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# Complex numbers

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### Complex numbers

1. 1. Complex Numbers December 6, 20051 Introduction of Complex Numbers1.1 DeﬁnitionA complex number z is a pair of real numbers z = (a, b) (1)Deﬁnition of addition: (a, b) + (c, d) = (a + b, c + d) (2)Deﬁnition of multiplication: (a, b) · (c, d) = (ac − bd, ad + bc) (3)Deﬁnition of multiplication by a scalar (a, b) · c = (ac, bc) (4)Note that multiplying by a scalar c yeilds the same result as muliplying by thecomplex number (c, 0).2 Notation2.1 The zeroAdding (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. 2. 2.2 The unityMultiplying a complex number by (1, 0) leaves the former unchanged, so (1, 0)is the "1" of multiplication. ³ ´ The multiplicative inverse of (a, b) is a2 +b2 , a2−b 2 since (try it!) a +b µ ¶ a −b (a, b) · , = (1, 0) (6) a2 + b2 a2 + b22.3 The complex conjugateThe complex conjugate of z = (a, b) denoted by z = (a, b), is deﬁned to be ¯(a, −b): z = (a, −b) A complex number times its complex conjugate has 0 for the second compo-nent ¡ ¢ (a, b) · (a − b) = a2 + b2 , 0 (7)2.4 Real and Imaginary partsTwo real valued functions, Re and Im, are deﬁned on the ﬁeld of complexnumbers Re (a, b) = a (8) Im (a, b) = b (9)2.5 Absolute valueAnother real valued function, the absolute value – denoted by |(a, b)|, is deﬁnedfor complex numbers: p |(a, b)| = a2 + b2 (10)This is the length of the vector (a, b) in the 2D sense.2.6 ArgumentThe 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 itfalls in (−π, π]. The function is denote by α = Arg (a, b).2.7 The i notationIn order to remember the complicated and seemingly arbitrary deﬁnition ofmultiplication, use the following pneumonic rule. Write (a, b) = a + ib (11) 2
3. 3. and, when multiplying, pretend that it is real addition and multiplication by iand that i2 = −1. Then (a, b) · (c, d) = (a + ib) (c + id) (12) = ac + i2 bd + i (ad + bc) (13) = ac − bd + i (ad + bc) (14) = (ac − bd, ad + bc) (15)so it works.3 Connection to trigonometry3.1 The basic relationshipRecall that sin (α + β) = sin α cos β + cos α sin β (16) cos (α + β) = cos α cos β − sin α sin β (17)Therefore, (cos α + i sin α) (cos β + i sin β) = cos (α + β) + i sin (α + β) (18)3.2 Euler’s notationDeﬁne eiα according to eiα = cos α + i sin α (19)Then eiα eiβ = ei(α+β) (20)so the exponentiation formally works. Everything that you might guess works actually works: ¡ iα ¢−1 e = e−iα (21) iα de = ieiα (22) dα3.3 A pretty relationThe equation eiπ = −1 (23)expresses a relationship among the most fundamental constant e, π, 1 and, ifyou believe that i is a number, i. 3
4. 4. 3.4 Roots of unityThe equation xN = 1 (24)always has exactly N roots in complex numbers. They can be expressed as 2πn x = ei N , 0 ≤ n < N. (25)The equation xN = eiα (26) αalso has exactly N roots. They are the same roots as 1 except "turned" by ei N α+2πn x = ei N (27)4 Polar form of the complex number4.1 DeﬁnitionA compex number z = a + ib can be rewritten as p µ ¶ a b z = a + ib = a2 + b2 √ + i√ (28) a2 + b2 a2 + b2 √Let r = a2 + b2 and α be the angle between −π and π such that a cos α = √ (29) a2 + b2 b sin α = √ (30) a2 + b2You recognize that r = |z| and α = Arg (z). Then z = reiα (31)This is the polar form of the complex number.4.2 MultiplicationLet z1 = r1 eiα1 and z2 = r2 eiα2 . Then z1 z2 = r1 eiα1 r2 eiα2 (32) = r1 r2 ei(α1 +α2 ) (33)Multiplying two complex numbers is equivalent to multiplying their absolutevalues and adding their agruments. If z = reiα , then 1/z = (1/r) e−iα . 4
5. 5. 5 Analytic functions5.1 IntroductionFor the sake of this writeup, a complex function f of z is called analytic if itis a "valid" expression purely in terms of z. By "valid" we mean (let a be acomplex constant) az n (34) eaz (35) ln z (36) cos z, sin z (37)and sums, products, and compozition thereof. "Invalid" are the following ex-pressions z ¯ (38) Arg z (39) |z| (40) Re z, Im z (41)Analytic functions will have a number of very many attractive and useful prop-erties.5.2 DeﬁnitionsLetting z = x + iy or, in polar form, z = reiα . Then zn = rn (cos nα + i sin nα) (42) n zn = (x + iy) (43) ez = ex (cos y + i sin y) (44) ln z = ln r + iα (45) eiz + e−iz cos z = (46) 2 eiz − e−iz sin z = (47) 2 cz = ez ln c (48)6 Cauchy-Riemann EquationsSuppose that z = x + iy and that f (z) is analytic and f (z) = u (x, y) + iv (x, y) (49) 5
6. 6. Then ∂u ∂v = (50) ∂x ∂y ∂u ∂u = − (51) ∂y ∂xIt follows (see Exercises) that both u and v are harmonic: uxx + uyy = 0 (52) vxx + vyy = 0 (53)7 Exercises1. Show that eln z = z and that ln ez = z. In other words, ez and ln z are, infact, the inverses of each other. 2. 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, ez , and ln z satisfy the Cauchy-Rieman equations. 4. Suppose that f1 (z) and f2 (z) are analytic according to the formal deﬁn-ition. Show that a. f1 (z) + f2 (z) is analytic (easy) b. cf1 (z) is analytic, where c = a + ib is a complex constant (easy) c. f1 (z) f2 (z) is analytic (more diﬃcult). Note that b. is a special case of c. d. f1 (f2 (z)) is analytic. You have now shown that any "valid" expression of z is an analytic function. ¡ ¢For example, cos ez + z 2 is analytic and its real an imaginary parts (bothexeedingly cumbersome) are harmonic. 5. Reduce to the form a + bi: π ei 2 (54) eiπ (55) eπ/6 (56) i5 (57) 5i (58) ii (59) i ii (60) 6. a. Derive an expression for arccos z. b. arccos i (61) arccos 1 (62) arccos 2 (63) 6