LeLing13: Polynomials and complex numbers. Division of polynomials. Euclides s algorithm and multiple roots. R ecommended exercises: Geoling 15.

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1 LeLing13: Polynomials and complex numbers. C ontents: Polynomials and non-polynomial expressions. The roots of x + 1 = 0: the complex numbers. The inverse 1 and the conjugate. z Roots of polynomials. Conjugate roots. The formula b± b 4ac. a Euclides s algorithm and the golden ratio Division of polynomials. Euclides s algorithm and multiple roots. R ecommended exercises: Geoling 15. Polynomials A polynomial p(x) in X is an expression p(x) = a 0 + a 1 X + + a n X n, where a 1, a,, a n are numbers. If a n 0 the number n is called the degree of the polynomial p(x), and we write deg(p(x)) = n. When all a 1, a,, a n are real numbers, p(x) is called a real polynomial. When instead a 1, a,, a n belong to a field K then p is said to have coefficients in K, so we write p(x) K[X]. Clearly, p(x) R[X] means p(x) is a real polynomial. Usually one thinks of a polynomial as a function and one writes P (x), where the x means we have substituted the number x where X was. Classical examples are the line p(x) = ax + b and the parabola p(x) = ax + bx + c. There are naturally maps f(x) that do not arise from substituting x to X in a polynomial. The typical examples include cos(x) and sin(x). In fact, this follows from the fact that sin(x) and cos(x) have infinitely many roots, whereas a polynomial p(x) has not more than deg(p(x)) roots 1. Sometimes we could want to substitute to X a matrix or any other object that is possible to add and multiply. ( For ) example, if p(x) = X 3 + X + 1, putting in the place 0 1 of X the matrix J = we may compute p(j), which will be a matrix This is proven using the division algorithm (or Ruffini s theorem). Ingegneria dell Autoveicolo, LeLing13 1 Geometria

2 From J 3 = J in fact, ( we) have p(j) = J + ( J + 1 = ) J + 1. Since 1 represents X 0, it follows J 0 = Id = and thus p(j) = Hence the matrix J is a root of the equation X + 1 = 0. Complex numbers It is completely obvious that the equation x + 1 = 0 canot be solved using the real numbers. Mathematicians have therefore invented the imaginary number i. In other words one declares i = 1, and naturally also ( i) = 1. Once i is defined as a number, we must make sure we can compute i, 4 i, 1, i i 5, 4 i + i 3 5 and the like. So mathematicians have invented the number i, but this in turn immediately generates many other numbers, namely those that can be written as a 0 +a 1 i +a i + +a n i n b 0 +b 1 i +b i + +b m i m with a 1,, a n, b 1,, b m real. Once i is born, a whole set C, called set of complex numbers a 0+a 1 i +a i + +a n i n b 0 +b 1 i +b i + +b m i m with a 1,, a n, b 1,, b m comes along. Notice R is a subset of C, for if a R then a = a i C, and R C. An important theorem is the following. Theorem 0.1. If z C, i.e. z = a 0+a 1 i +a i + +a n i n b 0 +b 1 i +b i + +b m i m, a 1,, a n, b 1,, b m, then z can be written uniquely as: z = x + i y with x, y R. The real number x is called the real part of z, Re(z) := x, while the real number y is the imaginary part and we write Im(z), so that z = Re(z) + i Im(z). proof. From i = 1 it follows that z = a 0+a 1 i +a i + +a n i n b 0 +b 1 i +b i + +b m i m R. Besides, z = ( A+B i C+D i x = AC+BD and y = BC AD C +D C +D z = x + y i = x + y i and y y, we would have = A+B i, where A, B, C, D C+D i C +D. Therefore z = x + y i, with C D i )( ) = (AC+BD)+(BC AD) i C D i. Uniqueness is easy to prove. If there were z such that i = x x y y which implies i is real! Contradiction. Hence y = y and necessarily x = x. Sometimes the symbol 1 is used to indicate i, but be careful because ( 1)( 1) is not the same as 1 1. Ingegneria dell Autoveicolo, LeLing13 Geometria

3 With this result in the bag we may think of a complex number as a pair z = x + y i of real numbers. We can then imagine the complex number as being represented by the point (x, y) of the plane R. The inverse 1 z and the conjugate z Let z denote a complex number z = x + i y. The previous proof provides us with the idea of how to find the inverse 1 of a complex number. It is convenient to define first the z conjugate z = x y i of z as the complex number whose imaginary part is opposite to the imaginary part of z. Geometrically the conjugate is the symmetric point to z with respect to the x-axis. Two important properties are: Proposition 0.. Let z, w be complex numbers. Then: zw = zw z + w = z + w Using the first property repeatedly we have z n = z n. Observe z = z iff Im(z) = 0, i.e. iff z is real. Now the important remark: z.z = x + y The product of a number by its conjugate equals the distance squared of the point from the origin, i.e.the square of the modulus of the vector z. Since the modulus is z, z.z = z. From this the inverse is easy. Ingegneria dell Autoveicolo, LeLing13 3 Geometria

