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CHAPTER 8. COMPLEX NUMBERS Why Do We Need Complex Numbers? First of All, a Simple Algebraic Equation Like X2 = −1 May Not Have

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CHAPTER 8. COMPLEX NUMBERS Why Do We Need Complex Numbers? First of All, a Simple Algebraic Equation Like X2 = −1 May Not Have CHAPTER 8. COMPLEX NUMBERS Why do we need complex numbers? First of all, a simple algebraic equation like x2 = 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. 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 be- come a power tool for answering many extremely difficult questions in mathematics and theoretical physics, and also finds its usefulness in many areas of engineering and com- munication technology. 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 Vall´ee Poussin in 1896 by using the complex theory developed at that time. A widely quoted statement by Jacques Hadamard says: “The shortest path between two truths in the real domain passes through the complex domain”. The basic idea for complex numbers is to introduce a symbol i, called the imaginary unit, which satisfies i2 = 1. − In doing so, x2 = 1 turns out to have a solution, namely x = i; (actually, there − is another solution, namely x = i). We remark that, sometimes in the mathematical − literature, for convenience or merely following tradition, an incorrect expression with correct understanding is used, such as 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. So here we use that letter “i” to stand nothing other than the imaginary unit. A general complex number is of the form c = a + bi (or c = a + ib), where a, b are real numbers. We call a the real part of c and b the imaginary part of c. Sometimes they are denoted by Re c and Im c respectively. Question 8.1. What is wrong by saying that “√2 is a real number and hence it is not a complex number”? Addition and multiplication of complex numbers can be carried out in a straightfor- 1 ward manner, keeping in mind that i2 = 1. Indeed, given complex numbers c = a +b i − 1 1 1 and c2 = a2 + b2i, the sum c1 + c1 and the product c1c2 are given by c1 + c2 =(a1 + b1i)+(a2 + b2i)=(a1 + a1)+(b1 + b2)i 2 c1c2 =(a1 + b1i)(a2 + b2i)= a1a2 + b1b2i + a1b2i + a2b1i =(a a b b )+(a b + a b )i 1 2 − 1 2 1 2 2 1 To give a quick example, letting c = 1 2i, c = 2+3i, we have c + c = 3+ i, 1 − 2 1 2 c c = 1 5i and c c =8+ i. 1 − 2 − − 1 2 Exercise 8.2. In each of the following cases, find c + c , c c and c c . Present 1 2 1 − 2 1 2 your answers as simple as possible. (a) c = i, c = 1 i; (b) c = 2+√3i, c = 2 √3i; 1 2 − 1 2 − (c) c =1+ i, c 1 i. 1 2 − − Exercise 8.3. Find the product c1c2 of c1 = cos α1 +i sin α1 and c2 = cos α2 +i sin α2. Present your answers as simple as possible. The expression a + bi is called the Cartesian representation of c, (vs the polar representation described in the future), since geometrically it is represented by the point or the vector in the Cartesian xy–plane with coordinates (a, b). (See the following figure.) The magnitude of the vector representing c is called the absolute value or the modulus of c, and is denoted by c . Thus, if c = a + bi, then c = √a2 + b2: | | | | a + bi = √a2 + b2 (8.1) | | The xy–plane for representing complex numbers is called the complex plane or Argand plane and is denoted by C. An alternative way to write “c is a complex number” is “c C”, read as “c belongs to C”. ∈ 2 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. The complex conjugate of c = a + bi, denoted by c, is defined to be a bi: − a + bi = a bi. (8.2) − For examples, 2 + 3i = 2 3i, i = i, 2 = 2, 1 i =1+ i, etc. − − − Question 8.4. In each of the following cases, what is the complex conjugate z and the modulus z of the given complex number z? (a) z = i, (b) z = 3, (c) z =4+3i | | − − (d) z = cos θ + i sin θ. Notice that, for a complex number z = a + bi, we have