable for a given natural number m with integers a, b if and only if, in the prime decomposition of m, all primes p 3 mod appear with even multiplicity. In other words, m can be factored as m = s g c where g has only prime factors congruent 3 mod, and (.) c = has only prime factors congruent 1 mod. N ν=1 Theorem.3. The number of different ways one can decompose m = s g c into squares of integers a, b such that m = a + b is τ(c). Here τ(c) = r sν ν N (1 + s ν ) ν=1 is the number of divisors of c. The numbers a + ib from the solutions are exactly the Gaussian integers (.9) a + ib := i u (1 + i) s g N (p ν + iq ν ) σν (p ν iq ν ) τν ν=1 All possible choices are obtained with u = 0, 1,, 3 and σ ν, τ ν 0 with σ ν + τ ν = s ν for ν = 1... N. For example m = 65 is decomposed as 65 = 5 13 and has 9 divisors. The possible 13

values of a + ib as solution of (.9) with unit factor i u = 1 are (1 + i) ( 3 + i) = ( 3 + i)(5 1i) = 33 + 56i, (1 + i) ( 3 + i)( 3 i) = ( 3 + i) 13 = 39 + 5i, (1 + i) ( 3 i) = ( 3 + i)(5 + 1i) = 63 16i, (1 + i)(1 i)( 3 + i) = 5 (5 1i) = 5 60i, (1 + i)(1 i)( 3 + i)( 3 i) = 5 13 = 65, (1 + i)(1 i)( 3 i) = 5 (5 + 1i) = 5 + 60i, (1 i) ( 3 + i) = ( 3 i)(5 1i) = 63 + 16i, (1 i) ( 3 + i)( 3 i) = ( 3 i) 13 = 39 5i, (1 i) ( 3 i) = ( 3 i)(5 + 1i) = 33 56i, As one sees there are still pairs of conjugate complex solutions, and one real solution since 65 is a perfect square. Remark. If m is not a perfect square, there are τ(c) essentially different solutions of m = a + b with 1 a b. If m is a perfect square, there are τ(c) 1 essentially different solutions of m = a + b with 1 a b. 10 Problem.9. Prove that a Gaussian integer a + ib with prime decomposition (.9) satisfies gcd (a, b) = 1 if and only if s = 0 or s = 1; g = 1; either σ ν = 0 or τ ν = 0 for all ν = 1,... N. Answer. A Gaussian integer has relatively prime real and imaginary parts if and only if it is not divisible by any real prime. But the three conditions just imply: a + ib is not divisible by ; a + ib is not divisible by any prime factor congruent to 3 mod ; a + ib is not divisible by any real prime congruent to 5 mod. Proof of Proposition.1 and Theorem.3. For any given natural number m, let d be the divisor of m assembling all prime factors congruent 3 mod. Let c be the divisor of m assembling all prime factors congruent 1 mod. The factor c is completely decomposed into its prime factors as in formula (.). In the ring of Gaussian integers, the usual prime factoring of m will split further into irreducible factors (.10) m = s dc = i 3s (1 + i) s d N (p ν + iq ν ) sν (p ν iq ν ) sν ν=1 1

where the real factoring of d is not needed here. Assume that m = a +b. Because of (a+ib)(a ib) = a +b = m, we conclude that a + ib is a divisor of m. The uniqueness of prime factorization implies equation (.9) holds for some values u {0, 1,, 3} and 0 σ ν s ν, 0 τ ν s ν for ν = 1... N (Why?). We now multiply equation (.9) with the conjugate complex equation and get a + b = s g N ν=1 (p ν + iq ν ) σν+τν (p ν iq ν ) σν+τν We compare all factors with those in equation (.10) and use the uniqueness of prime factorization. Hence g = d and σ ν + τ ν = s ν for ν = 1... N, as claimed. Too, we see that no equation m = a + b can hold unless the factor d is a perfect square. Too, the uniqueness of the prime factoring implies that all solutions given by equation (.9) are different. It is straightforward to check that equation (.9) yields solutions of the equation m = a + b. Hence we get exactly solutions. τ(c) = N (1 + s ν ) ν=1 10 Problem.10. Assemble different examples to illustrate Proposition.1 and Theorem.3. Examples. For different values of m, here is a table listing the factors s, d, c, the prime decomposition of c and the number of divisors τ(c). Furthermore, we list the possible values of a + ib form solution (.9). For abbreviation, I put the unit factor i u = 1, and ignore the obvious conjugate complex ones. m s d c factor c τ(c) list of a + ib 5 1 1 5 5 1 + i 1 1 1 1 + i 9 1 9 1 1 1 3 10 1 5 5 (1 + i)(1 + i) = 1 + 3i 6 1 13 13 (1 + i)( 3 + i) = 5 i 0 1 5 5 + i 36 9 1 1 1 6i 39 1 3 13 13 no solution 5 1 13 13 6i 65 1 1 65 5 13 (1 + i)( 3 ± i) = 7 i, 1 i 130 1 65 5 13 (1 + i)(1 + i)( 3 ± i) = 3 11i, 9 7i 60 1 65 5 13 i(1 + i)( 3 ± i) = 1i, 16 + i 15

