Chapter 1. Complex Numbers Weiqi Luo (骆伟祺) School of Software Sun Yat-Sen University Email:weiqi.luo@yahoo.com Office:# A313 Textbook: James Ward Brown, Ruel V. Churchill, Complex Variables and Applications (the 8th ed.), China Machine Press, 2008 Reference: 王忠仁 张静 《工程数学 - 复变函数与积分变换》高等教育出 版社,2006 2 School of Software Numbers System Natural Numbers Zero & Negative Numbers Integers Fraction Rational numbers Irrational numbers Real numbers Imaginary numbers Complex numbers … More advanced number systems Refer to: http://en.wikipedia.org/wiki/Number_system 3 School of Software Chapter 1: Complex Numbers Sums and Products; Basic Algebraic Properties Further Properties; Vectors and Moduli Complex Conjugates; Exponential Form Products and Powers in Exponential Form Arguments of Products and Quotients Roots of Complex Numbers Regions in the Complex Plane 4 School of Software 1. Sums and Products Definition Complex numbers can be defined as ordered pairs (x, y) of real numbers that are to be interpreted as points in the complex plane y Note: The set of complex numbers Includes the real numbers as a subset (x, y) (0, y) imaginary axis Real axis O (x, 0) x Complex plane 5 School of Software 1. Sums and Products Notation It is customary to denote a complex number (x,y) by z, x = Rez (Real part); y = Imz (Imaginary part) y z=(x, y) (0, y) z1=z2 iff 1. Rez1= Rez2 2. Imz1 = Imz2 O (x, 0) x 6 Q: z1<z2? School of Software 1. Sums and Products Two Basic Operations Sum (x1, y1) + (x2, y2) = (x1+x2, y1+y2) Product (x1, y1) (x2, y2) = (x1x2 - y1y2, y1x2+x1y2) 1. when y1=0, y2=0, the above operations reduce to the usual operations of addition and multiplication for real numbers. 2. Any complex number z= (x,y) can be written z = (x,0) + (0,y) 3. Let i be the pure imaginary number (0,1), then z = x (1, 0) + y (0,1) = x + i y, x & y are real numbers i2 =(0,1) (0,1) =(-1, 0) i2=-1 7 School of Software 1. Sums and Products Two Basic Operations (i2 -1) Sum (x1, y1) + (x2, y2) = (x1+x2, y1+y2) (x1 + iy1) + (x2+ iy2) = (x1+x2)+i(y1+y2) Product (x1, y1) (x2, y2) = (x1x2 - y1y2, y1x2+x1y2) (x1 + iy1) (x2+ iy2) = (x1x2+ x1 iy2) + (iy1x2 + i2 y1y2) = (x1x2+ x1 iy2) + (iy1x2 - y1y2) = (x1x2 - y1y2) +i(y1x2+x1y2) 8 School of Software 2. Basic Algebraic Properties Various properties of addition and multiplication of complex numbers are the same as for real numbers Commutative Laws z1+ z2= z2 +z1, z1z2=z2z1 Associative Laws (z1+ z2 )+ z3 = z1+ (z2+z3) (z1z2) z3 =z1 (z2z3) e.g. Prove that z1z2=z2z1 (x1, y1) (x2, y2) = (x1x2 - y1y2, y1x2+x1y2) = (x2x1 - y2y1, y2x1 +x2y1) = (x2, y2) (x1, y1) 9 School of Software 2. Basic Algebraic Properties For any complex number z(x,y) z + 0 = z; z ∙ 0 = 0; z∙1=z Additive Inverse -z = 0 – z = (-x, -y) (-x, -y) + (x, y) =(0,0)=0 Multiplicative Inverse when z ≠ 0 , there is a number z-1 (u,v) such that z z-1 =1 , then (x,y) (u,v) =(1,0) xu-yv=1, yu+xv=0 u x x y 2 2 ,v y x y 2 2 z 10 1 ( x x y 2 2 , y x y 2 2 ), z 0 School of Software Homework pp. 5 Ex. 1, Ex.4, Ex. 8, Ex. 9 11 School of Software 3. Further Properties If z1z2=0, then so is at least one of the factors z1 and z2 Proof: Suppose that z1 ≠ 0, then z1-1 exists z1-1 (z1z2)=z1-1 0 =0 z1-1 (z1z2)=( z1-1 