AMA4101 Complex Numbers PDF

SMA 305: COMPLEX ANALYSIS AMA4101: BackgroundCOMPLEX InformationNUMBERS AMA4101: COMPLEX NUMBERS Complex Numbers In the general quadratic equation ax 2 + bx + c = 0 -b ± b 2 - 4ac x= 2a When b - 4ac < 0 we cannot get the root of the negative numbers direc...

SMA 305: COMPLEX ANALYSIS AMA4101: BackgroundCOMPLEX InformationNUMBERS AMA4101: COMPLEX NUMBERS Complex Numbers In the general quadratic equation ax 2 + bx + c = 0 -b ± b 2 - 4ac x= 2a When b - 4ac < 0 we cannot get the root of the negative numbers directly. 2 We thus define i 2 = -1 Þ i = -1. This definition gives a set of new numbers called Complex numbers. A complex number can be defined as z = a + bi where a is the real part and b is the imaginary part. If z = a + bi , the conjugate of z is written as z = a - bi. z z = ( a - bi )( a - bi ) = a 2 - abi + abi - b 2i 2 = a2 + b2 Addition of Complex Numbers If z1 = a + bi and z2 = c + di , z1 + z2 = ( a + bi ) + ( c + di ) a + c + ( b + d ) i real part imaginary part Subtraction of Complex numbers z1 + z2 = ( a + bi ) - ( c + di ) = ( a + bi ) - c - di ) = ( a - c )+ ( b + d ) i real part imaginary part Multiplication of Complex numbers z1 ´ z2 = ( a + bi )( c + di ) = ac + adi + bci + bdi 2 But i 2 = -1 \ z1 ´ z 2 = ac + adi + bci - bd = ( ac - bd ) + ( ad + bc ) i Division of Complex numbers z1 a + bi = z2 c + di We can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator. z (a + bi) ´ ( c - di ) i.e 1 = z2 (c + di ) ( c - di ) ac - adi + bci - bdi 2 = 2 c - cdi + dci - d 2i 2 ac - adi + bci + bd = c2 + d 2 1  (ac + bd ) + (bc - ad )i = c2 + d 2 ac + bd æ bc - ad ö = 2 +ç ÷i c + d 2 è c2 + d 2 ø Example 1: z1 If z1 = 3 + i, z2 = 5 - i find, (a) z1 + z2 (b) z1 - z2 (c) z1 z2 (d) z2 Solution: (a) z1 + z2 = ( 3 + i ) + ( 5 - i ) = ( 3 + 5 ) + ( i - i ) = 8 (b) z1 - z2 = ( 3 + i ) - ( 5 - i ) = 3 + i - 5 + i = -2 + 2i (c) z1 z2 = ( 3 + i )( 5 - i ) = ( 3 )( 5 ) + 3 ( -i ) + i ( 5 ) - i 2 = 15 - 3i + 5i + 1 = 16 + 2i z 3 + i ( 5 + i ) 15 + 3i + 5i + i 2 (d) 1 = ´ = z2 5 - i ( 5 + i ) 52 + 11 15 + 8i - 1 = 26 14 + 8i 7 + 4i = = 26 13 Argand diagram A complex number a + bi can be considered as an ordered pair of real numbers. We can represent such numbers by points in an x,y-plane called the complex plane or argand diagram. Example 2: Let z = 3 + 4i.Represent z in the complex plane. Solution: y 4 z=(3,4) Imaginary 3 axis 2 1 1 2 3 x Real axis 2 Modulus-argument form Let z = a + bi ,we can represent z in the following diagram. b z(a,b) q a x b b tan q = Þ q = tan -1 a a Now, z = a + bi Þ z = a 2 + b 