. Roots Definition. (argument). The argument of a complex number z 0 is the set of angles which in polar coordinates yield z: arg z = {θ R : z = z e iθ } The principle value of the argument Arg z of a complex number z 0 is the unique value for the argument lying in the interval ( π, +π], including +π but excluding π. Remark. All values of the argument differ by integer multiples of π. Hence there always exists a unique principle argument. Nevertheless, the choice of a principle argument is just an arbitrary convention. Especially, the set of arguments arg z depends continuously on z 0, but the principle argument does not depend continuously on z. Lemma.1 (Principle value). The principle argument of a complex number z = x + iy 0 is given by the formula arctan y if x > 0; x π if x = 0 and y > 0; (.11) Arg z = π if x = 0 and y < 0; π + arctan y if x < 0 and y > 0; x π + arctan y if x < 0 and y < 0 x Remark. Several computer languages provide Arg (x + iy) as a two variable function arctan(x, y). 10 Problem.11. Give simple examples to illustrate all cases in Lemma.1. Proof of Lemma.1. The principle argument Arg z of any complex number z 0 is defined by restricting the argument of z = z e iarg z to the interval Arg z ( π, π]. Separating real- and imaginary parts yields x = z cos Arg z, y = z sin Arg z Hence y = tan Arg z unless x = 0. But since both cos(θ ± π) = cos θ and sin(θ ± π) = x sin θ, the tangent function has period π. Hence the value of tan Arg z determines the argument only modulo π not yet modulo π. The given cases result from comparing the signs of sin and cos with those of x and y. 16

10 Problem.1. The angle α of the tangent to the graph of a differentiable function y = f(x) with the x-axis is always given by α = arctan f (x) = arctan dy dx Why does one not distinguish the cases occurring in Lemma.1? Answer. The angle between the tangent and the x-axis is a directed angle between two lines, and hence always lies in the interval ( π, π ). But the principle argument of a complex number z is a directed angle between two rays the positive x-axis and the ray from the origin 0 to the point z. Hence it can take any value in the interval ( π, π], including +π and excluding π. Definition.3 (right semi-plane). Define the right semi-plane as the set RS = {z = x + iy C : x > 0 or (x = 0 and y 0) } Remark. The right semi-plane is neither open nor closed. I use the word half plane only for an open set. 10 Problem.13. What is the exact interval to which the principle value of the argument for any z 0 in the right semi-plane is restricted. Answer. For any z 0 in the right semi-plane, the principle value of the argument lies in the half open interval Arg z ( π, π]. Conversely, Arg z ( π, π ] implies z RS. Definition. (Principle value). The principle argument Arg z of any complex number z 0 is defined by restricting the argument of z = z e iarg z to the interval Arg z ( π, π]. The principle n-th root is define as P.V. n z := n z e iarg z n The principle logarithm of any complex number z 0 is defined as Ln z := ln z + iarg z 10 Problem.1. Prove that for all a, b RS and all n Arg ab = Arg a + Arg b for a, b RS and a, b 0 P.V. n ab = P.V. n a P.V. n b 17

Answer. For the product of any nonzero a, b RS, the principal values Arg ab ( π, π], and Arg a + Arg b ( π, π], too. Hence Arg ab = Arg a + Arg b + πk holds with k = 0, since both sides do not differ by an integer multiple of π. Now the second assertion easily follows P.V. n ab = n Arg ab i ab e n = P.V. n a P.V. n b = n Arg a+arg b i ab e n = n a e i Arg a n n b e i Arg b n Here are some examples of principle third roots. (a) (b) (c) P.V. 3 i = P.V. 3 i = 1 + i P.V. 3 3 + i 3 i = 1 + i (d) P.V. 3 1 = 1 + i 3 Note that the principal third root of any negative real, is not the negative real value. 10 Problem.15. In which examples does equality hold for the principle values. Show enough of your calculations. (a) P.V. 3 i P.V. 3 i? = P.V. 3 1 (b) P.V. 3 i P.V. 3 i? = P.V. 3 1 (c) P.V. 3 1+i P.V. 3 1+i [ ] =? P.V. 3 1+i [ ] (d) P.V. 3 1+i P.V. 3 1+i? = P.V. 3 1+i 1

