# Home_1 - Homework I Stephen Taylor Page 3 6 Let R z be a...

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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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Home_1 - Homework I Stephen Taylor Page 3 6 Let R z be a...

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