Example: marketing

1. CARTESIAN COMPLEX NUMBERS - Weebly

UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 1. CARTESIAN COMPLEX NUMBERS introduction Try to solve this quadratic equation : 0522=++xx By using quadratic formula : the discriminant , 16)5)(1(4)2(422 = = = acb the solution : )1(216)2( =x but it is not possible to evaluate 1 however if an operator j is defined as then the solution may be expressed as : 12 =j 21242)1(216)2(jjx = = = 21j+ and are known as COMPLEX NUMBERS . 21j Both solutions are of the form : = x1 2j COMPLEX number Real part Imaginarypart z = ajb this form is known as the CARTESIAN COMPLEX NUMBERS ( ALGEBRAIC FORM ) E2 - 1 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 EXAMPLES EXAMPLE 1 : Solve the quadratic equation , 042=+x 240422jxxx = ==+ EXAMPLE 2 : Solve the quadratic equation , 2 x 2 + 3 x + 5 = 0 2343463436344093)2(2)5)(2(4332jxjxxxx = = = = = E2 - 2 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 POWERS OF j j 0 0)1( 1 j 1 1 j j 2 )1)(1( -1 j 3 j2j = (-1)j -j j 4 j2j2 = (-1)(-1) 1 j 5 j4j = (1)j j In genera

4.1 INTRODUCTION Let a complex number Z = a + jb as shown in the Argand Diagram below. Let the distance OZ be r and the angle OZ makes with the positive real axis be θ. Imaginary axis Z r jb θ Real axis O a A From the trigonometry of right angle triangle :

Fullscreen Download

Tags:

  Introduction, Cartesian

Information

Domain:

Source:

Link to this page:

Please notify us if you found a problem with this document:

Other abuse

Advertisement

Transcription of 1. CARTESIAN COMPLEX NUMBERS - Weebly

1 UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 1. CARTESIAN COMPLEX NUMBERS introduction Try to solve this quadratic equation : 0522=++xx By using quadratic formula : the discriminant , 16)5)(1(4)2(422 = = = acb the solution : )1(216)2( =x but it is not possible to evaluate 1 however if an operator j is defined as then the solution may be expressed as : 12 =j 21242)1(216)2(jjx = = = 21j+ and are known as COMPLEX NUMBERS . 21j Both solutions are of the form : = x1 2j COMPLEX number Real part Imaginarypart z = ajb this form is known as the CARTESIAN COMPLEX NUMBERS ( ALGEBRAIC FORM ) E2 - 1 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 EXAMPLES EXAMPLE 1 : Solve the quadratic equation , 042=+x 240422jxxx = ==+ EXAMPLE 2 : Solve the quadratic equation , 2 x 2 + 3 x + 5 = 0 2343463436344093)2(2)5)(2(4332jxjxxxx = = = = = E2 - 2 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 POWERS OF j j 0 0)1( 1 j 1 1 j j 2 )1)(1( -1 j 3 j2j = (-1)j -j j 4 j2j2 = (-1)(-1) 1 j 5 j4j = (1)j j In general we can bring the power to the nearest multiplication of 4.

2 J 4 p + 0 = 1 j 4 p + 1 = j j 4 p + 2 = -1 j 4 p + 3 = - j where p Z DOMAIN The domain of the COMPLEX number is C where R is an element of C R C R C E2 - 3 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 THE ARGAND DIAGRAM A COMPLEX number may be represented graphically on rectangular or CARTESIAN axes . The horizontal ( or x ) axis is used to represent the real axis and the vertical ( or y ) axis is used to represent the imaginary axis . Such a diagram is called an ARGAND DIAGRAM . EXAMPLES : Represent Argand points A = 3 + j2 , B = -2 + j4 , C = -3 j3 , D = 2 j2 Real Axis Imaginary Axis D C B A 2 -2 -3 -3 -2 4 2 3 E2 - 4 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 2.

