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 March 19th, 2011, 07:58 AM #1 Newbie   Joined: Feb 2011 Posts: 1 Thanks: 0 open ball Let $|x|=|y|=r$, with $x \neq y$. If $0, prove that $|(1-t)x+ty|.
 March 19th, 2011, 08:23 AM #2 Member   Joined: Nov 2009 From: France Posts: 98 Thanks: 0 Re: open ball We have 0<1-t<1 and by triangular inequality we have $|tx+(1-t)y|\leq |t||x|+|1-t||y|\leq r$. Now you have to show that the inequality is strict.
March 20th, 2011, 07:03 PM   #3
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 Originally Posted by retep Let $|x|=|y|=r$, with $x \neq y$. If $0, prove that $|(1-t)x+ty|.

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. (1-t)x+ty, $0, is the segment which connects y to x. |(1-t)x+ty|, $0, 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 |(1-t)x+ty|<r in an intuitive manner.

But intuition is not a proof, so:

By the triangular inequality:

$|(1-t)x+ty|\leq |(1-t)|\cdot|x| + |t|\cdot|y|=(1-t)\cdot r + t\cdot r = r$, $0 : inequality ( 1 )

to prove that the inequality is strict you can calculate the values of t which for the equality holds. Make $x= r\cdot e^{i\alpha}$ and $y= r \cdot e^{i\beta}$.

$|(1-t)x+ty|=r$

$|(1-t)r \cdot e^{i\alpha}+t \cdot r \cdot e^{i\beta}|=r$

$|(1-t) \cdot e^{i\alpha}+t \cdot e^{i\beta}|=1$ : expression ( i )

To hold, in expression ( i ) $1-t= 0$ or $t=0$. But $0 and than, the inequality ( 1 ) is strict.

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