This preview has intentionally blurred sections. Sign up to view the full version.
View Full 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
 Three '09
 Cong
 Math, Complex number, Conic section, max β2, x2 − b2

Click to edit the document details