Answer. (a) (b) (c) [ ] P.V. 3 i P.V. 3 3 + i i = = 1 + i 3 [ ] P.V. 3 i P.V. 3 3 i i = = 1 i 3 [ ] 1 + i 1 + i 1 + i P.V. 3 P.V. 3 = = i = P.V. 3 1 = 1 + i 3 P.V. 3 1 = 1 + i 3 (d) 1 + i 1 + i P.V. 3 P.V. 3 = [ P.V. 3 e πi P.V. 3 [ 1 + i ] = P.V. 3 i = ] = [ e πi ] = e πi 6 = 3 i [1 ] + i P.V. 3 = P.V. 3 i = e πi 6.5 The solution of the reduced cubic equation Given is the equation (.1) x 3 + b x + c = 0 Since the quadratic term is lacking, it is called a reduced or depressed cubic. We want to find all three solutions, including the complex ones. Putting x := u b leads to the 3u equation ( ) 3 b (.13) u 3 + c = 0 3u Hence z = u 3 satisfies the quadratic resolvent equation (.1) z + cz b3 7 = 0 The two solutions of equation (.1) are (.15) z 1, = c ± c + b3 7 We get a resolvent equation with real or complex solutions, depending on whether the discriminant (.16) D = c + b3 7 19

is positive or negative. In the first case, the original cubic has only one or (exceptionally) two real solutions. The second case with complex solutions of the resolvent equation (.1) is called the casus irreducibilis. In that case, the original cubic has three real solutions. Hence one is forced to go through some complex arithmetic just in the case where all final results are real. The solution procedure turns out to be different for these two cases. Case 1: D 0 The cubic has one real solution or in the special case D = 0 two real solutions. From Viëta s formula (.17) z 1 z = b3 7 b 3 3 = z 3 z 1 Hence the real solution of the cubic (.1) is (.1) x 1 = u b 3u = 3 z 1 + 3 z = 3 c + c + b3 7 + 3 c c + b3 7 which is just Cardano s formula. The two complex solutions are (.19) x = 3 z 1 ω + 3 z ω x 3 = 3 z 1 ω + 3 z ω where (.0) ω = 1 + i 3 = cos π 3 + i sin π 3 is the primitive third root of unity. Simplifying (.1) and (.19) yields (.1) x,3 = 3 z 1 + 3 z ± 3 z 1 3 z 3 i Case : D < 0 The cubic has three real solutions. In this case, the resolvent equation (.1) has complex solutions. As already Rafael Bombelli discovered (157), this leads to a third root of a complex number in Cardano s formula (.1). Actually, this case can only occur for b < 0. To cover the general case, extraction of the third root has to be done via polar coordinates. Viëta s formula (.17) and 0

equation (.15) imply (.) z 1 = z 1 z = b3 7 z 1, = c b ± i 3 7 c b 3 = (cos θ ± i sin θ) 7 The argument θ can be calculated more simple from the real parts. One gets cos θ, and finally θ: (.3) c b = 3 7 cos θ c θ = arc cos b 3 7 Hence equations (.) and (.0) imply (.) 3 z1 = b 3 3 z1 ω = b 3 ( cos θ 3 + i sin θ ) 3 ( cos θ + π + i sin θ + π ) 3 3 and equations (.17) and (.1) imply that b x 1 = 3 cos θ 3 (.5) b θ + π x = cos 3 3 b θ π x 3 = cos 3 3 are the three real solutions of the original cubic. 10 Problem.16. Calculate the roots in the limiting case D 0+. Answer. x 1 = 3 c, x = x 3 = x 1 10 Problem.17. Calculate the roots in the limiting case D 0 in terms of c. Distinguish the cases c < 0 and c > 0. In both cases b < 0. 1

Answer. In the approach D 0, one gets c = b3 7, 3 c = c b = 3 7 Distinguish the cases c < 0 and c > 0. c b = 3 7 cos θ b 3 Case c < 0. Case c > 0. One gets cos θ = 1, and hence θ = 0. The result (.5) implies x 1 = 3 c, x = x 3 = x 1 One gets cos θ = 1, and hence θ = π. The result (.5) implies x = 3 c, x 1 = x 3 = x 1 10 Problem.1. Use Cardano s formula and Bombelli s method to get one solutions of x 3 30x 36 = 0. Answer. Cardano s formula for a solution of x 3 + bx + c = 0 yields in the present case (.6) x 1 = 3 c + c + b3 7 + 3 = 3 1 + 6i + 3 1 6i c c + b3 7 To calculate the third root, let 3 1 + 6i = u+iv and 1+6i = (u+iv) 3. We compare real- and imaginary parts and get the system 1 = u 3 3uv = u(u 3v ) 6 = 3u v v 3 = (3u v )v With the help of the factoring, one finds the integer solution u + iv = 3 + i. Hence Cardano s formula gives x 1 = 6. 10 Problem.19. Find the two other solutions of the equation x 3 30x 36 = 0. Answer. The simplest way is division of the polynomial. One gets (x 3 30x 36)/(x 6) = x + 6x + 6. Hence the two other solutions are x,3 = 3 ± 3.

