Complex Numbers. December 6, Introduction of Complex Numbers. (a, b) (c, d) =(ac bd, ad + bc) (3)

Transcription

1 Complex Numbers December 6, Introduction of Complex Numbers 1.1 Definition A complex number z is a pair of real numbers Definition of addition: Definition of multiplication: Definition of multiplication by a scalar z =(a, b) (1) (a, b)+(c, d) =(a + b, c + d) () (a, b) (c, d) =(ac bd, ad + bc) (3) (a, b) c =(ac, bc) (4) Note that multiplying by a scalar c yeilds the same result as muliplying by the complex number (c, 0). Notation.1 The zero Adding (0, 0) to a complex number leaves the latter unchanged, so (0, 0) is the "0" of addition. The additive inverse of z =(a, b) is ( a, b) since (a, b)+( a, b) =(0, 0) (5) 1

2 . The unity Multiplying a complex number by (1, 0) leaves the former unchanged, so (1, 0) is the "1" of multiplication. ³ a b The multiplicative inverse of (a, b) is a +b, a +b since (try it!).3 The complex conjugate µ a (a, b) a + b, b a + b =(1, 0) (6) The complex conjugate of z =(a, b) denoted by z = (a, b), isdefined to be (a, b): z =(a, b) A complex number times its complex conjugate has 0 for the second component (a, b) (a b) = a + b, 0 (7).4 Real and Imaginary parts Two real valued functions, Re and Im, are defined on the field of complex numbers.5 Absolute value Re (a, b) = a (8) Im (a, b) = b (9) Another real valued function, the absolute value denoted by (a, b), isdefined for complex numbers: (a, b) = p a + b (10) This is the length of the vector (a, b) in the D sense..6 Argument The argument of the comlex number also comes from the vector intepreation. It is the angle formed by the vector and the positive x-axis quoted so that it falls in ( π, π]. The function is denote by α =Arg(a, b)..7 The i notation In order to remember the complicated and seemingly arbitrary definition of multiplication, use the following pneumonic rule. Write (a, b) =a + ib (11)

3 and, when multiplying, pretend that it is real addition and multiplication by i and that i = 1. Then so it works. (a, b) (c, d) = (a + ib)(c + id) (1) = ac + i bd + i (ad + bc) (13) = ac bd + i (ad + bc) (14) = (ac bd, ad + bc) (15) 3 Connection to trigonometry 3.1 The basic relationship Recall that sin (α + β) = sinα cos β +cosαsin β (16) cos (α + β) = cosα cos β sin α sin β (17) Therefore, (cos α + i sin α)(cosβ + i sin β) =cos(α + β)+isin (α + β) (18) 3. Euler s notation Define e iα according to Then e iα =cosα + i sin α (19) e iα e iβ = e i(α+β) (0) so the exponentiation formally works. Everything that you might guess works actually works: e iα 1 de iα dα = e iα (1) = ie iα () 3.3 A pretty relation The equation e iπ = 1 (3) expresses a relationship among the most fundamental constant e, π, 1 and, if you believe that i is a number, i. 3

4 3.4 Roots of unity The equation x N =1 (4) always has exactly N roots in complex numbers. They can be expressed as x = e i πn N, 0 n<n. (5) The equation x N = e iα (6) also has exactly N roots. They are the same roots as 1 except "turned" by e i α N x = e i α+πn N (7) 4 Polar form of the complex number 4.1 Definition Acompexnumberz = a + ib can be rewritten as z = a + ib = p µ a + b a a + b + i b a + b (8) Let r = a + b and α be the angle between π and π such that cos α = a a + b (9) sin α = b a + b (30) You recognize that r = z and α =Arg(z). Then This is the polar form of the complex number. 4. Multiplication Let z 1 = r 1 e iα 1 and z = r e iα.then z = re iα (31) z 1 z = r 1 e iα 1 r e iα (3) = r 1 r e i(α 1+α ) (33) Multiplying two complex numbers is equivalent to multiplying their absolute values and adding their agruments. If z = re iα,then1/z =(1/r) e iα. 4

5 5 Analytic functions 5.1 Introduction For the sake of this writeup, a complex function f of z is called analytic if it isa"valid"expressionpurelyintermsofz. By "valid" we mean (let a be a complex constant) az n (34) e az (35) ln z (36) cos z, sin z (37) and sums, products, and compozition thereof. "Invalid" are the following expressions z (38) Arg z (39) z (40) Re z, Im z (41) Analytic functions will have a number of very many attractive and useful properties. 5. Definitions Letting z = x + iy or, in polar form, z = re iα.then z n = r n (cos nα + i sin nα) (4) z n = (x + iy) n (43) e z = e x (cos y + i sin y) (44) ln z = lnr + iα (45) cos z = eiz + e iz (46) sin z = eiz e iz (47) c z = e z ln c (48) 6 Cauchy-Riemann Equations Suppose that z = x + iy and that f (z) is analytic and f (z) =u (x, y)+iv (x, y) (49) 5

6 Then u x = v y (50) u = u y x (51) It follows (see Exercises) that both u and v are harmonic: u xx + u yy = 0 (5) v xx + v yy = 0 (53) 7 Exercises 1. Show that e ln z = z and that ln e z = z. In other words, e z and ln z are, in fact, the inverses of each other.. Show that if u (x, y) and v (x, y) satisfy the Cauchy-Riemann equations, then both u (x, y) and v (x, y) are harmonic. 3. Show that all basic analytic functions z, e z,andln z satisfy the Cauchy- Rieman equations. 4. Suppose that f 1 (z) and f (z) are analytic according to the formal definition. Show that a. f 1 (z)+f (z) is analytic (easy) b. cf 1 (z) is analytic, where c = a + ib is a complex constant (easy) c. f 1 (z) f (z) is analytic (more difficult). Note that b. is a special case of c. d. f 1 (f (z)) is analytic. You have now shown that any "valid" expression of z is an analytic function. For example, cos e z + z is analytic and its real an imaginary parts (both exeedingly cumbersome) are harmonic. 5. Reduce to the form a + bi: e i π (54) e iπ (55) e π/6 (56) i 5 (57) 5 i (58) i i (59) i ii (60) 6. a. Derive an expression for arccos z. b. arccos i (61) arccos 1 (6) arccos (63) 6