z1) z2 =1 z2 = z2 Associative Laws Therefore we have z2=0 12 School of Software 3. Further Properties Other two operations: Subtraction and Division Subtraction: z1-z2=z1+(-z2) (x1, y1) - (x2, y2) = (x1, y1)+(-x2, -y2) = (x1 -x2, y1-y2) Division: z1 z2 z1 z2 ( x1 , y1 )( 1 z1 z 2 ( z 2 0) x2 2 x2 + y2 2 , y2 2 x2 + y2 )( 2 13 x1 x 2 y1 y 2 2 x2 + y2 2 , y1 x 2 x1 y 2 2 x2 + y2 2 School of Software ) 3. Further Properties An easy way to remember to computer z1/z2 z1 z2 ( x1 iy1 ) ( x 2 iy 2 ) ( x1 iy1 )( x 2 iy 2 ) ( x 2 iy 2 )( x 2 iy 2 ) commonly used ( x 2 iy 2 )( x 2 iy 2 ) x 2 y 2 R 2 Note that 2 For instance 4i 2 3i (4 i )(2 3i ) (2 3 i )(2 3 i ) 14 5 14 i 13 5 13 14 i 13 School of Software 3. Further Properties Binomial Formula n ( z1 z 2 ) n C k n k z1 z 2 nk , n 1, 2, ... k 0 Where C k n n! k !( n k ) ! , k 0,1, 2, ..., n 15 School of Software 3. Further Properties pp.8 Ex. 1. Ex. 2, Ex. 3, Ex. 6 16 School of Software 4. Vectors and Moduli Any complex number is associated a vector from the origin to the point (x, y) y y z1=(x1, y1) | z 1 | x1 2 y1 2 z1+z2 z2=(x2, y2) | z 2 || z1 | O z1 z2 O x Sum of two vectors The moduli or absolute value of z is a nonnegative real number | z | x y 2 x z1 z 2 ( x1 x 2 ) i ( y1 y 2 ) 2 Product: refer to pp.21 17 School of Software 4. Vectors and Moduli Example 1 The distance between two point z1(x1, y1) and z2(x2, y2) is |z1-z2|. Note: |z1 - z2 | is the length of the vector representing the number z1-z2 = z1 + (-z2) y |z1 - z2 | Therefore z2 z1 -z2 z 1 z 2 ( x1 x 2 ) i ( y1 y 2 ) z1 - z2 O | z 1 z 2 | x 18 ( x1 x 2 ) ( y 1 y 2 ) 2 School of Software 2 4. Vectors and Moduli Example 2 The equation |z-1+3i|=2 represents the circle whose center is z0 = (1, -3) and whose radius is R=2 y x O z0(1, -3) 19 Note: | z-1+3i | = | z-(1-3i) | =2 School of Software 4. Vectors and Moduli Some important inequations Since | R e Z | | Im Z | | Z | 2 2 2 we have R e Z | R e Z | | Z |; Im Z | Im Z | | Z | Triangle inequality y z1=(x, y) | z 1 | x2 y2 O x y | z1 z 2 | | z1 | | z 2 | z1+z2 z1 z2 O 20 School of Software x 4. Vectors and Moduli | z1 z 2 | || z1 | | z 2 || Proof: when |z1| ≥ |z2|, we write Triangle inequality | z 1 | | ( z 1 z 2 ) ( z 2 ) | | z1 z 2 | | ( z 2 ) | | z1 z 2 | | z 2 | | z1 z 2 | | z1 | | z 2 | || z1 | | z 2 || Similarly when |z2| ≥ |z1|, we write | z 2 | | ( z 1 z 2 ) ( z 1 ) | | z1 z 2 | | ( z1 ) | | z1 z 2 | | z 1 | | z1 z 2 | | z 2 | | z1 | || z1 | | z 2 || 21 School of Software 4. Vectors and Moduli | z1 z 2 | | z1 | | z 2 | | z1 z 2 | || z1 | | z 2 || || z1 | | z 2 || | z1 z 2 | | z1 | | z 2 | | z1 z 2 ... z n | | z1 | | z 2 | ... | z n | 22 School of Software 4. Vectors and Moduli Example 3 If a point z lies on the unit circle |z|=1 about the origin, then we have y | z 2 | || z | 2 | 1 z | z 2 | | z | 2 3 O 23 1 School of Software 2 x 4. Homework pp. 12 Ex. 2, Ex. 4, Ex. 5 24 School of Software 5. Complex Conjugates Complex Conjugate (conjugate) The complex conjugate or simply the conjugate, of a complex number z=x+iy is defined as the complex number x-iy and is denoted by z y Properties: z(x,y) O | z | | z | z z x z (x,-y) 25 School of Software 5. Complex Conjugates If z1=x1+iy1 and z2=x2+iy2 , then z1 