Remark. Too, we can find the two other roots by introducing the third root of unity ω = 1+i 3 into Cardano s formula. This has to be done in a consistent way and produces the solutions x = ω 3 1 + 6i + ω 3 1 6i = 1 + i 3 x 3 = ω 3 1 + 6i + ω 3 1 6i = 1 i 3 which is the same answer as above. (3 + i) + 1 i 3 (3 + i) + 1 + i 3.6 The square root of a complex number (3 i) = 3 3 (3 i) = 3 + 3 Proposition.. A branch with Re z 0 for the square root of any complex number z = x + iy is (.7) x + iy = x + y + x + i sign (y) x + y x The square root w = z of any complex number z = x + iy has two branches. The second branch is w = w. Reason. Name the square root in question x + iy =: u + iv. Squaring yields x + iy = (u + iv). Separate the real- and imaginary part to get The absolute value squared is Add and substrate x = u v, y = uv x + y = x + iy = u + iv = (u + v ) u + v = x + y u v = x u x + y = + x v x + y = x Of the last two expression, one takes real square roots. u 0 has been assumed. One needs still to determine the sign of v. But y = uv and u > 0 imply sign y = (sign u)(sign v) = sign v. In the special case u = 0, both signs of v give a correct result for the square root. This special case corresponds to y = 0 and x 0 the negative numbers z 0. 3

(a) 10 Problem.0. Calculate the square roots: 3 + i, 1 + 16i, + 6i 6i, 3 i, 1 96i (b) (c) 3i, 16 1i, 6 (d) 6 + i, + 3i, 96 + 1i (e) i, i, 3i A Gaussian integer that is the square of another Gaussian integer is called a perfect square. Which of the numbers under the square roots are perfect squares, which are not?.7 Pythagorean triples Any three integers such that a + b = c are called a Pythagorean triple. 10 Problem.1. Show that for any Gaussian integer p + iq, the real and imaginary parts and the absolute value of its square are a Pythagorean triple. Answer. Let the square be a+ib := (p+iq). Its absolute value is c := a+ib = p+iq. Clearly a + b = c holds. 10 Problem.. Show that for any Gaussian integer a+ib 0, it is impossible that a + ib and b + ia are both perfect squares. Theorem. (Pythagorean triples). A nonzero Gaussian integer a + ib = (p + iq) is a perfect square if and only if the following three conditions hold (i) the sum a + b is a perfect square c ; (ii) gcd (a, b) =: g is either (a) a perfect square: g = d, or (b) twice a perfect square: g = d ; (iii) assuming t, but not t+1 divides d, we get two cases: (iii a) if g = d, then t a is odd and t b is even (iii b) if g = d, then t 1 a is even and t 1 b is odd. Corollary. Exactly one of the Gaussian integers a+ib = (p+iq) and b+ia = i(p iq) is a perfect square if and only if (i) the real and imaginary parts and the absolute value of a+ib are a Pythagorean triple; (ii) the greatest common divisor gcd (a, b) is either a perfect square or twice a perfect square.

Remark. It is always true that the imaginary part b of a perfect square is even, but this condition is sufficient only in the special case gcd (a, b) = 1. Remark. Condition (iii) in theorem. is only needed to determine which one of the two numbers either a + ib or b + ia is a perfect square. In Problem., we have already checked that these two numbers cannot be both perfect squares, unless they are both zero. Example.1. These are perfect squares, corresponding to t = 0, 1, in (iii a) or (iii b): (a) 3 + i = 1 + i, 1 + 16i = + i, + 6i = + i,... (b) 6i = 1 + 3i, 3 i = + 6i, 1 96i = + 1i,... but these are not perfect squares: ( 1 + 3i) 3i =, 16 1i = ( 1 + 3i), 6 = ( + 6i),... 6 + i = (1 + i), + 3i = ( + i), 96 + 1i = ( + i),... Here is the most simple example: (a) =, 16 =, 6 =,... (b) i = 1 + i, i = + i, 3i = + i,... Proof of necessity in theorem.. Necessity is easier to prove. We assume that a+ib = (p + iq) is a perfect square. Squaring the absolute values gives c = a + b = a + ib = p + iq = (p + q ) Separating real- and imaginary parts yields the Pythagorean triple a = p q, b = pq, c = p + q Hence condition (i) holds. To check (ii), let d := gcd (p, q). Clearly d divides both a and b, and hence d divides g := gcd (a, b). Let By these divisions we get p := p d, q := q d, a := a d, b := b d, c := c d, a = p q, b = p q, c = p + q, gcd (p, q ) = 1 We show that gcd (a, b ) =: h can only be 1 or. Indeed, any odd prime dividing b needs to divides either p or q, but not both. Hence it cannot divide a = p q. 5

