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Lecture38-rev

# Lecture38-rev - MAT A26 Lecture 38 1 Simple Functions of a...

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M AT A26 Lecture 38 1 Simple Functions of a Complex Variable If P ( x ) = a 0 + a 1 x + a 2 x 2 + ... + a n x n is a polynomial with real or complex coefficients, then the polynomial P is a perfectly well defined function function with domP = C taking values in C . Also, if f ( x ) = P ( x ) Q ( x ) where P and Q are polynomials with real or complex coefficients then the rational function f is a perfectly well defined function with domain C -{ z C | Q ( x ) = 0 } taking values in C 1.1 Examples 1. Find the domain of f ( z ) = z z 2 +1 . SOLUTION: We must determine where the denominator is zero, 1.e., we must solve z 2 + 1 = 0. Now z 2 + 1 = ( z + i )( z - i ) = 0 if and only if z = i or z = - i . Thus domf = C - {- i, i } 2. Let P ( z ) = i + (3 + 4 i ) z 2 Calculate P (2 - 3 i ). Find the real and imaginary parts of P ( z ). SOLUTION: P (2 - 3 i ) = i + (3 + 4 i )(2 - 3 i ) 2 = i + (3 + 4 i )(4 - 12 i + 9( i ) 2 ) = i + (3 + 4 i )(4 - 12 i - 9) = i + (3 + 4 i )( - 5 - 12 i ) = i + ( - 15 - 36 i - 20 i - 48( i ) 2 ) = i + ( - 15 - 56 i + 48) = 33 - 55 i To find the real and imaginary parts of P ( z ), evaluate P ( x + iy ) P ( x + iy ) = i + (3 + 4 i )( x + iy ) 2 = i + (3 + 4 i )( x 2 + 2 ixy - y 2 ) = i + 3 x 2 - 3 y 2 - 8 xy + 4 x 2 i - 4 y 2 i = 3 x 2 - 3 y 2 - 8 xy + i (1 + 4 x 2 - 4 y 2 ) Thus Re ( P ( z )) = 3 x 2 - 3 y 2 - 8 xy and Im P ( z ) = 1 + 4 x 2 - 4 y 2 .

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