1. Complex numbers A complex number z is defined as an ordered pair z = (x,y) , where x and y are a pair of real numbers. In usual notation, we write z = x + iy , where i is a symbol. The operations of addition and multiplication of complex numbers are defined in a meaningful manner, which force i2 = −1. The set of all complex numbers is denoted by C. Write Re z = x, Im z = y. 1 Since complex numbers are defined as ordered pairs, two complex numbers (x1,y1) and (x2,y2) are equal if and only if both their real parts and imaginary parts are equal. Symbolically, (x1,y1) = (x2,y2) if and only if x1 = x2 and y1 = y2 . A complex number z = (x,y), or as z = x + iy, is defined by a pair of real numbers x and y; so does for a point (x,y) in the x-y plane. We associate a one-to-one correspondence between the complex number z = x + iy and the point (x,y) in the x-y plane. We refer the plane as the complex plane or z-plane. 2 Polar coordinates x = r cos θ and y = r sin θ Modulus of z : |z| = r = x2 + y2. q y z = (x, y) |z| arg z x Vectorical representation of a complex number in the complex plane 3 Obviously, Re z ≤|z| and Im z ≤|z|; and z = x + iy = r(cos θ + i sin θ), where θ is called the argument of z, denoted by arg z. The principal value of arg z, denoted by Arg z, is the particular value of arg z chosen within the principal interval (−π,π]. We have arg z = Arg z + 2kπ k any integer, Arg z ∈ (−π,π] . Note that arg z is a multi-valued function. 4 Complex conjugate The complex conjugate z of z = x + iy is defined by z = x − iy . In the complex plane, the conjugate z = (x, −y) is the reflection of the point z = (x,y) with respect to the real axis. Standard results on conjugates and moduli z1 z1 (i) z1 + z2 = z1 + z2, (ii) z1z2 = z1 z2, (iii) = , z2 z2 z1 |z1| (iv) |z1z2| = |z1| |z2| (v) = . z |z | 2 2 5 Example Find the square roots of a + ib, where a and b are real constants. Solution Let u + iv be a square root of a + ib; and so (u + iv)2 = a + ib. Equating the corresponding real and imaginary parts, we have u2 − v2 = a and 2uv = b. By eliminating v, we obtain a fourth degree equation for u: 4u4 − 4au2 − b2 = 0 . The two real roots for u are found to be a + a2 + b2 u = ±v . u q u 2 t 6 From the relation v2 = u2 − a, we obtain a2 + b2 − a v = ±v . uq u 2 t Apparently, there are four possible values for u + iv. However, there can be only two values of the square root of a + ib. By virtue of the relation 2uv = b, one must choose u and v such that their product has the same sign as b. This leads to a + a2 + b2 b a2 + b2 − a u + iv = ± v + i v , u q uq u 2 |b|u 2 t t provided that b 6= 0. The special case where b = 0 is trivial. As a numerical example, take a = 3, b = −4, then 3 + 5 5 − 3 ± − i = ±(2 − i). s 2 s 2 7 De’ Moivre’s theorem Any complex number with unit modulus can be expressed as cos θ + i sin θ. By virtue of the complex exponential function∗, we have eiθ = cos θ + i sin θ. The above formula is called the Euler formula. As motivated by the Euler formula, one may deduce that (cos θ + i sin θ)n = (eiθ)n = einθ = cos nθ + i sin nθ, where n can be any positive integer. ∗ The complex exponential function is defined by ez = ex+iy = ex(cos y + i sin y), z = x + iy. 8 To prove the theorem, we consider the following cases: (i) The theorem is trivial when n = 0. (ii) When n is a positive integer, the theorem can be proven easily by mathematical induction. (iii) When n is a negative integer, let n = −m where m is a positive integer. We then have 1 1 (cos θ + i sin θ)n = = (cos θ + i sin θ)m cos mθ + i sin mθ cos mθ − i sin mθ = (cos mθ + i sin mθ)(cos mθ − i sin mθ) = cos mθ − i sin mθ = cos nθ + i sin nθ. 9 How do we generalize the formula to (cos θ + i sin θ)s, where s is a rational number? Let s = p/q, where p and q are irreducible integers. Note that the modulus of cos θ + i sin θ is