show that $x_{-} < \frac{1}{2} < x_{+}$

Question

If we have the equation $$f(x,y) = \left(\begin{matrix}x^2-y^2+\alpha\\2xy+\beta\end{matrix}\right)$$ where $\alpha$ and $\beta$ are constants. We wish to solve the equation $f(\boldsymbol{x}) = \boldsymbol{x}$

Show for all $\alpha$ and $\beta$ that it exists two solutions $\boldsymbol{x}_{\pm}$ that are like: $$\boldsymbol{x}_{\pm} = \left(\begin{matrix}x_{\pm}\\y_{\pm}\end{matrix}\right), x_{-} < \frac{1}{2} < x_{+}, y_{\pm} = \frac{\beta}{1-2x_{\pm}}$$

I managed to show that $y_{\pm} = \frac{\beta}{1-2x_{\pm}}$ but I don't understand how I can show that $x_{-} < \frac{1}{2} < x_{+}$

Thanks in advance!

Answer 1

Hint: Take $z=x+iy$ and $w=\alpha + i\beta$ then you get $f = (Re(z^2 + w) , Im(z^2+w))$. Finally the equation becomes

$$z^2 + w = z\to z_{\pm}=\frac{1\pm \sqrt{1-4w}}{2}.$$

Now substitute for $z_{\pm}$ and $w$ and decompose into real and imaginary parts. Notice that in square root you need to write polar representation of complex numbers, i.e. $a+ib \equiv R \exp[i\theta]$.

$\textbf{Proof of $x_- <0.5<x_+$}$

You can write $$\sqrt{1-w}= \sqrt{1-4\alpha -4i\beta }= \sqrt{R e^{i\theta}}=\sqrt{R} \exp\left(\frac{i\theta}{2}\right) , \quad 0 \le \theta < \pi$$ Notice that $\theta$ is the principal argument and is in the above interval. We don't need the exact form of $R$ and $\theta$, but notice that for all $\alpha,\beta\in \mathbb{R}$, then $R \ge 0$. Now roots are

$$z_{\pm}= x_{\pm} + iy_{\pm}=\frac{1}{2} \pm \frac{\sqrt{R}}{2}\exp\left(\frac{i\theta}{2}\right)$$ therefore

$$x_{\pm} = \frac{1}{2} \pm \frac{\sqrt{R}}{2}\cos\left(\frac{\theta}{2}\right)$$

Now for $0 \le \theta < \pi$ you have $0\le \theta/2 < \pi/2$ and in this interval $0 < \cos(\theta/2) \le 1$ therefore the second term is a positive constant that implies $x_+ > 1/2$ and $x_-< 1/2$.