z 2 ( x1 x 2 ) i ( y1 y 2 ) ( x1 iy1 ) ( x 2 iy 2 ) z 1 z 2 Similarly, we have z1 z 2 z1 z 2 z1 z 2 z1 z 2 z1 z2 z1 z2 , z2 0 26 School of Software 5. Complex Conjugates If z x iy , z x iy , then z z ( x iy ) ( x iy ) 2 x 2 R e z z z ( x iy ) ( x iy ) 2 yi 2 i Im z Re z zz , Im z zz 2 2i z z ( x iy ) ( x iy ) x y | Z | 2 2 27 2 School of Software 5. Complex Conjugates Example 1 1 3i 2i 1 3i 2i ? ( 1 3i )(2 i ) (2 i )(2 i ) 5 5i |2i| 2 5 5i 1 i 5 28 School of Software 5. Complex Conjugates a. | z1 | z2 proof :| | z1 | | z2 | z1 | 2 z2 z1 ( z2 z1 ) z2 | z1 z 2 | | z1 || z 2 | z1 z2 z1 z2 z1 z1 z2 z2 | z | | z | n | z1 | | z2 | 2 2 n Refer to pp. 14 Example 2 | z | 2 | z 3 z 2 z 1 | | z 3 | | 3 z 2 | | 2 z | | 1 | | z |3 3 | z | 2 2 | z | 1 2 5 3 2 29 School of Software 5. Homework pp. 14 – 16 Ex. 1, Ex. 2, Ex. 7, Ex. 14 30 School of Software 6. Exponential Form Polar Form Let r and θ be polar coordinates of the point (x,y) that corresponds to a nonzero complex number z=x+iy, since x=rcosθ and y=rsinθ, the number z can be written in polar form as z=r(cosθ + isinθ), where r>0 θ y y z(x,y) argz: the argument of z Argz: the principal value of argz r θ O arg z A rgZ 2 n , n 0, 1, 2, ... z(x,y) r θ x O Θ A rgZ 1 x 31 School of Software 6. Exponential Form Example 1 The complex number -1-i, which lies in the third quadrant has principal argument -3π/4. That is A rg ( 1 i ) 3 4 It must be emphasized that the principal argument must be in the region of (-π, +π ]. Therefore, A rg ( 1 i ) 5 4 However, arg( 1 i ) arg( 1 i ) 3 5 2 n , n 0, 1, 2, ... 4 2 n , n 0, 1, 2, ... argz = α + 2nπ Here: α can be any one of arguments of z 4 32 School of Software 6. Exponential Form The symbol eiθ , or exp(iθ) e 1 e 1 x x 1 1! 1 (2 n ) ! e 1 (2 n ) ! x n0 ( 1) n 1 ( i ) 1 (2 n ) ! cosθ n 1 ( ) x i[ ( 1) 1 Why? Refer to Sec. 29 n! x n n0 2 n 1 (2 n 1) ! n 1 x ... n n! 1 2n x ... (2 n 1) ! 1 3 1 n0 2n cos i sin 3! 2n n0 x 2 2! n0 i 1 i Let x=iθ, then we have ( i ) n 1 2 n 1 (i ) 2n (2 n 1) ! ( ) (2 n ) ! n0 1 1 2 n 1 2n i[ ( i ) n 1 1 2n2 (2 n 1) ! ] sinθ 33 School of Software 2 n 1 ] 6. Exponential Form Example 2 The number -1-i in Example 1 has exponential form 1 i 2 (cos( 3 ) i sin( 4 1 i 2 (cos( 3 4 ) i sin( 3 )) 2e 3 i( ) 4 4 3 )) 2e 3 i ( 2 n ) 4 , n 0, 1, 2, ... 4 34 School of Software 6. Exponential Form z=Reiθ where 0≤ θ ≤2 π y y Reiθ θ Reiθ z θ R O z0 O x x z=z0 +Reiθ |z-z0 |=R 35 School of Software 7. Products and Powers in Exponential Form Product in exponential form i e 1e i 2 (co s 1 i sin 1 )(co s 2 i sin 2 ) (co s 1 co s 2 sin 1 sin 2 ) i (sin 1 co s 2 co s 1 sin 2 ) co s( 1 2 ) i sin ( 1 2 ) e z1 r1 e i 1 & z 2 r2 e z1 z 2 ( r1 e i 1 i 2 )( r2 e i 2 ) r1 r2 e ( z1 ) ( r1 e n z1 z2 r1 e r2 e i 1 i 2 r1 r2 i (1 2 ) e i (1 2 ) i (1 2 ) i 1 ) r1 e n n in 1 , n 0, 1, 2, ... 