4 Proposition 0.3. The inverse 1 z of z 0 is 1 z = z z = x x + y y x + y i Proof. From the product z z z = zz z = z z = 1 we see that 1 z = z z, by definition. Example 0.4. The inverse of i is 1 i = i The inverse 1, together with the sum and the product allows to see C as a number z field 3. Two complex numbers z = x + y i, w = a + b i are multiplied as follows: z.w = (x + y i)(a + b i) = xa + xb i +ya i +yb i = xa + (xb + ya) i +yb( 1) so z.w = (xa yb) + (xb + ya) i. A famous formula: z w = (x + y )(a + b ) = (xa yb) + (xb + ya) = zw. This was Euler s starting point for proving Fermat s theorem 4, which states that the natural numbers of the form 4k + 1 are the sum of two squares. Notice at last that the real and imaginary parts can be found using conjugates: Re(z) = z + z Im(z) = z z i The number z is said purely imaginary if Re(z) = 0. Thus z is purely imaginary iff z = y i, y R. The condition z = z is necessary and sufficient for z to be purely imaginary. 3 First noticed by the mathematician Raffaele Bombelli from Bologna, in di Fermat sulle somme di due quadrati Ingegneria dell Autoveicolo, LeLing13 4 Geometria

5 Roots of real polynomials, conjugate roots &c. Although the polynomial p(x) = x + 1 is real, it has two complex roots i and i. Notice i is (by definition) the conjugate of i. This occurs to any real polynomial; if z is a complex root of the real equation a 0 + a 1 x + a x + + a n x n = 0, also its conjugate z is a root. In fact, because a 0, a 1,, a n are real we have so: a 0 + a 1 z + a z + + a n z n = 0 = 0 a 0 + a 1 z + a z + + a n z n = 0, a 0 + a 1 z + a z + + a n z n = 0, a 0 + a 1 z + a z + + a n z n = 0 and the conjugate z is a root of the same equation. Proposition 0.5. Let p(x) R[X] be a polynomial. If the complex number z C satisfies p(z) = 0, then the conjugate z satisfies p(z) = 0. Moreover, if Im(z) 0 the quadratic polynomial q(x) = (X z)(x z) = X (z+z)x +zz = X Re(z)X + z is real, i.e. q(x) R[X], and q(x) divides p(z). This result is a generalised theorem of Ruffini of sorts. As complex roots come in pairs, the proposition tells us that an odd-degree polynomial will always admit a real root. This can be proven in the Calculus (Mathematical Analysis) course by observing that the limits at infinity have different signs. The above result does not prove this fact because we do not (yet) know that all roots of a polynomial are complex numbers... At any rate, with complex numbers we are always capable of solving the quadratic equation with real coefficients a, b, c: ax + bx + c = 0. The formula b± b 4ac a will always spit out two numbers: real ones if b 4ac 0, complex if b 4ac < 0, i.e. b±i 4ac b a. Example 0.6. The roots of x + x + 1 = 0 are 1 ± 1 4 = 1 ± 3 = 1 ± 3 i Ingegneria dell Autoveicolo, LeLing13 5 Geometria

6 Euclides s algorithm. Already 300 years B.C. Euclides could easily compute the greatest common divisor gcd(a, b) 5 of a, b using this simple observation: Suppose a < b: (i) we may subtract a from b an integer number of times q, with a remainder r: b = a.q + r, 0 r < a (ii) gcd(a, b) is equal to gcd(r, a). To find gcd(a, b) it suffices to determine gcd(r, a); then we can go back to point (i) and find gcd(r, a), which is intuitively easier because r is smaller than a. Example 0.7. Using Euclides s observation over and over we can determine gcd(5341, 315). Now, gcd(5341, 315) = gcd(315, 1801) because 1801 is the remainder of the division between 5341 and 315. Then again, gcd(315, 1801) = gcd(1801, 1414), and again gcd(1414, 387) = gcd(387, 53). The idea is clear by now, so gcd(5341, 315) = gcd(315, 1801) = gcd(1801, 1414) = gcd(1414, 387) = = gcd(387, 53) = gcd(53, 134) = gcd(134, 119) = gcd(119, 15) = gcd(15, 14) = 1 Thus gcd(5341, 314) is 1, making 5341 and 314 relatively prime. Ancient Greeks were computing gcd(a, b) for a, b not even integer: given a, b, they were seeking a common unit d, i.e. a number d such that a, b became integer multiples of d. But if a, b are no longer integers, the algorithm does not necessarily end. one could continue subtracting the remainder indefinitely, without ever arriving at a zero remainder. Take for instance a = 1 and x larger than a = 1 and satisfying x 1 = 1 x 1. 5 Think of d = gcd(a, b) as a unit for measuring both a, b: a, b are integer multiples of d, and the latter is the largest number with such property. Ingegneria dell Autoveicolo, LeLing13 6 Geometria

