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In a proof I'm trying to understand the following lemma was used:

If $x, y \in \mathbb{C}$, with $|y|<|x|$ and $|\operatorname{arg}(x) - \operatorname{arg}(y)| < \frac{\pi}{3n}$ for a positive integer $n$, then the inequality $|x-y| \leq |x|$ holds.

I tried to apply the cosine law, but didn't get a proof of this little lemma. Maybe one of you has an instant idea that works.

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  • $\begingroup$ The appearance of $n$ in the denominator on the right hand side is mysterious- does it appear any where else in the lemma? $\endgroup$ – Brian Borchers Nov 13 '16 at 21:31
  • $\begingroup$ It's the degree of a nonconstant polynomial. I also think that it's value doesn't matter, so maybe $n$ should be $1$. $\endgroup$ – Celsius Nov 13 '16 at 21:41
  • $\begingroup$ It seems to be a geometric property of triangles: If we consider one corner of a triangle, then the (in some cases more) point on the opposite edge of the corner with the largest distance is one of the remaining corners. For our problem, if we take the intersection point of the ray from the origin crossing $y$ and the circle with radius |x| around the origin, then this point forms with the origin and $x$ a triangle. We than have to consider one of the corners differing from $0$. If the angle at the origin is $< \frac{\pi}{3}$, then the two radial edges are the longest and hence, $|x-y|<|x|$. $\endgroup$ – Celsius Nov 13 '16 at 22:13
  • $\begingroup$ This should be a geometric explanation of the proof below. $\endgroup$ – Celsius Nov 13 '16 at 22:19
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If $|y| = 0 $, note that $|x-y| = |y|$. Let $ |y| >0 $ and $\phi = |arg(x) - arg(y)| < \frac{\pi}{3} $.

From Law of cosines, it follows that

$$ \frac{|x|^2 + |y|^2 - |x-y|^2}{2|x||y|} = \cos \phi > \frac{1}{2} \implies |x|^2 - |x-y|^2 >|y|(|x| - |y|). $$

Since $ |x|>|y|$, we conclude that $ |x| > |x-y|$.

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  • $\begingroup$ OK, this is a very simple proof. Thank you. :-) $\endgroup$ – Celsius Nov 13 '16 at 22:16

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