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Unformatted text preview: 9 FUNCTIONS OF A COMPLEX VARIABLE 84 9 Functions of a Complex Variable 9.1 Complex Numbers The equation x 2 + 1 = 0 is not satisfied by any real number. For this reason mathematicians have extended the set of real numbers by introducing a new quantity, written j which does satisfy this equation The defining property of j is that : j 2 + 1 = 0 . That is j 2 = 1 . Given this new quantity we can develop the algebra of quantities involving j by applying all of the usual rules of algebra with the single additional rule j 2 = 1. We call any number of the form x + jy where x and y are real numbers a complex number. We call x the real part of the complex number and y the imaginary part of the complex number; and if z = x + jy we write Re( z )= x and Im( z )= y . (Also sometimes written ( z ) = x and ( z ) = y ). Here are examples of complex numbers with their real and imaginary parts: z 1 = 2 + 3 j , Re( z 1 ) = 2, Im( z 1 ) = 3 . z 2 = 5 . 324 j , Re( z 2 ) = 5 . 324, Im( z 2 ) = 1 . z 3 = √ 3 + 4 7 j , Re( z 3 ) = √ 3, Im( z 3 ) = 4 7 . z 4 = 8 e + πj , Re( z 4 ) = 8 e , Im( z 4 ) = π. z 5 = 457 . 92 + 106 . 1 j , Re( z 5 ) = 457 . 92, Im( z 5 ) = 106 . 1 . 9.1.1 Arithmetic operations with complex numbers To state the rules for operating with complex numbers we will use the con vention z = x + jy is a complex number and if z 1 is complex then z 1 = x 1 + jy 1 and z 2 = x 2 + jy 2 . Equality Two complex numbers z 1 , z 2 are equal if and only if Re( z 1 ) = Re( z 2 )and Im( z 1 )=Im( z 2 ). z 1 = z 2 iff x 1 = x 2 and y 1 = y 2 . Example Solve the equation 3 z = 6 9 j Solution Letting z = x + jy we have 3( x + jy ) = 6 9 j so 3 x + j 3 y = 6 9 j Equating real and imaginary parts gives 3 x = 6 and y = 9 Thus x = 2 and y = 3 so z = 2 3 i 9 FUNCTIONS OF A COMPLEX VARIABLE 85 Addition and Subtraction z 1 + z 2 = ( x 1 + jy 1 ) + ( x 2 + jy 2 ) = ( x 1 + x 2 ) + j ( y 1 + y 2 ) . z 1 z 2 = ( x 1 + jy 1 ) ( x 2 + jy 2 ) = ( x 1 x 2 ) + j ( y 1 y 2 ) . Examples (a) (2 5 j ) + ( 4 + j ) = (2 4) + j ( 5 + 1) = 2 4 j (b) ( 3 + 0 . 7 j ) (4 2 j ) = ( 3 4) + j (0 . 7 ( 2)) = 7 + 2 . 7 j Multiplication The important new element here is j 2 = 1 . When we apply this to multiplication of complex numbers we obtain z 1 .z 2 = ( x 1 + jy 1 ) . ( x 2 + jy 2 ) = x 1 .x 2 + jx 1 .y 2 + jy 1 .x 2 + j 2 y 1 y 2 z 1 .z 2 = x 1 .x 2 y 1 y 2 + j ( x 1 .y 2 + x 2 y 1 ) Examples (a) (3 + 2 j )(2 5 j ) = 6 10 j 2 + j (4 15) = 6 + 10 11 j = 16 11 j (b) ( √ 2 + j )( √ 2 j ) = ( √ 2) 2 j 2 + j ( √ 2 √ 2) = 2 + 1 = 3 Note that we can compute the successive powers of j : j 3 = j 2 j = 1 .j = j, j 4 = j 2 j 2 = 1 . 1 = 1 , j 5 = j and so on....
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This note was uploaded on 10/23/2009 for the course HES 2340 taught by Professor Tomedwards during the Three '09 term at Swinburne.
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