7 0.1 Factorisation Geometria Lingotto. Since x > 1, 0 < x 1 < 1 so the remainder r of the division between a = 1 and x is r = 1 x: x 1 = 1 r. the ratio x over 1 equals the ratio 1 over the remainder r. But { b = b, a a b = q.a + r, 0 r < a = { a = a, r r b = q.a + r, 0 r < a so the division will go on forever, always having to divide 1 by a remainder standing in the same initial ratio x = 1 ; The example itself was conceived so that the procedure 1 r would not stop. Solving x x 1 = 0 we have x = 1 + 5, which represents the length od a regular pentagon of side length 1 6. The remainder r of the division between x and 1 is the diagonal of the inner pentagon drawn by the diagonals, whose side is 1 r. This proves that the division doesn t end, because there will always be a smaller pentagon (formed by the larger pentagon s diagonals). As the algorithm of Euclides applied to 1 and a rational number p q we have the following: definitely ends, Theorem 0.8. The numbers 1 and 1+ 5 are not commensurable, i.e is irrational Factorisation When a, b are integers, to find gcd(a, b) one must factorise a and b and then take the common prime factors raised to the minimum power. Mathematically, 6 The number x = 1+ 5 is called the golden ratio. 7 Discovered by Hippasus of Metapontum, disciple of the Pythagorean school; the legend has it that he was drowned at sea for breaking the pledge to secrecy about this fact. Ingegneria dell Autoveicolo, LeLing13 7 Geometria

8 0.1 Factorisation Geometria Lingotto. if a = i p a i i and b = i p b i i then 8 gcd(a, b) = i p min{a i,b i } i where p i is the sequence of prime numbers, p 1 =, p = 3, p 4 = 5, &c. Hence if the numbers a, b are easy to factorise the foregoing formula is useful to find gcd(a, b). But factorising a number is, generally speaking, hard, so we need a more efficient method to compute the gcd(a, b). Greatest common divisor of polynomials. The idea of Euclides can be employed to find gcd(p (X), Q(X)) for two given polynomials P (X), Q(X) 9 ; assuming deg(p (x)) deg(q(x)), if R(X) denotes the remander of the division P (X)/Q(X) we have: gcd(p (X), Q(X)) = gcd(q(x), R(X)). Since the degree of R(X) is smaller than the degree of P (X), after a finite number of steps the gcd(p (X), Q(X)) will be attained. Example 0.9. gcd(x 5 3X + 1, X 3 + X ) = gcd(x 3 + X, X + 4X 3) = = gcd( X + 4X 3, 15X 14) = gcd(15x 14, 31 5 ) = 1, making X 5 3X + 1 and X 3 + X relatively prime. 8 the power 0 indicates that the prime does not divide the given number, so that no factorisation is possible. 9 This goes back to the Persian mathematician Omar Khayyam ( ). Ingegneria dell Autoveicolo, LeLing13 8 Geometria

9 0.1 Factorisation Geometria Lingotto. Multiple roots of P (X) and gcd(p (X), P (X)). Let P (X) = a 0 + a 1 X + + a n X n = 0 be an algebraic equation of degree n. Ruffini s theorem 10 says that if one knows λ is a root of the previous equation, P (λ) = 0, then X λ divides P (X), and the other way around. Suppose then that for some reason we already know a root λ of P (X), and we need to find a second root. Ruffini s theorem reduces the problem to an equation of degree n 1. Its moral is, in fact, that if λ solvs the algebraic equation P (X) = 0, we may factorise P (X) = (X λ)q(x), so we are left with Q(X) = 0. The latter Q(X) is the quotient of P (X) by X λ. That s all. A root λ of P (X) = 0 is called multiple if λ annihilates, apart from P (X), also Q(X); this means (X λ) divides P (X). The largest exponent r such that (X λ) r divides P (X) is known as multiplicity of the root. A root of multiplicity one is called simple. It is striking that deciding whether P (X) has or not multiple roots does not requite the determination of said roots. Theorem A polynomial P (X) has one multiple root iff P (X) and its derivative P (X) are not relatively prime, i.e. gcd(p (X), P (X)) 1. Moreover, the multiple roots of P (X) are precisely the roots of gcd(p (X), P (X)). The next two examples explain the relevance of the theorem. Example Does the polynomial P (X) = X 5 5X 3 + 4X 1 admit multiple roots? The answer is no. After a few divisions one sees that gcd(x 5 5X 3 + 4X 1, 5X 4 15X + 4) = 1. On the other hand, Example 0.1. The polynomial P (X) = 4 + 8X + X 5X 3 X 4 + X 5 has multiple roots, for gcd(p (X), P (X)) is 3X + X 3, hence P (X) and P (X) are not coprime polynomials. The multiple roots are 1,, as one can easily check. 10 Contrary to conventional wisdom, Ruffini s theorem does not tell how to find a root of the algebraic equation P (X) = a 0 + a 1 X + + a n X n = 0. Ingegneria dell Autoveicolo, LeLing13 9 Geometria

1 Complex numbers and the complex plane

1 Complex numbers and the complex plane L1: Complex numbers and complex-valued functions. Contents: The field of complex numbers. Real and imaginary part. Conjugation and modulus or absolute valued. Inequalities: The triangular and the Cauchy.