3 ADDITION AND SUBTRACTION - ALGEBRAIC FORM Two COMPLEX NUMBERS are added / subtracted by adding / subtracting separately the two real parts and two imaginary parts . Given two COMPLEX number Z = a + j b and W = c + j d IDENTITY If two COMPLEX NUMBERS are equal , then their real parts are equal and their imaginary parts are equal . Hence , two COMPLEX NUMBERS are identical , Z = W if : a = c and b = d EXAMPLE : Solve the COMPLEX equations ; (a) 36)(2jjyx =+ SOLUTION 3622jyjx =+ Therefore [ Re ] 32662===xxx [ Im ] 2332 = =yy E2 - 5 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 (b) ()()jbajj+= +3221 SOLUTION jbajjbajjbajj+= += ++ += +74)34()62()32)(21( Therefore : [ Re ] : a = 4 and [ Im ] : b = - 7 ADDITION & SUBTRACTION The sum of two COMPLEX number , Z + W )()()()(dbjcawzjdcjbawz+++=++++=+ EXAMPLE Given : and 32jz+=41jw = jwzjwz =+ ++ +=+1)]4(3[)]1(2[ The difference of two COMPLEX number , Z - W )()()()(dbjcawzjdcjbawz + = + += EXAMPLE Given.

4 And 32jz+=41jw = 73)]4(3[)]1(2[jwzjwz+= + = E2 - 6 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 The addition and subtraction of COMPLEX NUMBERS may be achieved graphically in the Argand diagram . Represent Example 1 and Example 2 in the Argand diagram . IMAGINARYAXIS3 2 -1-4w z Addition jwz =+1 REAL AXIS Subtraction 73jwz+= IMAGINARY AXIS 73zz w-4 -1 32 REAL AXIS E2 - 7 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 SCALAR MULTIPLICATION If Z = a + j b and k R , where k is a scalar ; then k Z , jkbkakzjbakkz+=+=)( EXAMPLE 3 : Given Z1 = 2 + j4 and Z2 = 3 - j Determine : (a) 4Z1 = 168)42(4jj+=+ (b) 5 Z2 = 515)3(5jj = E2 - 8 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 3.

5 MULTIPLICATION AND DIVISION - ALGEBRAIC FORM MULTIPLICATION Multiplication of COMPLEX NUMBERS is achieved by assuming all quantities involved are real and using j 2 = -1 to simplify : Given two COMPLEX NUMBERS : Z = a + jb and W = c + jd The product of two COMPLEX number , Z . W ))((jdcjbawz++= by using F O I L method )()(2bcadjbdacwzbdjbcjadacwzbdjjbcjadacw zjjbdjbcjadacwz++ = ++= +++= +++= Z . W = )()(bcadjbdac++ EXAMPLE : multiply the following COMPLEX number (a) ( 3 + j2 )( 4 - j5 ) = 1222)5)(2()4)(12()5)(3()4)(3(jjjjj = ++ + (b) ( -2 + 5j )( -5 + 2j ) = 29)5)(5()2)(5()2)(2()5)(2(jjjjjj =++ + E2 - 9 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 COMPLEX CONJUGATE The COMPLEX conjugate of a COMPLEX number is obtained by changing the sign of the imaginary part.

6 Hence the COMPLEX conjugate of : Z = a + j b is Z= a - j b W = c - j d is W= c + j d EXAMPLE : Let Z = 2 + j5 1. The COMPLEX conjugate of Z , is 52jz = 2. Calculate ZZ. : 2252+=zz= 49+ = 13 CONCLUSION : The product of the COMPLEX number and its conjugate ZZ. is always a real number. EXAMPLE : Let Q = 1 + j2 and R = 3 + j4 1. Calculate RQ+ Solution 64)43()21(jjjRQ+=+++=+ Therefore 64jRQ =+ or 644321jRQjRjQ =+ = = E2 - 10 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 2. Calculate QR Solution 105)64()83()43)(21(jQRjQRjjQR = + = = or 105105)64()83()43)(21(jQRjQRjQRjjQR =+ =++ =++= 3. Calculate 2Q Solution 43441)21(22222jQjjQjQ =+ = = or 4343441)21(222222jQjQjjQjQ =+ =++=+= From the previous examples , we can conclude that the : PROPERTIES OF the COMPLEX CONJUGATES ()nnzzwzzwwzwz= =+=+ E2 - 11 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 The geometric interpretation of the COMPLEX conjugate ( shown below ) Z is the reflection of Z in the real axis.

