1 Introduction We introduce the integer numbers because otherwise the equation a + x = b cannot always be solved. We introduce the rational numbers because otherwise the equation ax = b cannot always be solved. We introduce the irrational numbers because otherwise equations like x2 = 2 cannot always be solved. We introduce the complex numbers because otherwise equations like x2 = −1 cannot always be solved. 2 Imaginary unit We postulate that there is a number j such that j2 = −1. This number is called the imaginary unit. We assume that j obeys the usual arithmetic rules. Practical rule 1. Expressions containing j can be manipulated like ex- pressions containing numbers and/or variables with the added bonus that powers of j can be simplified. 3 Imaginary numbers An imaginary number is a product of j, the imaginary unit, with a real number, e.g. j2, j3.1, jπ, j(−1). Imaginary numbers can be multiplied, as in j2 × j5 = j × j(2 × 5) = j210 = (−1)10 = −10 1 The square of an imaginary number is always negative: (ja)2 = j2a2 = −a2 < 0 j and −j are both square roots of −1 because j2 = 1, (−j)2 = (−1)2j2 = 1(−1) = −1. We shall see later that the imaginary unit is sufficient to express all kinds of roots we need. For example, just as for positive numbers we can find two square roots of any negative number as in √ √ √ √ √ −2 = −1 2 = j 2 or − j 2 4 Powers of j j2 = −1 (by definition) j3 = j2j = −1j = −j j4 = j2j2 = 1 j0 = 1 j1 = j j5 = j4j = j (Omitted) Any positive integer n can be written as n = 4q + r, where q is a non-negative integer, r = 0, 1, 2 or 3. Hence 1 when r = 0 j when r = 1 j4q+r = −1 when r = 2 −j when r = 3 5 Complex numbers The sum of a real and an imaginary number is called a complex number, e.g. √ 1 1 + j, −1 + j, −2 − j3, 3 − j . 2 In general, let a and b be real numbers. Then the complex number z = a+jb is said to be in standard (or cartesian) form. The real part of z is a, the imaginary part of z is b. Re(z) = a, Im(z) = b. 2 5.1 Argand diagram 4 quadrants, show a number in each complex plane 5.2 Notational conventions The terminology is natural but if you are baffled by questions like: “Is the complex number z real?” or “Is the complex number z purely imaginary?”, you may find the following explanations useful. The real numbers are those complex numbers whose imaginary part is 0. We normally write 2 rather than 2 + j0. For emphasis such numbers are often said to be purely real. The imaginary numbers are those complex numbers whose real parts are 0. We normally write j3 rather than 0 + j3. A complex number whose real part is zero is also said to be purely imaginary. 0 + j0 is both purely real and purely imaginary and is normally written as 0. We also write −j rather than j(−1), etc. 6 Arithmetic with complex numbers Let z1 = a1 + jb1, z2 = a2 + jb2. We apply Practical rule 1 to arrive at appropriate definitions for the basic arithmetic operations. There is no need to remember the formulas. Simply expand and collect terms as appropriate for each operation. Only the division is somewhat tricky. Notice the way to remove the imaginary unit from the denominator. Let z1 = 1 + j3, z2 = 2 − j. Introduce the following operations first using z1 and z2 and then using general letters. • z1 + z2 = (a1 + a2) + j(b1 + b2) • z1 − z2 = (a1 − a2) + j(b1 − b2) • z1z2 = (a1a2 − b1b2) + j(a1b2 + b1a2) 3 • Division z a + jb 1 = 1 1 z2 a2 + jb2 (a + jb )(a − jb ) = 1 1 2 2 (a2 + jb2)(a2 − jb2) (a1 + jb1)(a2 − jb2) = 2 2 a2 − (jb2) (a1 + jb1)(a2 − jb2) = 2 2 a2 + b2 (a1a2 + b1b2) + j(b1a2 − a1b2) = 2 2 a2 + b2 (a1a2 + b1b2) (b1a2 − a1b2) = 2 2 + j 2 2 a2 + b2 a2 + b2 • z−1. This is the same as 1/z. 6.1 Solving quadratic equations Any quadratic equation has two roots, possibly complex. The familiar for- mula applies. Example. Solve the equation z2 − 2z + 5 = 0. Solution: √ √ √ √ 2 ± 4 − 4 × 5 2 ± 4 − 20 2 ± −16 2 ± j 4 z = = = = = 1 ± j2. 