Chapter 1: Complex Numbers

Transcription

1 Chapter 1: Complex Numbers Why do we need complex numbers? First of all, a simple algebraic equation like X 2 = 1 may not have a real solution. Introducing complex numbers validates the so called fundamental theorem of algebra: every polynomial with a positive degree has a root. Next, a basic problem in linear algebra is to find eigenvalues and eigenvectors of a given matrix. Consider the following simple matrix [ ] Its characteristic equation is λ = 0 and hence we need complex numbers for its eigenvalues. However, the usefulness of complex numbers is much beyond such simple applications. Nowadays, complex numbers and complex functions have been developed into a rich theory called complex analysis and become a power tool for answering many extremely difficult questions in mathematics and theoretical physics, and also finds its usefulness in many areas. For example, a famous result called the prime number theorem, which was conjectured by Gauss in 1849, and defied efforts of many great mathematicians, was finally proven by Hadamard and de la Vallée Poussin in 1896 by using the complex theory developed at that time. Jacques Hadamard said: The shortest path between two truths in the real domain passes through the complex domain. We assume that the reader knows all basics about complex numbers. In this section we give a quick review of them, emphasizing those which will be important for our future chapters. Also, we describe certain point of view which are suggestive for further development outside the realm of complex numbers. There are two basic ways to represent a complex number z algebraically: Cartesian form: z = x+iy (or x+yi) polar form: z = re iθ r(cosθ +isinθ) with r 0. Geometrically, z is represented as a vector with (x,y) as its coordinates in a Cartesian plane, called complex plane. (This vector picture of a complex number is called an Argand diagram. But we often only draw a point to simplify this picture.) The magnitude of this 1

2 vector is called the absolute value or the modulus of z and is denoted by z. It is equal to r given in its polar form: z = r = x 2 +y 2. The angle between this vector and the x axis is give by θ. The value of r is uniquely determined by z, but the value of θ is not. For example, we have e πi = cosπ +isinπ = 1 as well as e πi = cos( π)+isin( π) = 1. Thus, both π and π are possible values of θ for z = 1. When a complex number is represented in Cartesian form, say z = x + iy, the real number x is called the real part of z and the real number y is called the imaginary part of z. We write x = Re z, y = Im z. Two complex numbers are equal if and only if both of their real parts and their imaginary parts are equal. Thus For z = x+iy and z = x +iy, we have z = z x = x and y = y. Using the Cartesian form, addition and multiplication of complex numbers are straightforward, as long as we keep in mind that i 2 = 1: (a+bi)+(c+di) = (a+c)+(b+d)i (a+bi)(c+di) = (ac bd)+(ad+bc)i. (1.1) The usual algebraic identities still hold for complex numbers, such as (z + w)(z w) = z 2 w 2 and (z +w) 2 = z 2 +2zw+w 2. 2

3 Example 1.1. Find the square roots of 5 12i. Solution. We have to solve the equation z 2 = i. Write z = x + iy. Then z 2 = x 2 y 2 +2xyi. Thus we have x 2 y 2 = 5 and 2xy = 12. On the other hand, x 2 +y 2 = z 2 = z 2 = 5 12i = = 169 = 13. Together with x 2 y 2 = 5 we obtain 2x 2 = 5+13 = 18 and hence x = ±3. On the other hand, 2xy = 12 gives y = 6/x = ±2. Thus z = ±(3+2i) are the square roots of 5+12i. The relation between the Cartesian form and the polar form is given by x = rcosθ, y = rsinθ. If we write P for the point in the plane with Cartesian coordinates (x,y), then r is the length of the line segment OP and θ is the angle between the line OP and the x axis. The angle θ in the polar form z = re iθ is called the argument of z. It is not uniquely determined by z. It depends on our choice: when r z = 0, any real number would do; when r 0, any θ satisfying cosθ = x/r and sinθ = y/r would do. You may see θ = arg z in some old fashioned textbooks. Strictly speaking, this is incorrect. (Sometimes, for convenience or merely following tradition, we use an incorrect expression with correct understanding. A typical example is writing 1 for i so that we can reserve the letter i for other purposes. But we try to avoid incorrect usage as much as possible.) Notice that, in the polar form z = re iθ, we use the identity e iθ = cosθ +isinθ. (1.2) Here we consider cosθ +isinθ as the definition of the expression e iθ. Anyway, there is a name for this identity: Euler s formula. One of the most amazing things in complex numbers is the following so called addition formula: e i(α+ β) = e iα e iβ. (1.3) Using Euler s formula, we can rewrite it as cos(α+β)+isin(α+β) = (cosα+isinα)(cosβ +isinβ). Multiplying out the right hand side and comparing the real and imaginary parts of both sides, we have arrived at cos(α+β) = cosαcosβ sinαsinβ, sin(α+β) = cosαsinβ +sinαcosβ, 3

