Home_1 - Homework I Stephen Taylor April 28, 2005 Page 3 :...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This note was uploaded on 10/20/2009 for the course MATH 814 taught by Professor Cong during the Three '09 term at University of Adelaide.

Page1 / 4

Home_1 - Homework I Stephen Taylor April 28, 2005 Page 3 :...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online