7 Im Zaj b=+Zajb= j -j O Re DIVISION Division of COMPLEX NUMBERS is achieved by multiplying both numerator and denominator by the COMPLEX conjugate of the denominator . Given two COMPLEX NUMBERS : Z = a + jb and W = c + jd The quotient of two COMPLEX number , ; Z / W 22)()())(())(()()(dcadbcjbdacwzjdcjdcjdc jbazjdcjbawzw+ ++= + +++== 2222)()(dcadbcjdcbdaczw+ +++= E2 - 12 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 EXAMPLE : evaluate the following 4352jj+ 25232514252314169)815()206()4()3()]4)(2( )3)(5[()]4)(5()3)(2[(22jwzjwzjwzjwz = =+ + =+ + += E2 - 13 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 4. THE TRIGONOMETRIC FORM AND THE POLAR FORM OF A COMPLEX NUMBER introduction Let a COMPLEX number Z = a + jb as shown in the Argand Diagram below.

8 Let the distance OZ be r and the angle OZ makes with the positive real axis be . Imaginary axis Z r jb Real axis O a A From the trigonometry of right angle triangle : a = r cos and b = r sin Hence : Z = a + jb = cosr + sinjr = )sin(cos jr+ TRIGONOMETRIC FORM AND POLAR FORM Z = r ( cos + j sin ) known as the TRIGONOMETRIC FORM is usually abbreviated to Z = [ r , ] or Z = r which is known as the POLAR FORM of a COMPLEX number . MODULUS / MAGNITUDE r is called the modulus or magnitude of Z and is written as mod Z or Z r is determined by using Pythagoras Theorem on triangle OAZ : mod Z = Z = r = ab22+ E2 - 14 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 ARGUMENT / AMPLITUDE is called the argument or amplitude of Z and is written as arg Z.

9 By trigonometry on triangle OAZ : arg Z = = arctan ab algebraic form Z = + j b a Z = r cos + j r sin trigonometric form Z = )sin(cos jr+ polar form Z = [], r or r EXAMPLE 1 : Determine the modulus and argument of the COMPLEX number Z = 2 + j3 and express Z (i) in trigonometric form and (ii) in polar form Solution Find r and , 13943222=+=+=r == (i) trigonometric form ) (cos13 + =jz (ii) Polar form = E2 - 15 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 EXAMPLE 2 Express the following COMPLEX NUMBERS in (i) in trigonometric form and (ii) in polar form Represent each COMPLEX NUMBERS on the Argand diagram SOLUTION Im 3 z j4 (i) 43jz+= +=+= abr Re Therefore; ) (cos5 + =jz ' ReIm -3 j4 (ii) 43jz+ = = ===+=+ = '5251694)3(12 r Therefore.

10 = = ) (cos5 + =jz E2 - 16 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 (iii) 43jz = ' Im -j4 -3 '525169)4()3(122= ===+= + = r Re Therefore the actual = + = ) (cos5 + =jz (iv) 43jz = '525169)4()3(122 = ===+= += r Therefore ; )) () (cos(5 + j=z ' Im Re3 z - j4 or = = )) () (cos(5 + j=z E2 - 17 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 EXAMPLE 3 : Convert the following COMPLEX NUMBERS into a + jb form , correct to 4 significant figures . (a) Z = 4 30 )30sin30(cos4jzjz+=+= (b) Z = 7 -145 )145sin145(cos7jzjz = + = E2 - 18 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 5. MULTIPLICATION AND DIVISION TRIGONOMETRIC / POLAR FORM TRIGONOMETRIC FORM Given two COMPLEX NUMBERS .

Show more