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Chapter 5 Complex Numbers

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Chapter 5 Complex Numbers Chapter 5 Complex numbers Why be one-dimensional when you can be two-dimensional? ? 3 2 1 0 1 2 3 − − − ? We begin by returning to the familiar number line, where I have placed the question marks there appear to be no numbers. I shall rectify this by defining the complex numbers which give us a number plane rather than just a number line. Complex numbers play a fundamental rˆolein mathematics. In this chapter, I shall use them to show how e and π are connected and how certain primes can be factorized. They are also fundamental to physics where they are used in quantum mechanics. 5.1 Complex number arithmetic In the set of real numbers we can add, subtract, multiply and divide, but we cannot always extract square roots. For example, the real number 1 has 125 126 CHAPTER 5. COMPLEX NUMBERS the two real square roots 1 and 1, whereas the real number 1 has no real square roots, the reason being that− the square of any real non-zero− number is always positive. In this section, we shall repair this lack of square roots and, as we shall learn, we shall in fact have achieved much more than this. Com- plex numbers were first studied in the 1500’s but were only fully accepted and used in the 1800’s. Warning! If r is a positive real number then √r is usually interpreted to mean the positive square root. If I want to emphasize that both square roots need to be considered I shall write √r. ± When the discriminant of a quadratic equation is strictly less than zero, we know that it has no real roots. In this section, we shall show that in this case the equation has two complex roots. This will mean that quadratic equations will always have two roots. The key step is the following We introduce a new number, denoted by i, whose defining property is that i2 = 1. We shall assume that in all other respects it satisfies the usual− axioms of high-school algebra. This assumption will be justified later. We shall now explore the consequences of this definition which turns out to be a profound one for mathematics. It follows that i and i are the two missing square roots of 1. In all other respects the number i will− behave like a real number. Thus if b is any real number then bi is a number, and if a is any real number then a + bi is a number. A complex number is a number of the form a + bi where a, b R. We ∈ denote the set of complex numbers by C. Complex numbers are sometimes called imaginary numbers. This is not such a good term: they are not figments of our imagination like unicorns or dragons. Like all numbers they are, however, products of our imagination: no one has seen the complex number number i but, then again, no one has seen the number 2. If z = a + bi then we call a the real part of z, denoted Re(z), and b the complex or imaginary part of z, denoted Im(z). Two complex numbers a + bi and c + di are equal precisely when a = c and b = d. In other words, when their real parts are equal and when their complex parts are equal. 5.1. COMPLEX NUMBER ARITHMETIC 127 We can think of every real number as being a special kind of complex number because if a is real then a = a + 0i. Thus R C. Complex numbers of the form bi are said to be purely imaginary. ⊆ Now we show that we can add, subtract, multiply and divide complex numbers. Addition, subtraction and multiplication are all easy. Let a+bi, c+di C. To add these numbers means to calculate (a+bi)+ (c+di). We assume∈ that the order in which we add complex numbers doesn’t matter and that we may bracket sums of complex numbers how we like and still get the same answer and so we can rewrite this as a + c + bi + di. Next we assume that multiplication of complex numbers distributes over addition of complex numbers to get (a + c) + (b + d)i. Thus (a + bi) + (c + di) = (a + c) + (b + d)i. The definition of subtraction is similar and justified in the same way (a + bi) (c + di) = (a c) + (b d)i. − − − To multiply our numbers means to calculate (a + bi)(c + di). We first assume complex multiplication distributes over complex addition to get (a + bi)(c + di) = ac + adi + bic + bidi. Next we assume that the order in which we multiply complex numbers doesn’t matter to get ac + adi + bic + bidi = ac + adi + bci + bdi2. Now we use the fact that i2 = 1 to get ac + adi + bci + bdi2 = ac + adi + bci bd. We now rearrange the terms− to get the following definition of multiplication− (a + bi)(c + di) = (ac bd) + (ad + bc)i. − Examples 5.1.1. Carry out the following calculations. 