Neither can divide both a and b. Indeed, assuming divides b, implies one number among p and q is even, the other one is odd. Hence a is odd. Hence we see that h = 1 or h = are the only possibilities left. Therefore gcd (a, b) = d gcd (a, b ) = hd is either a perfect square or twice a perfect square, as claimed by condition (ii). Finally, we need to confirm that either case (iii a) or (iii b) occurs. Again, t 0 be the integer for which t, but not t+1 divides d. These two cases can occur: (iii a): g = d and h = 1 = gcd (a, b ) = gcd ((p q )(p + q ), p q ). Hence p ± q are both odd, and hence one number among p and q is odd, but the other one is even. Hence a is odd, but divides b. Similarly, t a is odd and t b is even, since these are odd multiples of a and b. (iii b): g = d and h = gcd (a, b ) = gcd ((p q )(p + q ), p q ) =. Hence p ± q are both even, and different from case (a) p and q are both odd. Hence a is divisible by, but b mod. Hence, t 1 a is even and t 1 b is odd, since these are odd multiples of a and b. Proof of sufficiency theorem.. We assume that the integers a and b satisfy the four conditions (i) through (iii). Let g = gcd (a, b) = hd with h = 1 or h = and as above In cases (a) and (b) one gets a := a d, b := b d, c := c d (iii a) if g = d, then gcd (a, b ) = 1, and a is odd, b is even. Hence c is odd, and c ±a are both integers. (iii b) if g = d, then gcd (a, b ) =, a is even and b is odd. Hence c is odd. As explained in Proposition., one obtains a square root of a complex number by the formula c + a (.7) a + ib = c a + i sign (b) c In both cases (iii a) and (iii b), ±a and b are integers. Furthermore a + b = c implies [ ] b = c + a c a The two factors are relatively prime: [ c + a gcd ], c a = 1 6

Indeed, the greatest common divisor of c ±a is a divisor of h = gcd (a, b ), which is either 1 or. But is the case (b) where h =, both numbers c ±a turn out to be odd. It is a (too easy) exercise to show that uniqueness of the prime factorization implies Lemma.. If the product of two relatively prime factors is a perfect square, then both factors are perfect squares. Hence we conclude that both roots (E:doit) p = c + a and q = sign (b) c a are integers. In other word, we have shown that a + ib = p + iq is a Gaussian integer. Multiplying by d, we get that a + ib = p + iq is a Gaussian integer, too, as to be shown. 10 Problem.3. Generalize the Lemma. to the product of three or more factors. Are the following conjectures true or false: If the product of three pairwise relatively prime factors is a perfect square, then all three factors are perfect squares. If the product of three relatively prime factors is a perfect square, then the product of any two of them is a perfect square. Remark. Under the additional assumption that gcd (a, b), indeed only the following possibilities occur for perfect squares a + ib = (p + iq) : p q a mod b mod c mod h odd even 1 0 1 1 even odd 3 0 1 1 odd odd 0. Roots of unity 10 Problem.. Find all different powers exactly, expressed using square roots: (a) ω = 1 + i 3. 5 1 + i 10 + 5 (b) z =. (c) w = 1 + i. 7

You need not show all calculations. Simplify your answers by using the conjugate complex. What is remarkable about these expressions? Answer. It is remarkable that the different powers can be obtained simply by some sign changes of some square roots: (a) (b) (c) ω = 1 + i 3, ω = 1 i 3 = ω, ω 3 = 1 5 1 + i 10 + 5 z =, z = 5 1 + i 10 5, z 3 = z, z = z w = 1 + i, w = i, w 3 = 1 + i, w = 1, w 5 = w 3, w 6 = i, w 7 = w An exact trigonometric table of cos 3 k and sin 3 k for all k = 1,..., 15 can be obtained using the three roots of unity ω, z and w. This trick goes in principle back to Ptolemy! (a) From the primitive third and eighth root of unity ω and w from Problem., we get a product of argument 15. ω 1 w 3 = e πi[ 1 3 + 3 ] = e πi = cos 15 + i sin 15 (b) We can calculate this product in terms of the exact square roots, obtained in Problem.. ω 1 w 3 = 1 i 3 1 + i = 1 + 3 + i( 3 1) (c) Finally, we compare the real- and imaginary parts obtained form (a) and (b), and use the common denominator. + 6 cos 15 = 6 sin 15 = (d) In this case, we get the table immediately. k 15 cos k 15 sin k 15 15 + 6 30 3 5 60 1 75 6 6 1 3 6+ 90 1 0

