March 19th, 2011, 07:58 AM  #1 
Newbie Joined: Feb 2011 Posts: 1 Thanks: 0  open ball
Let , with . If , prove that .

March 19th, 2011, 08:23 AM  #2 
Member Joined: Nov 2009 From: France Posts: 98 Thanks: 0  Re: open ball
We have 0<1t<1 and by triangular inequality we have . Now you have to show that the inequality is strict.

March 20th, 2011, 07:03 PM  #3  
Guest Joined: Posts: n/a Thanks:  Re: open ball Quote:
Considering x and y complexes. If their moduli are the same, they are different only in phase(they can be equal too). Both x and y lie in a circle of radius r in the complex plane. (1t)x+ty, , is the segment which connects y to x. (1t)x+ty, , is the distance of a point in that segment(that is a circle's chord) to the complex plane origin, excluding x and y. You can, that way, see that (1t)x+ty<r in an intuitive manner. But intuition is not a proof, so: By the triangular inequality: , : inequality ( 1 ) to prove that the inequality is strict you can calculate the values of t which for the equality holds. Make and . : expression ( i ) To hold, in expression ( i ) or . But and than, the inequality ( 1 ) is strict.  

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