814hw1sol - Math 814 HW 1 September 19, 2007 p.4, Exercise...

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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 i and w = e i with , [0 , 2 ) , this means that e 2 i ( - ) = 1 so that - = 0 or - = . 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 ....
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814hw1sol - Math 814 HW 1 September 19, 2007 p.4, Exercise...

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