10 Problem.5. Calculate cos 15 + i sin 15 directly by formulas cos θ = 1 + cos θ and confirm that you get the same values. Answer. Similarly, one gets sin 15 = cos 30 1 + cos 30 = 3, sin θ = 1 cos θ = + 3 = + 3. By squaring both sides, one checks that indeed + 3 = 6 +, 3 = 6 10 Problem.6. The following steps can be elaborated to obtain an exact trigonometric table of cos 9 k and sin 9 k for all k = 1,..., 0. (a) Find natural numbers p, q such that p 5 + q = 1 0. (b) With z, w from Problem., calculate the argument θ in z p w q = e πiθ (c) Hint: calculate 5 + 1 i 10 5 1 + i (d) Calculate z p w q with the exact square roots, obtained in Problem.. (e) Use Euler s (.) e πi π π 0 = cos + i sin 0 0 and get exact expressions for cos 9 and sin 9. Use the common denominator in your results. Answer. (a) The natural numbers p, q have to satisfy p + 5q = 1. solution is p = 3, q = 5. The smallest (b) With z, w from Problem., z p w q = e πi[ 3 5 + 5 ] = e πi 0 9

(d) With the same z, w from Problem., using z 5 = 1 and w = i, we can simplify and get 5 + 1 i 10 5 z 3 w 5 = z w = 1 + i 10 + + 5 5 10 + 5 5 = + i (e) Comparing real- and imaginary part from parts (b) and (c) yields 10 + + 5 5 (.9) cos 9 = 10 + 5 5 (.30) sin 9 = 10 Problem.7. Give the exact expression for the perimeter of a regular 0- gon, inscribed into the unit circle. Compare with π. How large is the deviation? Answer. The circumference of a regular 0-gon is [ 10 0 sin 9 = 5 + 5 ] 5 6.573760 Bur π 6.315307 so only one decimal point. But the relative error of less than a percent is a beginning. In a similar way, we can use the twelfth and fifth root of unity, and get sin and cos for the multiples of 6. (a) The natural numbers p, q have to satisfy 1p 5q = ±1. The smallest solution is p = 3, q = 7. Dividing by 60 yields 3 5 7 1 = 1 60 (b) With the twelfth and fifth roots of unity τ and z we get z 3 τ 7 = e πi[ 3 5 7 1] = e πi 60 = cos 6 + i sin 6 (d) We can calculate this product with the exact values of the square roots, simplify and get 5 + 1 + i 10 5 3 i z 3 τ 7 = z 3 τ 1 = 15 + 3 + 10 5 = + i 5 1 + 3 10 5 30

(e) Comparing real- and imaginary part from parts (b) and (c) yields (.31) (.3) cos 6 = 15 + 3 + 10 5 sin 6 = 5 1 + 3 10 5 10 Problem.. Give the exact expression for the perimeter of a regular 30- gon, inscribed into the unit circle. Compare with π. How large is the deviation? Answer. The circumference of a regular 30-gon is 60 sin 6 = 15 [ 5 1 + 3 10 ] 5 6.71707796 But π 6.315307 so just one decimal point coincides. 10 Problem.9. Prove that 5 + 1 + 3 10 5 (.33) cos = 15 + 3 10 5 (.3) sin = Here is a table assembling the information we have gathered so far. θ cos θ sin θ 6 15+ 3+ 10 5 5 1+ 3 10 5 9 10+ + 5 5 10+ 5 5 15 + 6 1 5+ 5 5+1+ 3 30 3 36 5+1 5 10 5 6 5 1 15+ 3 10 5 1 5 5 10 Problem.30. An exact trigonometric table of cos 3 k and sin 3 k for all k = 1,..., 15 can be obtained with the same trick which goes in principle back to Ptolemy! (a) Calculate 1 3 1 5 1. With ω, z and w from Problem., get the argument θ in ωz 1 w 1 = e πiθ = cos πθ + i sin πθ 31

(b) Calculate ωz 1 w 1 with the exact square roots, obtained in Problem.. Get a readable result, do not distribute every term. (c) Get exact expressions for cos 3 and sin 3. Solution. (a) Since 1 3 1 5 1 = 1, we get the argument 10 ωz 1 w 1 = e πi[ 1 3 1 5 1 ] π π = cos + i sin 10 10 (b) With the exact square roots from Problem., [ ωz 1 w 1 1 + i ] [1 ] 3 i 5 1 i 10 + 5 = ωwz = = 3 1 + i( 3 + 1) = ( 3 1)( 5 1) + ( 3 + 1) 10 + 5 5 1 i 10 + 5 + i ( 3 + 1)( 5 1) ( 3 1) 10 + 5 (e) We get exact expressions ( 3 1)( 5 1) + ( 3 + 1) 5 + 5 cos 3 = 16 ( 3 + 1)( 5 1) ( 3 1) 5 + 5 sin 3 = 16 10 Problem.31. Calculate the circumference of a regular 60-gon. Which approximation of π do we get? How many decimal points are correct? Answer. The circumference of an 60-gon is 10 sin 3 = 15 One gets two correct decimal points. [ ( 3 + 1)( 5 1) ( 3 1) 5 + ] 5 π 6.315307 6.03179 10 Problem.3. Gather the entire table with cos k3 and sin k3 for all k = 1... 30. Go on as far as you get. 3

