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!
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!