proving $|\lambda_1|<1<|\lambda_2|$.

Question

Let $\lambda_{1,2} = \frac{1}{2}(a+1+\rho \pm \sqrt{(a+1+\rho)^2-4a})$, where $a,\rho$ are complex numbers, $|a|=1, \rho\neq 0$. I want to prove the statement $|\lambda_1|<1<|\lambda_2|$. To do this I prove $(|\lambda_1|^2-1)(|\lambda_2|^2-1)<0$. After using parallelogram rule I am stuck at $$|a^2|-\frac{1}{2}(|(a+1+\rho)^2|+|(a+1+\rho)^2-4a)|+1 <0.$$ Can anyone give a hint on how to continue? I also do not see why I need $|a|=1$.

Answer 1

Let's start by assuming that $\lambda_1 = \lambda_2$. For this to hold we must find some $a$ and $\rho$ with $4a = (a+1+\rho)^2 = a^2 + 2(1 +\rho)a + (1 +\rho)^2$, which gives $a_{1,2} = -1 +\rho \pm \sqrt{(-1 +\rho)^2 - (1 +\rho)^2} = -1 +\rho \pm \sqrt{-4 \rho}$. Let $ i \sqrt \rho =q$ then $a_{1,2} = -1 - q^2 \pm 2 q = - (1 \pm q)^2$. However, since $q \neq 0$, this makes it impossible that $|a| =1$. Hence we have a contradiction, and $\lambda_1 \neq \lambda_2$.

Since $|\lambda_1||\lambda_2| =1$, as pointed out by @dxiv, there is one option to be considered (to be excluded) for $|\lambda_1|<1<|\lambda_2|$ to hold, namely $|\lambda_1| = |\lambda_2| =1$ but $\lambda_1 \neq \lambda_2$. This is the case if $\lambda_1 = e^{i \phi}$ and $\lambda_2 = e^{i \psi}$ with $\phi \ne \psi$ (and no multiples of $2 \pi$). If we had that, then $a = e^{i (\phi + \psi)}$ and $\lambda^2 +a = (a+1+\rho) \lambda $ would give us, for $\lambda_1$: $e^{i 2 \phi} +e^{i (\phi + \psi)} = (e^{i (\phi + \psi)}+1+\rho) e^{i \phi} $ or $$ \rho = - e^{i (\phi + \psi)}-1 +e^{i \phi} +e^{i \psi} $$
which is the same for $\lambda_2$. Checking again with the orginal equation we obtain $$\lambda_{1,2} = \frac{1}{2}(a+1+\rho \pm \sqrt{(a+1+\rho)^2-4a})\\ = \frac{1}{2}(e^{i \phi} +e^{i \psi}\pm \sqrt{(e^{i \phi} +e^{i \psi})^2-4e^{i (\phi + \psi)}})\\ = \frac{1}{2}(e^{i \phi} +e^{i \psi}\pm {(e^{i \phi} -e^{i \psi})}) $$ So this is indeed correct, and this case has to be excluded in the task decription.

Here is my earlier solution that, when excluding the case $|\lambda_1|= |\lambda_2|$, we have $|\lambda_1|<1<|\lambda_2|$:

$\lambda_{1,2} = \frac{1}{2}(a+1+\rho \pm \sqrt{(a+1+\rho)^2-4a})$ are the two solutions of the quadratic equation $\lambda^2 +a = (a+1+\rho) \lambda $. The complex conjugate of this is ${\lambda^*}^2 + a^*= (a^*+1+\rho^*) \lambda^* $. Multiplying the two gives $|\lambda|^4 + |a|^2 + 2 Re (a \lambda^2 )= |a+1+\rho|^2 |\lambda|^2 $. Let $|\lambda^2| = x$. We equivalently need to establish $x_1 <1< x_2$.

We have $$ x^2- |a+1+\rho|^2 x +1 = - 2 Re (a \lambda^2 ) = - 2 |a||\lambda^2 |\cos \alpha = - 2 x\cos \alpha $$ ,with $\alpha$ the angle between $a$ and $\lambda^2$ in the complex plane. This gives $$ x^2+ (2 \cos \alpha -|a+1+\rho|^2) x +1 = 0 $$ which has the solutions $$ x_{1,2} = \frac12|a+1+\rho|^2 - \cos \alpha \pm \sqrt{(\frac12|a+1+\rho|^2 - \cos \alpha )^2 -1} $$ for $x =|\lambda^2 |$ which is real and positive. Since it does have real solutions, we have that $(\frac12|a+1+\rho|^2 - \cos \alpha )^2> 1$, and since $x$ is positive, it follows that $\frac12|a+1+\rho|^2 - \cos \alpha > 1$.

Hence the larger solution is evidently $>1$, whereas for the smaller solution to be $<1$ it is required that $$ \frac12|a+1+\rho|^2 - \cos \alpha - \sqrt{(\frac12|a+1+\rho|^2 - \cos \alpha )^2 -1} <1 $$ or $$ (\frac12|a+1+\rho|^2 - \cos \alpha -1)^2 < {(\frac12|a+1+\rho|^2 - \cos \alpha )^2 -1} $$ or $$ -2(\frac12|a+1+\rho|^2 - \cos \alpha) +1 < -1 $$ or $$ \frac12|a+1+\rho|^2 - \cos \alpha >1 $$ but we have already established that. This completes the proof. $\qquad \Box$