[ ] k [ ω k z k w k = e πik 3 1 + i( 3 + 1) 5 1 i 10 + 5 10 = [ ω z w = e πi 3 + i 10 = ] [ 5 1 i 10 ] 5 ] k = 15 + 3 + 10 5 i 5 i + i 3 10 5 [ ω z w = e πi 1 i ] [ ] 3 5 1 + i 10 + 5 10 = ωz = = 5 1 + 3 10 + 5 i 15 + i 3 + i 10 + 5 = [ ω 7 z 7 w 7 = e 1πi 10 = ωwz 3 1 + i( ] [ 3 1) = 5 1 i 10 ] 5 ( 3 + 1)( 5 + 1) + ( 3 1) 5 5 ( 3 + 1)( 5 + 1) + ( 3 + 1) 5 5 + i 16 16 Further values were obtained by trial and error for the signs of the roots. 33

θ cos θ sin θ 3 ( 3 1)( 5 1)+( 3+1) 5+ 5 ( 3+1)( 5 1) ( 3 1) 5+ 5 16 16 6 15+ 3+ 10 5 5 1+ 3 10 5 9 10+ + 5 5 10+ 5 5 1 5 1+ 3 10+ 5 15+ 3+ 10+ 5 15 + 6 1 5+ 5 1 ( 3+1)( 5+1)+( 3 1) 16 5+1+ 3 10 5 7 10 + 5+ 5 30 3 33 ( 3 1)( 5 1)+( 3+1) 16 36 5+1 5 5 5+ 5 39 ( 3+1)( 5+1)+( 3+1) 5 5 16 15 3+ 10+ 5 5 5+1+ 3 10+ 5 51 ( 3+1)( 5+1) ( 3 1) 5 5 16 5 5 5 57 ( 3+1)( 5 1)+( 3 1) 16 60 1 63 10+ + 5+ 5 66 15+ 3 10 5 69 ( 3+1)( 5+1)+( 3+1) 7 5 1 75 6 7 15+ 3+ 1 10+ 5 1+ 3 16 10+ 5 5 5 10 5 5+ 5 5 5 6 5 1 ( 3+1)( 5+1)+( 3+1) 5 5 16 15+ 3 10 5 10+ + 5+ 5 1 ( 3+1)( 5 1)+( 3 1) 5+ 5 16 5 5 ( 3+1)( 5+1) ( 3 1) 5 5 16 5+1+ 3 10+ 5 15 3+ 10+ 5 ( 3+1)( 5+1)+( 3+1) 5 5 16 5+1 ( 3 1)( 5 1)+( 3+1) 5+ 5 16 3 10 + 5+ 5 5+1+ 3 10 5 ( 3+1)( 5+1)+( 3 1) 5 5 16 5+ 5 6+ 5 1+ 3 10+ 5 10+ + 5 5 15+ 3+ 10 5 ( 3 1)( 5 1)+( 3+1) 5+ 5 16 7 ( 3+1)( 5 1) ( 3 1) 5+ 5 16 90 1 0 3

3 Polynomials 3.1 Rational zeros of polynomials Proposition 3.1. Assume x is a real or complex zero of a monic 1 integer polynomial x r + a r 1 x r 1 + + a 1 x + a 0 = 0 where a r 1,... a 1, a 0 are integers. If x is rational, then it is even an integer, and a divisor of the constant term a 0, too. Proof. Assume x = m n yields where m and n are relatively prime integers. Multiplying by nr m r n r + a m r 1 r 1 n + + a m r 1 1 n + a 0 = 0 m r + a r 1 m r 1 n + + a 1 mn r 1 + a 0 n r = 0 m r = ( a r 1 m r 1 + + a 1 mn r + a 0 n r 1) n Now we argue the same way as in the proof of Proposition 1.6: Assume that any prime p divides n. By the last line, the same prime divides m r. Hence, by Euclid s Lemma p divides m. This cannot occur after cancellation of the fraction m. Hence n = 1, and n the rational root x is an integer. To get the last claim, we reuse the formulas from above, now with n = 1, and get m r + a r 1 m r 1 + + a 1 m + a 0 = 0 m ( m r 1 + a r 1 m r + + a 1 ) = a0 Hence any zero m has to be a divisor of the constant coefficient a 0. Proposition 3.. Assume x is a real or complex zero of any polynomial a r x r + a r 1 x r 1 + + a 1 x + a 0 = 0 where a r 0, a r 1,... a 1, a 0 are integers. If the root x = m is rational with m and n n relatively prime, then m is a divisor of a 0 and n is a divisor of a r. Proof. a r m r + a r 1 m r 1 n + + a 1 mn r 1 + a 0 n r = 0 ( a r m r 1 + a r 1 m r n + + a 1 n r 1) m = a 0 n r a r m r = ( a r 1 m r 1 + + a 1 mn r + a 0 n r 1) n 1 A polynomial is called monic if the coefficient of the leading term x r is one. Hence n is a divisor of m r and hence n = 1, since n and m are relatively prime. 35