2 z is called the modulus of the complex number and q is called the argument. If we let r = z = a 2 + b 2 , then r 2 = a 2 + b 2 - (1) But from the diagram above, b = a tan q Þ b 2 = a 2 tan 2 q. Substituting this in 1, we get r 2 = a 2 + a 2 tan 2 q ( = a 2 1 + tan 2 q ) r = a sec q 2 2 2 a2 r2 = cos 2 q Þ a 2 = r 2 cos2 q Þ a = r cos q Similarly, b = r sin q We can replace a and b in z = a + bi by a = r cos q , b = r sin q to get z = r cos q + ir sin q - = r ( cosq + i sin q ) -modulus-argument form But cosq + i sin q = eiq \ z = reiq If r = 1, z = eiq Example 3: Express the following complex numbers in polar form. 3 (a) z = -5 + 5i (b) z = - 6 - 2i Solution: (a) z = -5 + 5i; a = -5, b = 5 b 5 tan q = = = -1; tan -1 1 = 450 a -5 Tangent is negative and q is in 2nd quadrant \q = 1350 iy 5 1350 q x -5 ( -5 ) 2 z =r= + 52 = 50 = 5 2 \ z = r ( cosq + i sin q ) ( = 5 2 cos1350 + i sin1350 ) - 2 (b) z = -6 - 2i i q = tan -1 = 300 - 6 iy - 6 q x - 2 In the 3rd quadrant, q = (180 + 30 ) = 210 0 0 (- 6 ) + (- 2 ) 2 2 r= z = = 8=2 2 ( \ z = 2 2 cos 2100 + i sin 2100 ) 4 Exercise 1 z 1. Find 1 , where z1 = 2 + i and z2 = 3 - 2i. z2 De moivre’s theorem If z1 = a1 + ib1 = r1 ( cos q1 + i sin q1 ) and z2 = a2 + ib2 = r2 ( cos q 2 + i sin q 2 ) then z1 z2 = r1 ( cos q1 + i sin q1 ) r2 ( cos q 2 + i sin q 2 ) = r1r2 [ cos q1 cos q 2 + i sin q 2 cos q1 + i sin q1 cos q 2 - sin q1 sin q 2 ] = r1r2 [ cos q1 cos q 2 - sin q1 sin q 2 + i (sin q 2 cos q1 + sin q1 cos q 2 ] But cosq1 cos q 2 - sin q1 sin q 2 = cos (q1 + q 2 ) and sin q 2 cos q1 + sin q1 cos q 2 = sin (q1 + q 2 ) \ z1 z2 = r1r2 éë cos (q1 + q 2 ) + i sin (q1 + q 2 ) ùû …(1) A generalization of (1) leads to z1 z2 K zn = r1r2 K rn {cos (q1 + q 2 + Kq n ) + i sin (q1 + q 2 + Kq n )} …(2) If z1 = z 2 = K zn = z , equation (2) becomes zz K zn = rr K rn {cos (q + q + Kq n ) + i sin (q + q + Kq n )} i.e z n = r n {cos nq + i sin nq } …(3) When r = 1, z n = cos nq + i sin nq Equation (3) is called De-Moivre’s theorem. Example 4 If z = 1 + 3i , find z 5 Solution iy ( 3) 2 z = r = 12 + = 4=2 q 3 3 1 z q = tan -1 = 600 1 \ z = 2 ( cos 600 + i sin 600 ) ( ) ( z 5 = 25 éë cos 5 ´ 600 + i sin 5 ´ 600 ùû ) = 3 2 ( cos300 0 + i sin 3000 ) Example 5 Prove the identities , cos 5q = 16cos5 q - 20cos3 q + 5cos q Solution z n = ( cos q + i sin q ) = cos nq + i sin nq n When r=1 When n=5 z 5 = cos 5q + i sin 5q Þ ( cos q + i sin q ) = cos 5q + i sin 5q 5 5 Expanding the L.H.S ,we get ( cos q + i sin q ) = cos 5 q + 5(i sin q ) cos 4 q + 10 ( i sin q ) cos3 q + 10 ( i sin q ) cos 2 q + 5 ( i sin q ) cos q + ( i