1 , z2 0 z2 36 1e r2 e i0 i 2 1 e i 2 r2 School of Software , z2 0 7. Products and Powers in Exponential Form Example 1 In order to put only write ( 3 i ) (2 e 7 i / 6 ( 3 i) ) 2 e 7 7 7 in rectangular form, one need i7 /6 2 (cos 7 7 6 37 i sin 7 ) 64( 3 +i) 6 School of Software 7. Products and Powers in Exponential Form Example 2 de Moivre’s formula i ( e ) (cos i sin ) cos n i sin n , n 0, 1, 2, ... n n (cos i sin ) cos 2 i sin 2 2 (cos i sin ) cos sin i (2 sin cos ) 2 2 2 pp. 23, Exercise 10, 11 38 School of Software 8. Arguments of products and quotients Ifz1 r1 e i 1 & z 2 r2 e i 2 , then z1 z 2 ( r1 e i 1 )( r2 e i 2 ) r1 r2 e i (1 2 ) θ1 is one of arguments of z1 and θ2 is one of arguments of z2 then θ1 +θ2 is one of arguments of z1z2 arg(z1z2)= θ1 +θ2 +2nπ, n=0, ±1, ±2 … argz1z2= θ1 +θ2 +2(n1+n2)π = (θ1 +2n1π)+ (θ2 +2n2π) = argz1+argz2 Q: Argz1z2 = Argz1+Argz2? Here: n1 and n2 are two integers with n1+n2=n 39 School of Software 8. Arguments of products and quotients Example 1 ≠ When z1=-1 and z2=i, then Arg(z1z2)=Arg(-i) = -π/2 but Arg(z1)+Arg(z2)=π+π/2=3π/2 Note: Argz1z2=Argz1+Argz2 is not always true. 40 School of Software 8. Arguments of products and quotients Arguments of Quotients arg( z1 z2 1 1 ) arg( z1 z 2 ) arg( z1 ) arg( z 2 ) arg( z1 ) arg( z 2 ) 41 School of Software 8. Arguments of products and quotients Example 2 In order to find the principal argument Arg z when z 2 1 i 3 observe that arg z arg( 2) arg(1 since A rg ( 2) A rg (1 3i ) 3i ) 3 argz ( 3 ) 2 n 2 3 42 2 n Argz 2 3 School of Software 8. Homework pp. 22-24 Ex. 1, Ex. 6, Ex. 8, Ex. 10 43 School of Software 9. Roots of Complex Numbers Two equal complex numbers z1 r1 e i 1 z 2 r2 e i 2 At the same point z1 z 2 If and only if r1 r2 & 1 2 2 k for some integer k 44 School of Software 9. Roots of Complex Numbers Roots of Complex Number z 0 r0 e Given a complex number the number z, s.t. z n z 0 Let z re i then z n ( re i ) n i 0 r e n , we try to find all in r0 e i 0 thus we get r r0 & n 0 2 k , k 0, 1, 2, ... n r n r0 & 0 n 2k , k 0, 1, 2, ... n The unique positive nth root of r0 45 School of Software 9. Roots of Complex Numbers The nth roots of z0 are z n r0 exp[ i ( 0 2k n )], k 0, 1, 2, ... n Note: 1. All roots lie on the circle |z|; 2. There are exactly n distinct roots! ck n r0 exp[ i ( 0 n 2k )], k 0,1, 2, ..., n 1 n |z| 46 School of Software 9. Roots of Complex Numbers ck n ck n Let r0 exp[ i ( r0 exp( i 0 n 0 w n exp( i 2k )], k 0,1, 2, ..., n 1 n ) exp( i n 2k ), k 0,1, 2, ..., n 1 n 2 w n exp( i k then ) n Therefore c k c 0 w n , k 0,1, 2, ..., n 1 where c0 2k ) n k n r0 exp( i 0 ) exp( i n 2 0 n ) n r0 exp( i 0 ) n Note: the number c0 can be replaced by any particular nth root of z0 47 School of Software 10. Examples Example 1 Let us find all values of (-8i)1/3, or the three roots of the number -8i. One need only write 8 i 8 exp[ i ( 2 k )], k 0, 1, 2, ... 2 To see that the desired roots are c k 2 exp[ i ( 6 2k 2i )], k 0,1, 2 3 3i 48 3i School of Software 10. Examples Example 2 To determine the nth roots of unity, we start with 1 1 exp[ i (0 2 k )], k 0, 1, 2, ... 1 And find that n=3 1n n 1 exp[ i ( 0 n 2k n n=4 49 )] exp( i 2k ), k 0,1, 2, ..., n 1 n n=6 School of Software 10. Examples Example 3 the two values ck (k=0,1) of ( 3 i ) , which are the square roots of 3 i , are found by writing 1/ 2 3 i 2 exp[ i ( 2 k )], k 0, 1, 2, ... 