JUST THE MATHS SLIDES NUMBER 6.3 COMPLEX NUMBERS 3 (The polar - PDF document

“JUST THE MATHS” SLIDES NUMBER 6.3 COMPLEX NUMBERS 3 (The polar & exponential forms) by A.J.Hobson 6.3.1 The polar form 6.3.2 The exponential form 6.3.3 Products and quotients in polar form

UNIT 6.3 - COMPLEX NUMBERS 3 THE POLAR AND EXPONENTIAL FORMS 6.3.1 THE POLAR FORM ✯ P( x, y ) y ✟ ✟✟✟✟✟✟✟✟✟✟✟ ✻ r θ ✲ x O From the diagram, x r = cos θ and y r = sin θ ; x = r cos θ, y = r sin θ ; x + jy = r (cos θ + j sin θ ) . 1

x + jy is called the “rectangular form” or “cartesian form” . r (cos θ + j sin θ ) ( r � θ for short) is called the “polar form” . θ may be positive, negative or zero and may be expressed in either degrees or radians. EXAMPLES √ 1. Express the complex number z = 3 + j in polar form. Solution √ | z | = r = 3 + 1 = 2 and Arg z = θ = tan − 1 1 3 = 30 ◦ + k 360 ◦ , √ where k may be any integer. Alternatively, using radians, Arg z = π 6 + k 2 π, where k may be any integer. Hence, in polar form, z = 2(cos[30 ◦ + k 360 ◦ ]+ j sin[30 ◦ + k 360 ◦ ]) = 2 � [30 ◦ + k 360 ◦ ] 2

Alternatively, z =  π  π  π          + j sin  = 2 �  . 2  cos 6 + k 2 π 6 + k 2 π 6 + k 2 π  2. Express the complex number z = − 1 − j in polar form. Solution √ √ | z | = r = 1 + 1 = 2 and Arg z = θ = tan − 1 (1) = − 135 ◦ + k 360 ◦ , where k may be any integer. Alternatively Arg z = − 3 π 4 + k 2 π, where k may be any integer. Hence, in polar form, √ 2(cos[ − 135 ◦ + k 360 ◦ ] + j sin[ − 135 ◦ + k 360 ◦ ]) z = √ 2 � [ − 135 ◦ + k 360 ◦ ] = or √  − 3 π  − 3 π        + j sin z = 2  cos 4 + k 2 π 4 + k 2 π         √  − 3 π    . = 2 � 4 + k 2 π   3

Note: If it is required that the polar form should contain only the principal value of the argument, θ , then, provided − 180 ◦ < θ ≤ 180 ◦ or − π < θ ≤ π , the component k 360 ◦ or k 2 π of the result is simply omitted. 6.3.2 THE EXPONENTIAL FORM It may be shown that 1! + z 2 2! + z 3 3! + z 4 e z = 1 + z 4! + ... sin z = z − z 3 3! + z 5 5! − z 7 7! + ... cos z = 1 − z 2 2! + z 4 4! − z 6 6! + ... These are the definitions of e z , sin z and cos z . In the equivalent series for sin x and cos x , the value x (real), must be in radians and not degrees . 4

Deductions 1! + ( jθ ) 2 + ( jθ ) 3 + ( jθ ) 4 e jθ = 1 + jθ + ... 2! 3! 4! But j 2 = − 1, so 1! − θ 2 2! − jθ 3 3! + θ 4 e jθ = 1 + j θ 4! + ... Hence, e jθ = cos θ + j sin θ provided θ is expressed in radians and not de- grees. The complex number x + jy , having modulus r and ar- gument θ + k 2 π may thus be expressed not only in polar form but also in the exponential form , re jθ . 5

ILLUSTRATIONS 1. √ 3 + j = 2 e j ( π 6 + k 2 π ) . 2. √ 2 e j ( 3 π 4 + k 2 π ) . − 1 + j = 3. √ 2 e − j ( 3 π 4 + k 2 π ) . − 1 − j = Note: If it is required that the exponential form should contain only the principal value of the argument, θ , then, pro- vided − π < θ ≤ π , the component k 2 π of the result is simply omitted. 6

6.3.3 PRODUCTS AND QUOTIENTS IN POLAR FORM Let z 1 = r 1 (cos θ 1 + j sin θ 1 ) = r 1 � θ 1 and z 2 = r 2 (cos θ 2 + j sin θ 2 ) = r 2 � θ 2 . (a) The Product z 1 .z 2 = r 1 .r 2 (cos θ 1 + j sin θ 1 ) . (cos θ 2 + j sin θ 2 ) That is, z 1 .z 2 = r 1 .r 2 ([cos θ 1 . cos θ 2 − sin θ 1 . sin θ 2 ] + j [sin θ 1 . cos θ 2 + cos θ 1 . sin θ 2 ]) . z 1 .z 2 = r 1 .r 2 (cos[ θ 1 + θ 2 ]+ j sin[ θ 1 + θ 2 ]) = r 1 .r 2 � [ θ 1 + θ 2 ] . To determine the product of two complex num- bers in polar form, we construct the product of their modulus values and the sum of their argument values. 7

(b) The Quotient (cos θ 1 + j sin θ 1 ) z 1 = r 1 (cos θ 2 + j sin θ 2 ) . z 2 r 2 Multiplying the numerator and denominator by cos θ 2 − j sin θ 2 , z 1 = r 1 ([cos θ 1 . cos θ 2 + sin θ 1 . sin θ 2 ] z 2 r 2 + j [sin θ 1 . cos θ 2 − cos θ 1 . sin θ 2 ]) . z 1 = r 1 (cos[ θ 1 − θ 2 ] + j sin[ θ 1 − θ 2 ]) = r 1 � [ θ 1 − θ 2 ] . z 2 r 2 r 2 To determine the quotient of two complex num- bers in polar form, we construct the quotient of their modulus values and the difference of their argument values. 8

ILLUSTRATIONS 1. √ √ 3 + j ) . ( − 1 − j ) = 2 � 30 ◦ . 2 � ( − 135 ◦ ) ( √ 2 � ( − 105 ◦ ) . = 2 For all of the complex numbers in this example, includ- ing the result, the argument appears as the principal value. 2. √ √ 2 � 30 ◦ 3 + j 2 � 165 ◦ . √ − 1 − j = 2 � ( − 135 ◦ ) = For all of the complex numbers in this example, includ- ing the result, the argument appears as the principal value. 3. √ √ 2 � ( − 135 ◦ ) . 2 � ( − 150 ◦ ) ( − 1 − j ) . ( − 3 − j ) = √ 2 � ( − 285 ◦ ) . = 2 √ 2 � (75 ◦ ) if the This must be converted to 2 principal value of the argument is required. 9