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I have to prove the following property, but no idea how to start... I was told to solve it with polar coordinates, but I still don't know how.

Let $\tau$ be a complex number with positive imaginary part. Prove that there exists a $\delta >0$ satisfying

$|x+\tau y|\geq\delta \sqrt{x^2+y^2}$ $\forall x,y\in\mathbb{R}$.

Could somebody help me? Thanks in advance!

Marc
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    Usual question: what have you done so far? – MattAllegro Apr 25 '14 at 16:34
  • I've been thinking to prove it by contradiction, i.e. that there exist two real numbers $x,y$ satisfying $|x+\tau y|>\delta\sqrt{x^2+y^2}$ $\forall\delta >0$, but I realized that I really had no idea what I was trying to reach with that – Marc Apr 25 '14 at 16:42
  • Usual question: what is your reaction to the answers posted? – Did May 18 '15 at 17:29

2 Answers2

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Try $\delta=\dfrac{\Im(\tau)}{\sqrt{1+|\tau|^2}}$.

Did
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Call $\tau=re^{i\theta}$ with $\theta\in]0,\pi[$.

Then you have \begin{align*} |x+\tau y|^2=& |x+re^{i\theta}y|^2\\ =&(x+ry\cos\theta)^2+r^2y^2\sin^2\theta\\ =&r^2y^2+x^2+2rxy\cos\theta\\ \end{align*}

The statement is true iff \begin{align*} &r^2y^2+x^2+2rxy\cos\theta\geq\delta^2(x^2+y^2) \Longleftrightarrow\\ &y^2(r^2-\delta^2)+x^2(1-\delta^2)+2rxy\cos\theta\geq0 \end{align*}

Since $\theta\in]0,\pi[$ we have that $\cos\theta\in]-1,1[$. But the term in which $\cos\theta$ appears is $2rxy\cos\theta$, and $x,y$ are taken in the whole $\mathbb R$. Being $\theta$ fixed, $\cos\theta$ it's only constant. So you can write $2r\cos\theta:=\alpha\in\mathbb R$. Remeber: $\alpha$ is a constant and WLOG we can think it as positive.

Hence you want to search $\delta>0$ s.t. $$ y^2(r^2-\delta^2)+x^2(1-\delta^2)+\alpha xy\geq0\;\;\;\forall x,y\in\mathbb R\;. $$

Take first $\delta<\frac{1}{2}\min\{1,r\}$. So you are ok when $x$ and $y$ have same sign.

Suppose finally $y<0<x$ and try to argue by your own!

Joe
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  • I've been trying to find a $\delta$ that turns your last statement into something like $(ax+by)^2$, but without any result. It's really disappointing -.- – Marc Apr 25 '14 at 17:16