# 814hw1sol - Math 814 HW 1 p.4 Exercise 1 Prove that || z...

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 814 HW 1 September 19, 2007 p.4, Exercise 1. Prove that || z |-| w || ≤ | z- w | and give conditions for equality . It is equivalent to prove that || z | - | w || 2 ≤ | z- w | 2 , which, after multiplying out and cancelling, amounts to proving that- 2 | z || w | ≤ - ( z ¯ w + ¯ zw ) . (Note that z ¯ w + ¯ zw equals its own complex conjugate, hence is real.) Changing the signs and squaring, we have to prove that 4 | z | 2 | w | 2 ≥ ( z ¯ w ) 2 + 2 | z || w | + (¯ zw ) 2 , i.e., that ( z ¯ w ) 2- 2 | z || w | + (¯ zw ) 2 ≤ . (1) But ( z ¯ w ) 2- 2 | z || w | + (¯ zw ) 2 = ( z ¯ w- ¯ zw ) 2 and since z ¯ w- ¯ zw equals its negative complex conjugate, we have ( z ¯ w- ¯ zw ) 2 real and ≤ . Hence (1) holds, which was equivalent the desired inequality. We have equality iff z ¯ w- ¯ zw = 0 , iff z ¯ z = w ¯ w . Writing z = | z | e iα and w = e iβ with α, β ∈ [0 , 2 π ) , this means that e 2 i ( α- β ) = 1 so that α- β = 0 or α- β = ± π . Hence we have equality iff z and w are on the same line through . 1 p.4, Exercise 2. Let z 1 , . . . , z n be nonzero complex numbers, with n ≥ 2 . Show that | z 1 + ··· + z n | = | z 1 | + ··· + | z n | if and only if all the ratios z k /z ` are real and nonnegative. One direction is easy. If all the ratios are ≥ , we have z k = t k z 1 for each k , where t k ≥ . Let t = t 1 + t 2 + ··· + t n . Then | z 1 + ··· + z n | = t | z 1 | = t 1 | z 1 | + t 2 | z 1 | + ··· + t n | z 1 | = | z 1 | + ··· + | z n | . For the other direction, first assume n = 2 . Let w = z 1 /z 2 and write w = x + iy . We have | z 1 + z 2 | = | z 1 | + | z 2 | iff | 1+ w | = 1+ | w | . Squaring, this implies y = 0 , so w = x . If x < then | 1 + x | = 1- x = | 1- x | , which implies x = 0 , a contradiction. Hence w = x ≥ ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

814hw1sol - Math 814 HW 1 p.4 Exercise 1 Prove that || z...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online