4 which are well-known (but by no means obvious) identities in trigonometry. At present, identity (1.3) is quite mystifying. We will give a thorough explanation in the future. By induction, we can deduce form (1.3) that, for all integers n, e int = (e it ) n, or cosnt+isinnt = (cost+isint) n (1.4) which is usually called de Moivre s theorem. Though pretty well known among school teachers, this theorem does not look too impressive. In the future we will give some applications to re-assure its importance. Considerthespecialcasen = 3of(1.4). Bymeansofthewellknonwidentity (a+b) 3 = a 3 +3a 2 b+3ab 2 +b 3, we have cos3t+isin3t = (cost+isint) 3 = cos 3 t+3cos 2 t(isint)+3cost(isint) 2 +(isint) 3 = cos 3 t 3costsin 2 t+i(3cos 2 tsint sin 3 t). By comparing the real parts and the imaginary parts from both sides, we get cos3t = cos 3 t 3costsin 2 t = cos 3 t 3cost(1 cos 2 t) = 4cos 2 t 3cos 2 t; sin3t = 3cos 2 t sin 3 t = (3cos 2 t sin 2 t)sint = (4cos 2 t 1)sint. which are rewritten as cos3t = T 3 (cost) and sin3t = U 2 (cost)sint, where T 3 (x) = 4x 3 3x 2, a polynomial of degree 3, and U 2 (x) = 4x 2 1, a polynomial of degree 2. This suggests that, for any positive integer n, we can write cosnt = T n (cost) for some polynomial T n (x) ofdegreenand sinnt = U n 1 (cost) sint forsomepolynomial U n 1 (x) of degree n 1. We prove this by induction. We have already checked the case for n = 3. Thereadershouldcheck thecasesforn=1and n = 2. Nowmake theinductionhypothesis that e int = T n (x)+iu n 1 (x) sint with x = cost, where T n and U n 1 are polynomials of degrees n and n 1 respectively. So e i(n+ 1)t = e int e it = (T n (x)+iu n 1 (x) sint)(cost+isint) = T n (x)cost U n 1 (x)sin 2 t+i(t n (x)sint+u n 1 (x)sintcost) = T n (x)x+u n 1 (x)(1 x 2 )+i(t n (x)+u n 1 (x)x)sint. Thus we have e i(n+ 1)t = P n+ 1 (x)+iq n (x)sint, where T n+ 1 (x) = T n (x)x+(1 x 2 )U n 1 (x) U n (x) = T n (x)+xu n 1 (x). 4

5 The last two identities tells us how to generate polynomials T n (x) and U n (x) recursively. Polynomials T n (x) are called Chebyshev s polynomials and U n (x) are known as Chebyshev s polynomials of the second kind. They have extensive applications in umerical analysis and some extremal problems arising from electrical engineering. Every complex number z has a twin sister z, called the complex conjugate of z. The twins z and z do not quite have exactly the same look. They are more like mirror images to each other. We put the pair z and z together in the Cartesian form and the polar form as follows: z = x+iy z = x iy. (1.5) z = re iθ = r(cosθ +isinθ) z = re iθ = r(cos( θ)+isin( θ)). It is clear from the above identities that the complex conjugate of z is z. Also, z is a real number if and only if z = z. Example 1.2. We are asked to prove that if z = 1 and z 1, then w = i z 1 z +1 is a real number. To do this, it is enough to check that w = w. Notice that z = 1 gives zz = z 2 = 1. Since that complex congugates of i and z are i and ovz respectively, we have w = ( i) 1 z 1+z Hence w is a real number. z z = iz zz z +zz = i z 1 z +1 = i 1 z 1+z = w. One of the most useful identities about complex numbers is the following z 2 = zz The proof of this is simple: writing z = x+iy, we have zz = (x+iy)(x iy) = x 2 i 2 y 2 = x 2 +y 2 = z 2. The following example illustarte its usefulness. Example 1.3. Let a be a complex number with a < 1. We are asked to verify that if z is unit modulus, that is, z = 1, then so is w = z a 1 az. 5