zz =(a + bi)(a bi)= a2 b2i2 = a2 + b2 = z 2. − − | | So z 2 = zz. (8.3) | | This identity, which resembles v 2 = v v, is one of the most useful facts about complex | | · numbers, despite of its simplicity. Usual algebraic identities such as uv = vu (commutative law), (u + v)w = uw + vw (distributive Law) (8.4) 3 hold for complex numbers u, v and w. Furthermore, taking complex conjugates “goes along” with algebraic operations well: z + w = z + w, zw = z w, z/w = z/w. (8.5) Exercise 8.5. Check all identities in (8.4) and (8.5). Example. We are asked to verify the identity zw = z w (8.6) | | | || | for two complex numbers z and w. Notice that zw = z w is equivalent to | | | || | zw 2 = z 2 w 2 | | | | | | and hence it suffices to verify the latter. By (8.3), we have zw 2 = zw zw = zw z w = z zw w = z 2 w 2, | | | | | | which is what we want to show. Exercise 8.6. Use (8.6) to deduce (a2 + b2)(c2 + d2)=(ac + bd)2 +(ad bc)2, − where a, b, c, d are arbitrary real numbers. Which integers u and v satisfy u2 + v2 = (92 + 102)(102 + 112)? Exercise 8.7. Verify the following “parallelogram identity” z + w 2 + z w 2 = 2 z 2 + 2 w 2 | | | − | | | | | for arbitrary complex numbers z and w. An important use of complex conjugation is to divide complex numbers. Given com- plex numbers z and w with z = 0, to divide w by z, we multiply both the denominator 6 and the numerator of w/z by z, the complex conjugate of the denominator, to obtain w wz wz = = z zz z 2 | | 4 in order to convert the denominator of w/z into a real number. Then we can divide the real part and the imaginary part of wz by the real number z 2. For example, to divide | | 1 i by 1+ i, we proceed as follows, by noticing that the complex conjugateof 1+ i − is 1 i: − 1 i (1 i)(1 i) 1+ i2 2i − = − − = − = i. 1+ i (1 + i)(1 i) 11 + 12 − − In division of complex numbers, it is important to recognize the complex conjugate z of z = x + yi and to use zz = z 2 = x2 + y2. | | Exercise 8.8. Perform each of the following complex division: 8+ i 5+ i 75 + 223i cos θ + i sin θ (a) (b) (c) (d) . 2 i 3 2i 63 + 17i cos θ i sin θ − − − Example. We are asked to check that if z = 1 and if | | z a w = − , 1 a z − where a is any fixed complex number with a < 1, then w = 1. First, let us notice | | | | that w = 1 is the same as w 2 = 1, which is the same as ww = 1, (in view of (8.3) | | | | above). This leads us to compute w w. Now z a z a zz za az + aa 1 za az + aa w w = − − = − − = − − = 1, 1 a z 1 az 1 az az + az az 1 az za + aa − − − − − − in view of zz = z 2 = 1. Done. | | Problem 8.9. Verify that, if a is a real number with 0 <a< 1, then z a | − | = a z a 1 | − − | for all complex numbers z with z = 1. | | The real part of a complex number z, denoted by Re z, is given by z + z Re z = (8.7) 2 A complex number z is real if and only if z = z. Exercise 8.10. Check the last statement as well as identity (8.6). 5 Exercise 8.11. Verify that, if z =1 and z = 1, then | | 6 z + 1 w = i z 1 − is a real number. Take any complex number z = x+iy, with x as its real part and y as its imaginary part. Write r for the modulus of z: r = z = x2 + y2. Let θ be the angle between | | the vector representing z; (see the next figure). Thenp x = r cos θ and y = r sin θ. Hence z = r cos θ + ir sin θ, or z = r(cos θ + i sin θ). (8.8) The last expression is called the polar form of z. Multiplication of complex numbers is especially revealing when expressed in polar form. Let z1 = r1(cos θ1 + i sin θ1), z2 = r2(cos θ2 + i sin θ2) be complex numbers in polar form. Then z1z2 = r1r2(cos θ1 + i sin θ1)(cos θ2 + i sin θ2) = r r ((cos θ cos θ sin θ sin θ )+ i(cos θ sin θ + sin θ cos θ )) 1 2 1 2 − 1 2 1 2 1 2 = r1r2(cos(θ1 + θ2)+ i sin(θ1 + θ2)). (8.9) Here we have used the addition formula for sine and cosine in trigonometry cos(α + β) = cos α cos β sin α sin β, sin(α + β) = sin α cos β + cos α sin β.
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