Basic complex variable proposition 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!
 A: Try $\delta=\dfrac{\Im(\tau)}{\sqrt{1+|\tau|^2}}$.
A: 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!