6 We need to check that w = 1. It is enough to check that w 2 = 1, or equivalently, ww = 1. Now ww = z a 1 az z a 1 az = zz az az aa 1+az az +aazz. Notice that z = 1 gives zz = z 2 = 1. Thus, replacing zz = 1 in the last fraction, we see that its numerator and denominator are the same. Hence ww = 1. (The rest of the material in the present chapter is optional.) Here we mention some formal mathematical definitions of complex numbers. In some textbooks, complex numbers are defined as planar vectors of the form (x,y) at the beginning. After identifying (1,0) with 1 and (0,1) with i, the expression (x,y) is converted into a more preferable form: (x,y) = (x,0)+(0,y) = x(1,0)+y(0,1) = x+yi. This definition provides a rigorous way to build complex numbers, but somehow it is too artificial. A better way is defined the complex field C as the splitting field of X 2 +1 over the real field. This is a highly motivated definition since, after all, historically, complex numbers were introduced in order to give a solution to X = 0. But this approach needs too much background in field theory, which is an advanced course. Here we introduce an interesting approach called the doubling construction, using some tools from linear algebra. Recall that the matrix [ cosθ sinθ A θ sinθ cosθ ]. ( ) represents the rotation of the Cartesian plane (about the origin) through angle θ. It is a priori clear that if we rotate a vector by angle α, followed by a rotation through angle β, the result is the same as a single rotation with angle α+β. Putting this in mathematical symbols, wehavea α A β = A α+ β. Multiplyingoutthelefthandsidematricesoftheidentity and comparing both sides, we recapture the well known identities in trigonometry again: cos(α+β) = cosαcosβ sinαsinβ, sin(α+β) = cosαsinβ+sinαcosβ. Expression ( ) suggests the following approach: consider 2 2 matrices of the form [ a b Z = b a where a, b ar some real numbers. We can rewrite it in the form Z = a1+bi, where 1 = ], [ ] and i = [ ]

7 When a 2 +b 2 = 1, P = (a,b) becomes a point on the circle x 2 +y 2 = 1 and hence we can write a = cosθ and b = sinθ (here θ is the angle between OP and the x axis) and hence Z becomes A θ in ( ). In general, when Z O, we have r = a 2 +b 2 0 and the point (a/r,b/r) is a point on he circle x 2 + y 2 = 1. Consequently we can write a = cosθ and b = sinθ for some θ and hence [ ] [ ] [ ] a b a/r b/r cosθ sinθ Z = = r = r = r(cosθ 1+sinθ i). b a b/r a/r sinθ cosθ It is easy to check that i 2 = 1. Also, we we denote by Z the transpose of Z, then Z Z = (a 2 +b 2 )1 and Z +Z = 2a1. Also, the determinant and the trace of Z are given by detz = a 2 +b 2 and tr Z = 2a. Exercise 1.1. Do all of the checking mentioned above. Thus, if we identify the matrix Z = a1+bi, then: 1. addition of matrices corresponds to addition of complex numbers; 2. multiplication of matrices corresponds to multiplication of complex numbers; 3. the transpose Z of Z corresponds to the complex conjugate z of z; 4. the determinant detz is equal to z 2 and the identity Z Z = (a 2 + b 2 )1 corresponds to zz = z 2 ; 5. the trace tr Z is equal to 2Re z; the identity Z +Z = 2a1 corresponds to z +z = 2Re z, or Re z = (z +z)/2; and 6. the rotation A θ corresponds to the Eulerian expression e iθ. Drills 1. Find the complex conjugate of each of the following numbers: 2+3i, 4, 1 i, 4i 1, 3i, (2+5i) 99 1, 3 i 2 2. Express each of the following complex numbers in the polar form re iθ : 2i, 2, 1 i, i 3 1, 2cos π 7 2isin π 7, 2sin π 7 +2icos π 7 3. Find the value of each of the following products (a) (2+3i)(1 i) (b) (1+i) 4 (c) ( 3+i 2)( 2+i 3) (d) ( i) 2. 7

8 4. Find the value of each of the following quotients (a) 2+3i 1 i (b) 1 i 1+i (c) 3 i 2 2+i 3 (d) cos(π/8) isin(π/8) cos(π/8)+isin(π/8). 5. Find the square roots of each of the following complex numbers (a) 3+4i (b) 16+30i (c) 4 3i (d) 4+3i; (notice the relation to part (c)). 6. Let ABC be a triangle in the complex plane, where vertices A and B are located at i and 2 respectively. Find the complex numbers which give possible locations of C. 7. If youwalk fromtheoriginofthe complexplane, arriveat a point Arepresented by the complex number z, and then turn left (i.e. turning 90 o to the left) and walk the same distance to a point B, what makes you think that the complex number representing B is (1+i)z? Exercises 1.2. Find the roots of z 2 (3+i)z +4+3i = Prove the identity z w 2 + z+w 2 = 2 z 2 +2 w 2. Give a geometric interpretation of this identity. (Hint: Use z 2 = zz several times.) 1.4. Prove the identity Re zw = z +w 2 z w Prove that, if z = 1, w = 1 and z w, then z w 1 zw is a real number. 1.6*. (This is a hard problem) Prove the following so called Hlawka s inequality z 1 +z 2 + z 2 +z 3 + z 3 +z 1 z 1 + z 2 + z 3 + z 1 +z 2 +z 3. (Hint: Square both sides.) 8