sin q ) 5 2 3 4 5 (Using Binomial expansion/ Pascal’s triangle) = cos5 q + 5i sin q cos4 q - 10sin 2 q cos3 q - 10i sin 3 q cos 2 q + 5sin 4 q cos q + i sin 5 q Separating the real and imaginary parts, we get ( cos q + i sin q ) 5 { = cos5 q - 10sin 2 cos3 q + 5sin 4 q cos q + i 5sin q cos 4 q - 10sin 3 q cos2 q + sin 5 q } { \ cos 5q + i sin 5q = cos5 q - 10sin 2 q cos3 q + 5sin 4 q cos q + i 5sin q cos4 q - 10sin 3 q cos 2 q + sin 5 q } Equating the real and imaginary parts, we get cos 5q = cos5 q - 10sin 2 q cos3 q + 5sin 4 q cos q ( ) ( ) 2 = cos5 q - 10 1 - cos 2 q cos 3 q + 5 cos q 1 - cos 2 q = cos5 q - 10 cos 3 q + 10 cos 5 q + 5 cos q [1 - 2 cos 2 q + cos 4 q ) = cos5 - 10 cos3 q + 10cos5 q + 5cosq - 10cos3 q + 5cos5 q = 16 cos5 q - 20cos3 q + cos q Exercise 2 1. Prove the identities (a) sin 3q = 3sin q - 4sin 3 q (b) cos 4q = 8sin 4 q - 8sin 2 q + 1 (c) cos 3q = 4cos3 q - 3cos q (d) sin 5q = 16 cos4 q - 12cos 2 q + 1 Example 6 Solution (question (b) in exercise 2) (b) cos 4q = 8sin 4 q - 8sin 2 q + 1 Using Demoivre’s theorem when n=4, ( cos q + i sin q ) = cos 4q + i sin 4q 4 6 Expanding the L.H.S, we get ( cos q + i sin q ) = cos 4 q + 4i cos 3 q sin q + 6 cos 2 q ( i sin q ) + 4 cos q ( i sin q ) + ( i sin q ) 4 2 3 4 = cos4 q + 4i cos3 sin q - 6cos2 q sin 2 q - 4i cos q sin 3 q + sin 4 q = cos 4 q - 6 cos 2 q sin 2 q + sin 4 q + i ( 4 cos3 q sin q - 4 cos q sin 3 q ) Equating the real parts, we get cos 4q = cos 4 q - 6 cos2 q sin 2 q + sin 4 q ( ) - 6 cos q sin q + sin q 2 = cos 2 q 2 2 4 = (1 - sin q ) - 6 (1 - sin q ) sin q + sin 2 2 2 2 4 q = 1 - 2sin 2 q + sin 4 q - 6sin 2 q + 6sin 4 q + sin 4 q cos 4q = 8sin 4 q - 8sin 2 q + 1 Roots of Complex numbers 1 A number w is called an n root of a complex number z if w = z and we write w = z. From De th n n Moivre’s theorem, we can show that if n is a positive integer, 1 1 z n = {r ( cos q + i sin q )} n 1 ì æ q + 2p k ö æ q + 2p k ö ü = r n ícos ç ÷ + i sin ç ý î è n ø è n ÷ø þ k = 0, +1, ±2L 1 It follows that there are n different values of z n , i.e n different nth roots of z ,provided z ¹ 0. Example 7 Find all the values of z for which z 5 = -32 and locate these values in the complex plane. Solution: z 5 = -32 = -32 + 0i iy æ 0 ö 0 = tan -1 ç ÷=0 q x è -32 ø Acute angle is 0 (-32,0) q is in 2nd quadrant, thus q = 1800 ( -32 ) 2 r= + 02 = 32 z 5 = 32 éë cos (180 + 2p k ) + i sin (180 + 2p k ) ùû Taking the 5th root of both sides, we have 1 { z = 32 éë cos (180 + 2p k ) + i sin (180 + 2p k ) ùû } 5 7  1 ì æ 180 + 2p k ö æ 180 + 2p k ö ü = 32 ícos ç 5 ÷ + i sin ç ÷ý î è 5 