1. (7 i) + ( 6 + 3i). We add together the real parts to get 1; adding together− −i and 3i we get 2i. Thus the solution is 1 + 2i. − 2. (2 + i)(1 + 2i). First we multiply out the brackets as usual to get 2 + 4i + i + 2i2. We now use the fact that i2 = 1 to get 2 + 4i + i 2. Finally we simplify to get 0 + 5i = 5i. − − 2 1 i 3. − . Multiply out and simplify to get i. √2 − 128 CHAPTER 5. COMPLEX NUMBERS The final operation is division. We have to show that when a + ib = 0 the reciprocal 6 1 a + ib is also a complex number. We use an idea that can also be applied in other situations called rationalizing the denominator. It is convenient first to define a new operation on complex numbers. Let z = a + bi C. Define ∈ z¯ = a bi. − The numberz ¯ is called the complex conjugate of z. Why is this operation useful? Let’s calculate zz¯. We have zz¯ = (a + bi)(a bi) = a2 abi + abi b2i2 = a2 + b2. − − − Notice that zz¯ = 0 if and only if z = 0. Thus for non-zero complex numbers z, the number zz¯ is a positive real number. Let’s see how we can use the complex conjugate to define division of complex numbers. Our goal is to calculate 1 a + bi where a+bi = 0. The first step is to multiply top and bottom by the complex conjugate of6 a + bi. We therefore get a bi a bi 1 − = − = (a bi) . (a + bi)(a bi) a2 + b2 a2 + b2 − − Examples 5.1.2. Carry out the following calculations. 1+i 1. i . The complex conjugate of i is i. Multiply top and bottom of the i+1 − fraction to get − = 1 i. 1 − i 2. 1 i . The complex conjugate of 1 i is 1 + i. Multiply top and bottom − i(1+i) i 1 − of the fraction to get 2 = −2 . 4+3i 3. 7 i . The complex conjugate of 7 i is 7 + i. Multiply top and bottom − (4+3i)(7+i) −1+i of the fraction to get 50 = 2 . We shall need the following properties of the complex conjugate later on. Lemma 5.1.3. 5.1. COMPLEX NUMBER ARITHMETIC 129 1. z1 + ... + zn = z1 + ... + zn. 2. z1 . zn = z1 ... zn. 3. z is real if and only if z = z. We now introduce a way of thinking about complex numbers that enables us to visualize them. A complex number z = a + bi has two components: a and b. It is irresistible to plot these as a point in the plane. The plane used in this way is called the complex plane: the x-axis is the real axis and the y-axis is interpreted as the complex axis. z = a + ib ib a Although a complex number can be thought of as labelling a point in the complex plane, it can also be regarded as labelling the directed line segment from the origin to the point. By Pythagoras’ theorem, the length of this line is √a2 + b2. We define z = √a2 + b2 | | where z = a + bi. This is called the modulus1 of the complex number z. Observe that z = √zz.¯ | | We shall use the following important property of moduli. Lemma 5.1.4. wz = w z . | | | | | | Proof. Let w = a + bi and z = c + di. Then wz = (ac bd) + (ad + bc)i. Now wz = (ac bd)2 + (ad + bc)2 whereas w z = (−a2 + b2)(c2 + d2). But | | − | | | | (ac bdp)2 + (ad + bc)2 = (ac)2 + (bd)2 + (ad)2 + (pbc)2 = (a2 + b2)(c2 + d2). − Thus the result follows. 1Plural: moduli 130 CHAPTER 5. COMPLEX NUMBERS The complex numbers were obtained from the reals by simply throwing in one new number, i, a square root of 1. Remarkably, every complex number has a square root. − Theorem 5.1.5. Every nonzero complex number has exactly two square roots. Proof. Let z = a + bi be a nonzero complex number. We want to find a complex number w so that w2 = z. Let w = x + yi. Then we need to find real numbers x and y such that (x+yi)2 = a+bi. Thus (x2 y2)+2xyi = a+bi, and so equating real and imaginary parts, we have to solve− the following two equations x2 y2 = a and 2xy = b. − Now we actually have enough information to solve our problem, but we can make life easier for ourselves by adding one extra equation. To get it, we use the modulus function.
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