6 ck c0 2 exp[ i ( k )], k 0,1 12 2 exp( i 12 ) 2 (cos i sin 12 c1 c 0 50 School of Software 12 ) 10. Homework pp. 29-31 Ex. 2, Ex. 4, Ex. 5, Ex. 7, Ex. 9 51 School of Software 11. Regions in the Complex Plane ε- neighborhood The ε- neighborhood | z z 0 | of a given point z0 in the complex plane as shown below | z z0 | y z | z z0 | y ε z0 O | z z 0 | z O x Neighborhood ε z0 0 | z z 0 | x Deleted neighborhood 52 School of Software 11. Regions in the Complex Plane Interior Point A point z0 is said to be an interior point of a set S whenever there is some neighborhood of z0 that contains only points of S Exterior Point A point z0 is said to be an exterior point of a set S when there exists a neighborhood of it containing no points of S; Boundary Point (neither interior nor exterior) A boundary point is a point all of whose neighborhoods contain at least one point in S and at least one point not in S. The totality of all boundary points is called the boundary of S. 53 School of Software 11. Regions in the Complex Plane Consider the set S={z| |z|≤1} All points z, where |z|>1 are Exterior points of S; y S={z| |z|≤1-{1,0}} z0 ? z0 O z0 All points z, where |z|<1 are Interior points of S; x All points z, where |z|=1 are Boundary points of S; 54 School of Software 11. Regions in the Complex Plane Open Set A set is open if it and only if each of its points is an interior point. Closed Set A set is closed if it contains all of its boundary points. Closure of a set The closure of a set S is the closed set consisting of all points in S together with the boundary of S. 55 School of Software 11. Regions in the Complex Plane Examples S={z| |z|<1} ? Open Set S={z| |z|≤1} ? Closed Set S={z| |z|≤1} – {(0,0)} ? Neither open nor closed S= all points in complex plane ? Both open and closed Key: identify those boundary points of a given set 56 School of Software 11. Regions in the Complex Plane Connected An open set S is connected if each pair of points z1 and z2 in it can be joined by a polygonal line, consisting of a finite number of line segments joined end to end, that lies entirely in S. y O x The set S={z| |z|<1 U |z-(2+i)|<1} is open However, it is not connected. The open set 1<|z|<2 is connected. 57 School of Software 11. Regions in the Complex Plane Domain A set S is called as a domain iff 1. S is open; 2. S is connected. e.g. any neighborhood is a domain. Region A domain together with some, none, or all of it boundary points is referred to as a region. 58 School of Software 11. Regions in the Complex Plane Bounded A set S is bounded if every point of S lies inside some circle |z|=R; Otherwise, it is unbounded. y e.g. S={z| |z|≤1} is bounded S O S={z| Rez≥0} is unbounded R x 59 School of Software 11. Regions in the Complex Plane Accumulation point A point z0 is said to be an accumulation point of a set S if each deleted neighborhood of z0 contains at least one point of S. If a set S is closed, then it contains each of its accumulation points. Why? A set is closed iff it contains all of its accumulation points e.g. the origin is the only accumulation point of the set Zn=i/n, n=1,2,… The relationships among the Interior, Exterior, Boundary and Accumulation Points! An Interior point must be an accumulation point. An Exterior point must not be an accumulation point. A Boundary point must be an accumulation point? 60 School of Software 11. Homework pp. 33 Ex. 1, Ex. 2, Ex. 5, Ex. 6, Ex.10 61 School of Software