ø è 5 øþ = 2k = 0,1, 2, 3, 4 When k=0, z1 = 2 ( cos 360 + i sin 360 ) æ 3p 3p ö When k=1, z2 = 2 ç cos + i sin ÷ è 5 5 ø When k=2, z3 = 2 ( cos 5p + i sin 5p ) æ 7p 7p ö When k=3, z4 = 2 ç cos + i sin ÷ è 5 5 ø æ 9p 9p ö When k=4, z5 = 2 ç cos + i sin ÷ è 5 5 ø Note: For k=5,6,…as well as -1,-2,… repetitions of the above values are obtained. Hence these are the only roots/ solutions of the given equation. x The five roots are called the fifth roots of -32. Example 8 Find the third roots of -1+i Solution Let z=-1+i iy ( -1) + (1) 2 2 z =r= = 2 From the diagram, q in the 2nd quadrant. 1 q 1350 x 1 -1 tan -1 q = = -1 -1 acute angle is 450 \q = 1800 - 450 = 1350 8  ( \ z = 2 éë cos 1350 + 2p k + i sin 1350 + 2p k ùû) ( ) { )} 1 1 z = 3 ( 0 ) 2 éë cos 135 + 2p k + i sin 135 + 2p k ùû ( 0 3 é 1 = ( 2 ) ê cos 6 (135 0 + 2p k ) + i sin (135 0 + 2p k ) ù ú êë 3 3 úû K=0,1,2 1 When k=0, z1 = 2 6 ( cos 450 + i sin 450 ) æ 11 ö 1 11 When k=1, z2 = 2 6 ç cos p + i sin p ÷ è 12 12 ø æ 19 ö 1 19 When k=2, z3 = 2 6 ç cos p + i sin p ÷ è 12 12 ø Example 9 Find the fourth root of z = -2 3 - 2i Solution: ( -2 ) ( ) 2 z =r= + -2 3 = 4 + 12 = 16 = 4 2 -2 q = tan -1 -2 3 From the diagram, q is in the 3rd quadrant iy -2 1 tan q = = 2100 -2 3 3 p Acute angle is 300 = -2 3 -2 q x 6 q = 300 + 1800 = 2100 = ( \ z = 4 cos 210 0 + i sin 210 0 ) 1 1 é ì ï z = ê 4 ícos ( 2100 + 2p k ) 2100 + 2p k ( ) üïù 4 4 + i sin ýú ê ï 4 4 ïþúû ë î =for k=0,1,2,3 æ 7p 7p ö When k=2, z1 = 2 ç cos + i sin ÷ è 24 24 ø æ 19p 19p ö When k=1, z2 = 2 ç cos + i sin ÷ è 24 24 ø 9  æ 31p 31p ö When k=2, z3 = 2 ç cos + i sin ÷ è 24 24 ø æ 43p 43p ö When k=3, z4 = 2 ç cos + i sin ÷ è 24 24 ø Example 10 Find the square roots of -15 - 8i Solution Let z = -15 - 8i ( -8 ) + ( -15) 2 2 z =r= = 64 + 225 = 17 -8 q = tan -1 ; From the diagram below, q is in the 3rd quadrant -15 Acute angle is 280 \q = 1800 + 280 = 2080 2080 -8 x q -15 \q = 17 {cos (108 + 2p k ) + i sin ( 208 + 2p k )} 1 1 z = {17{cos ( 208 + 2p k ) + i sin ( 208 + 2p k )} 2 2 æ æ 208 + 2p k ö ( 208 + 2k ) ö = 17 ç cos ç ÷ + i sin ÷ è è 2 ø 2 ø For k=0,1 When k=0, z1 = 17 ( cos104 0 + i sin104 0 ) When k=1, z2 = 17 ( cos(104 + p ) + i sin(104 + p ) Exercise 3 1.Express in polar form (a) 2 - 2i (b) -1 + 3i (c) -i 3 3 (d) - i 2 2 10 2. Solve the equation 24 + 8i = 0 3. Solve the equation 26 + 1 = 3i 4. Find the square roots of (a) 5 - 12i (b) 8 + 4 5i 5. Find the cube roots of -11 - 2i Expansion of cos n q ,sin n q , cos nq and sin nq Let z = cos q + i sin q z - m = ( cos q + i sin q ) -m 1 = ( cos q + i sin q ) m 1 = (De Moivre’s) cos