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Unformatted text preview: Homework I Stephen Taylor April 28, 2005 Page 3 : 6. Let R ( z ) be a rational function of z . Show that R ( z ) = R (¯ z ) if all the coefficients in R ( z ) are real. Since R ( z ) is a rational function of z it is in the form a + a 1 z + ··· + a n z n b + b 1 z + ··· + b m z m where { n, m } ∈ N Since all a i , b j are real by hypothesis, we use (2.3) in quotient form to find the following: R ( z ) = a + a 1 z + ··· + a n z n b + b 1 z + ··· + b m z m = a + a 1 ¯ z + ··· + a n ¯ z n b + b 1 ¯ z + ··· + b m ¯ z m = R (¯ z ) Page 4 : 1. Prove the following and give necessary and sufficient condition for equality:  z    w  ≤  z w  Proof: Applying the second identity in Exercise 4 on page 3 we find:  z w  2 =  z  2 2Re[ z ¯ w ] +  w  2 ≥  z  2 2  z  w  +  w  2 = (  z    w  ) 2 Taking roots we find  z w  ≥  z    w  . Exchanging z and w and repeating the same argument we find a second inequality  z w  ≥  w    z  . Combing these two results we find the desired inequality  z w  ≥ ± (  z    w  ) =  z    w  Equality is achieved iff 2Re[ z ¯ w ] = 2  z  w  . Letting....
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 Three '09
 Cong
 Square Roots, Complex number, Complex Plane, Polar Representation

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