2 2 2 2 6.2 Complex conjugate If z = a + jb then the complex conjugate of z is denoted by z∗ orz ¯ and is obtained by reversing the sign of the imaginary part of z: z∗ = a − jb Show it on Argand diagram! Important properties: • z + z∗ = 2Re(z); • z − z∗ = j(2Im(z)); • zz∗ = (a + jb)(a − jb) = a2 + b2 = Re(z)2 + Im(z)2, i.e. it is real; • (z∗)∗ = z. 4 6.3 Equality of complex numbers Two complex numbers are equal if and only if they have the same real parts and the same imaginary parts. Symbolically: z1 = z2 if and only if Re(z1) = Re(z2) and Im(z1) = Im(z2). This rule gives us a simple device to find complex numbers with prescribed properties. Example. Find a complex number z such that 1 z + z∗ = 3 + j2. 2 Solution. Let the standard form of z be z = x+jy. Then the above becomes 1 x + jy + (x − jy) = 3 + j2, 2 we now expand the expression and collect the terms as appropriate to get 1 1 x + x + j(y − y) = 3 + j2, 2 2 which simplifies to 3 1 x + j( y) = 3 + j2. 2 2 The real and imaginary parts of the number to the left of the equality must be equal to the corresponding parts on the right. Hence, 3 1 x = 3, y = 2, 2 2 which gives x = 2, y = 4. Hence the answer is z = 2 + j4. We can verify this by substituting it into the equation above: 1 1 z + z∗ = 2 + j4 + (2 − j4) = 2 + j4 + 1 − j2 = 3 + j2, as required. 2 2 Note: There is no “less then” relation betwen complex numbers! 6.4 Modulus and argument 1) Show on the picture 2) define modulus, mention r > 0. Note that |z| = zz∗. 3) define argument, 4) discuss on z = 1 + j 5) mention principal value, (−π, π]; 6) advice to use radians rather than degrees. 5 6.5 Polar form of a complex number Say what it is, restriction on the parameters. Mention that it is unique only if the argument is restricted to an interval like (−π, π]. 6.5.1 Finding polar form from standard form 4 Example: z = −3 + j4. |z| = 5, tan(θ) = − 3 . So θ = 2.2143 or θ = −0.9272. We choose the first one because it gives the correct quadrant. So z = 5(cos 2.2143 + j sin 2.2143) Discuss particular cases: z = 1, j, −1, −j. 6.5.2 Finding standard form from polar form. Example. Given that |z| = 3 and arg(z) = π/3 write z in standard√ form. Solution: We have a = 3 cos(π/3) = 3/2, b = 3 sin(π/3) = 3 3/2. So, √ 3 3 3 z = + j . 2 2 6.6 Exponent of a complex number Let b be real. We define ejb by ejb = cos b + j sin b. Let z = a + jb, where a and b are real. We define ez by ez = ea(cos b + j sin b) = eaejb. and we see that the modulus of ez is equal to ea while arg(z) = b. The function ez has the same properties as its real counterpart, in par- ticular, ez1+z2 = ez1 ez2 , (ez)n = enz. The importance of the exponent cannot be overstated. One magnificient identity which follows from the definition above is ejπ = −1. It gives a simple relation between some fundamental mathematical constants. Other special cases with modulus one: ejπ/4, ejπ/2, e−jπ/2, ej3π/2, e2πj, etc. 6 6.7 Exponential form of a complex number The complex exponent provides another way to write the polar form of a complex number. Namely, z = r(cos θ + sin θ) = reiθ, where r is the modulus of z, θ its argument. Example. √ z = 1 + j = 2ejπ/4. 6.7.1 Multiplication and division in exponential form. jθ1 jθ2 If z1 = r1e , z2 = r2e , then jθ1 jθ2 jθ1 jθ2 j(θ1+θ2) z = z1z2 = r1e r2e = r1r2e e = (r1r2)e , i.e., to get the product we multiply the moduli and add the arguments. Similarly, for division, jθ1 jθ2 jθ1 −jθ2 j(θ1−θ2) z1/z2 = r1e /(r2e ) = r1r2e e = (r1r2)e , i.e., we divide the moduli and subtract the arguments. Geometrical representation of multiplication by j and other complex num- bers. 6.7.2 Addition in exponential form Example. Find the real and imaginary parts of z = ejπ/4 + 2ejπ/3 Now, 1 1 ejπ/4 = cos(π/4) + j sin(π/4) = √ + j √ 2 2 √ 1 3 ejπ/3 = cos(π/3) + j sin(π/3) = + j 2 2 Hence, √ ! 1 1 1 3 z = √ + j √ + 2 + j 2 2 2 2 1 1 √ = (1 + √ ) + j(√ + 3) 2 2 = 1.7071 + j2.4392 The exponent is extremely useful. We apply it below to derive the De Moivre’s formula. 7 6.7.3 De Moivre’s formula This is de Moivre’s formula: (cos(θ) + j sin(θ))n = cos(nθ) + j sin(nθ). Indeed, let z = cos(θ) + j sin(θ). Then zn = (cos(θ) + j sin(θ))n.