one, so does (cos θ + i sin θ)s. Hence, the polar representation of (cos θ + i sin θ)s takes the form cos φ + i sin φ for some φ. Now, we write cos φ + i sin φ = (cos θ + i sin θ)s = (cos θ + i sin θ)p/q. Taking the power of q of both sides cos qφ + i sin qφ = cos pθ + i sin pθ, 10 which implies pθ + 2kπ qφ = pθ + 2kπ or φ = , k = 0, 1, ··· , q − 1. q The value of φ corresponding to k that is beyond the above set of integers would be equal to one of those values defined in the equation plus some multiple of 2π. There are q distinct roots of (cos θ + i sin θ)p/q, namely, pθ + 2kπ pθ + 2kπ cos + i sin , k = 0, 1, ··· , q − 1. q ! q ! 11 nth root of unity By definition, any nth roots of unity satisfies the equation zn = 1 . By de’ Moivre’s theorem, the n distinct roots of unity are 2kπ 2kπ z = ei2kπ/n = cos + i sin , k = 0, 1,...,n − 1 . n n i2π/n 2 n−1 If we write ωn = e , then the n roots are 1, ωn, ωn,...,ωn . Alternatively, if we pick any one of the nth roots and call it α, then 2 n−1 the other n−1 roots are given by αωn, αωn,...,αωn . This is obvious since each of these roots satisfies k n n n k (αωn) = α (ωn) = 1, k = 0, 1, ··· ,n − 1. 12 In the complex plane, the n roots of unity correspond to the n vertices of a regular n-sided polygon inscribed inside the unit circle, with one vertex at the point z = 1. The vertices are equally spaced on the circumference of the circle. The successive vertices are obtained by increasing the argument by an equal amount of 2π/n of the preceding vertex. Suppose the complex number in the polar form is represented by r(cos φ + i sin φ), its nth roots are given by φ + 2kπ φ + 2kπ r1/n cos + i sin , k = 0, 1, 2,...,n − 1, n n where r1/n is the real positive nth root of the real positive number r. The roots are equally spaced along the circumference with one vertex being at r1/n[cos(φ/n)+ i sin(φ/n)]. 13 Triangle inequalities For any two complex numbers z1 and z2, we can establish 2 |z1 + z2| = (z1 + z2)(z1 + z2) = z1z1 + z2z2 + z1z2 + z2z1 2 2 = |z1| + |z2| + 2Re(z1z2) . By observing that Re(z1z2) ≤|z1z2|, we have 2 2 2 |z1 + z2| ≤ |z1| + |z2| + 2|z1z2| 2 2 2 = |z1| + |z2| + 2|z1||z2| = (|z1| + |z2|) . Since moduli are non-negative, we take the positive square root on both sides and obtain |z1 + z2|≤|z1| + |z2| . 14 To prove the other half of the triangle inequalities, we write |z1| = |(z1 + z2) + (−z2)|≤|z1 + z2| + |− z2| giving |z1|−|z2|≤|z1 + z2| . By interchanging z1 and z2 in the above inequality, we have |z2|−|z1|≤|z1 + z2| . Combining all results together |z1|−|z2| ≤|z1 + z2|≤|z1| + |z2|. 15 Example Find an upper bound for |z5 − 4| if |z|≤ 1. Solution Applying the triangle inequality, we get |z5 − 4|≤|z5| +4= |z|5 + 4 ≤ 1+4=5, since |z|≤ 1. Hence, if |z|≤ 1, an upper bound for |z5 − 4| is 5. In general, the triangle inequality is considered as a crude inequality, which means that it will not yield least upper bound estimate. Can we find a number smaller than 5 that is also an upper bound, or is 5 the least upper bound? Yes, we may use the Maximum Modulus Theorem (to be discussed later). The upper bound of the modulus will correspond to a com- plex number that lies on the boundary, where |z| = 1. 16 Example For a non-zero z and −π < Arg z ≤ π, show that |z − 1|≤ |z|− 1 +|z||Arg z|. Solution |z − 1| ≤ z −|z| + (|z|− 1) ≤ z −|z| + |z|− 1 = | z|| cos θ + i sin θ − 1| + |z|− 1 , θ = Arg z 2 2 = |z| (cos θ − 1) + sin θ + |z|− 1 q θ = |z| 2 sin + |z|− 1 ≤|z||Arg z| + |z|− 1 , 2 θ 2 θ θ θ since (cos θ−1)2+sin 2 θ = 1 − 2 sin2 − 1 + 4 sin 2 cos 2 = 4 sin2 2 2 2 2 θ θ and sin ≤ . 2 2 17 Take |z| > 1 as an illustration Geometric interpretation: Consider the triangle whose vertices are 1,z and |z|, by the Triangle Inequality, we have |z − 1|≤ |z|− 1 + z −|z| . The chord joining |z| and z is always shorter than the circular arc joining the same two points.