The second line implies that n is a divisor of a r m r. Since gcd (n, m) = 1, the Euclidean Property stated in Proposition 1.1 implies that n is a divisor of a r. The last line implies that m is a divisor of a 0 n r. Since gcd (n, m) = 1, the Euclidean Property stated in Proposition 1.1 implies that m is a divisor of a 0. Rule: The rational solutions of an integer polynomial a r x r + + a 0 = 0, are among the positive or negatives of the rational numbers top and bottom of which are divisors of top and bottom of the solution of a r X + a 0 = 0. 3. Gauss Lemma Proposition 3.3 (Linear Gauss Lemma). Suppose that an integer polynomial has a rational zero m, with n, m are relatively prime. Then it can be factored into integer n polynomials (3.1) a r x r + + a 0 = (nx m)(b r 1 x r 1 + + b 0 ) If additionally gcd (a r,..., a 0 ) = 1, then gcd (b r 1,..., b 0 ) = 1, too. Proposition 3. (Monic Linear Gauss Lemma). Suppose that a monic integer polynomial has a rational zero. Then the root is an integer, and the polynomial can be factored into monic integer polynomials x r + + a 0 = (x m)(x r 1 + + b 0 ) Proof. We may assume from the start that gcd (a r,..., a 0 ) = 1. The factoring of equation (3.1) holds with rational coefficients b r 1 = B r 1,..., b B 0 = B 0. We choose the B smallest possible denominator B. In that case, the assumption that gcd (a r,..., a 0 ) = 1 implies that gcd (B r 1,..., B 0 ) = 1, too. 3 After multiplying with the least common denominator B of the b 0,... b r 1, comparison of coefficients yields Ba 0 = mb 0 Ba 1 = nb 0 mb 1 Ba = nb 1 mb Ba r 1 = nb r mb r 1 Ba r = nb r 1 Assume towards a contradiction that a prime p divides all Ba 0,... Ba r. Either p does not divide n or p does not divide m. 3 Indeed, a common prime divisor of all B r 1,..., B 0 could not divide B and hence would be a common divisor of all a r,..., a 0, what we just have ruled out. 36

Take the second case. Successively, we conclude that p divides Ba 0 = mb 0, B 0, Ba 1 nb 0 = mb 1, B 1, Ba nb 1 = mb, B,..., B r 1, hence all B 0,... B r 1 contrary to the assumption. Hence we conclude gcd (Ba 0,... Ba r ) = 1. In the other case that p does not divide n, we get the same conclusion, now going through the system in reversed order: Successively, we see that the prime p divides Ba r = nb r 1, B r 1, Ba r 1 +mb r 1 = nb r, B r,..., Ba 1 + mb 1 = nb 0, B 0. In both cases, the conclusion gcd (Ba 0,... Ba r ) = 1 holds. Hence B = 1, and in the first place, the rational numbers b 0,... b r 1 need to be integers. Finally, we see that gcd (b r 1,..., b 0 ) = 1. Indeed, since B = 1, a prime p dividing all coefficients b r 1,..., b 0 would divides all coefficients a r,..., a 0, too, contrary to the assumption. Corollary (Gauss Lemma). If an integer polynomial can be factored over the rationals, it can be factored over the integers by adjusting a common factor. Especially, a monic integer polynomial that can be factored over the rationals, can even be factored over the integers into monic integer polynomials. Proof. Suppose that an integer polynomial of degree r can be factored into rational polynomials of degree s 1 and t 1. Multiplying with the least common denominators, we get the integer formula (3.) B (a r x r + + a 0 ) = A (b s x s + + b 0 ) ( ) c t x t + + c 0 where a r x r + + a 0 is obtained from the given integer polynomial by dividing through with the greatest common divisor of the coefficients. Because of using least common denominators everywhere, and cancellation, we get (3.3) gcd (A, B) = 1 gcd (a r,..., a 0 ) = 1 gcd (b s,..., b 0 ) = 1 gcd (c t,..., c 0 ) = 1 Question. Show that A = 1. Answer. Assume any prime number p divides A, and derive a contradiction. By formula (3.) this would imply that p divides all numbers Ba r,... Ba 0. Since gcd (a r,..., a 0 ) = 1, we conclude that p divides B. But this contradicts the assumption gcd (A, B) = 1. Lemma 3.1. If a prime p divides all coefficients of the product of two polynomials with integer coefficients (3.) (b s x s + + b 0 ) ( c t x t + + c 0 ), then p divides either gcd (b s,..., b 0 ) or gcd (c t,..., c 0 ). 37