mq + i sin mq = ( cos mq + i sin mq ) -1 = cos ( -mq ) + i sin ( -mq ) = cos mq - i sin mq When n=1, z = cos q + i sin q 1 z -1 = = cos q - i sin q z 1 z + = cos q + i sin q + cos q - i sin q z 1 z + = 2cos q z 1 Similarly, z - = cos q + i sin q - cos q + i sin q z 1 z - = 2i sin q z In the same way, z n = ( cos q + i sin q ) = cos nq + i sin nq and n 1 = ( cos q + i sin q ) = cos nq - i sin q -n n z Thus z n = ( cos q + i sin q ) = cos nq + i sin nq n 1 zn + = 2cos nq zn 1 zn - = cos nq + i sin nq - cos nq + i sin nq zn = 2i sin nq Example 11 Express cos6 q in multiple angles. Solution: 11  1 2cos q = z + z 6 2 3 4 5 6 æ 1ö æ1ö æ1ö æ1ö æ1ö æ1ö æ1ö \ ( 2 cos q ) = ç z + ÷ = z 6 + 6 z 5 ç ÷ + 15 z 4 ç ÷ + 20 z 3 ç ÷ + 15 z 2 ç ÷ + 6 z ç ÷ + ç ÷ 6 è zø èzø èzø èzø èzø èzø èzø æ 1ö æ 1 ö æ 1 ö = ç z 6 + 6 ÷ + 6 ç z 4 + 4 ÷ + 15 ç z 2 + 2 ÷ + 20 z è z ø è z ø è z ø 1 But z 6 + 6 = 2cos 6q z 1 z 4 + 4 = 2cos 4q z 1 z 2 + 2 = 2cos 2q z 6 æ 1ö \ ( 2 cos q ) = ç z + ÷ = 2 cos 6q + 12 cos 4q + 30 cos 2q + 20 6 è zø 1 cos6 q = {2cos 6q + 12cos 4q + 30cos 2q + 20} 32 Example 12 Express sin 5 q in multiple angles Solution 5 æ 1ö ( 2i sin q ) 5 =çz- ÷ è zø æ -1 ö æ -1 ö æ -1 ö æ -1 ö æ -1 ö æ -1 ö 0 2 3 4 5 = z ç ÷ + 5 z 4 ç ÷ + 10 z 3 ç ÷ + 10 z 2 ç ÷ + 5 z ç ÷ + ç ÷ 5 è z ø è z ø è z ø è 2 ø è z ø è z ø æ1ö æ 1 ö 1 5 = z 5 - 5 z 3 + 10 z - 10 ç ÷ + 5 ç 3 ÷ - èzø èz ø z 1 æ 1ö æ 1ö = z5 - - 5 ç z 3 - 3 ÷ + 10 ç z - ÷ z 5 è z ø è zø = 2i sin 5q - 5 ( 2i sin 3q ) + 10 ( 2i sin q ) 32i sin 5 q = 2i sin 5q - 10i sin 3q + 20i sin q 1 sin 5 q = {sin 5q - 5sin 3q + 10sin q } 16 12 Exercise 4 Express the following in multiple angles. 1. cos 4 q 2. sin 3 q 3. sin 5 q Absolute Value The absolute value or modulus of a complex number a + bi is defined as z = x + iy = x 2 + y 2 e.g z = -4 + 2i ( -4 ) 2 z = + 22 = 20 = 2 5 If z1 , z 2 , z3 ,K , zm are complex numbers, the following properties hold (1) z1 z2 = z1 z2 or z1 z2 K zm = z1 z2 K zm z1 z (2) = 1 ,if z2 ¹ 0 z2 z2 (3) z1 + z2 £ z1 + z2 (4) z1 + z2 ³ z1 - z2 Proof (1) z1 z2 = z1 z2 Let z1 = x1 + iy1 , z2 = x2 + iy2 z1 z 2 = ( x1 + iy1 )( x2 + iy2 ) = x1 x2 + ix1 y2 + iy1 x2 - y1 y2 = ( x1 x2 - y1 y2 ) + i ( x1 y2 + y1 x2 ) ( x1 x2 - y1 y2 ) + ( x1 y2 + y1 x2 ) 2 2 = ( x1 x2 ) - 2 x1 x2 y1 y2 + ( y1 y2 ) + ( x1 y2 ) + 2 x1 x2 y1 y2 + ( y1 x2 ) 2 2 2 2 = = x12 x2 2 - 2 x1 x2 y1 y2 + y12 y2 2 + x12 y2 2 + 2 x1 y2 y1 x2 + y12 x2 2 13 = x12 x22 + x12 y22 + y12 y2 2 + x22 y12 ( ) = x12 x2 2 + y2 2 + y12 y2 2 + x22 ( ) = (x1 2 + y12 )( x 2 2 + y2 2 ) = x12 + y12 x22 + y2 2 = z1 z2 3. z1 + z2 £ z1 + z2 14