If $|z + \frac{1}{z}| = a$ where $a$ is a positive real number, then why is
$$\frac{-a+\sqrt{a^2+4}}{2}\leq\lvert z\rvert\leq\frac{a+\sqrt{a^2+4}}{2} \quad?$$
This is a formula in a list of formulae given in my book, and I was wondering how it was derived. How do I get this result? I refer to the chapter complex numbers of the book 39 years chapterwise topicwise solved papers by Arihant.
$\textbf{Hint: }$$$$$ Use the polar form of $z$ by representing it as $re^{i\theta}$.
Also note that $|z|^2=z\bar z$.
Lastly note that $$|z|=|\bar z|=|-z|=|-\bar z|\Rightarrow \left|z+\dfrac 1z\right|=\left|\bar z+\dfrac 1{\bar z}\right|=\left|-\bar z-\dfrac 1{\bar z}\right|=\left|-z-\dfrac 1{z}\right|$$ $$$$ Hence, it suffices to consider only one of the numbers $z,\bar z, -\bar z, -z$, namely the one in the first quadrant. Therefore, let $z=re^{i\theta}, \theta\in[0,\pi/2]$ $$$$ Thus, $$\left|z+\dfrac 1z\right|=a\Rightarrow\left|r(\cos\theta+i\sin\theta)+\dfrac 1r(\cos\theta-i\sin\theta)\right|^2=a^2$$
$$\left(r\cos\theta+\dfrac 1r \cos\theta\right)^2+\left(r\sin\theta-\dfrac 1r \sin\theta\right)^2=a^2$$
After expanding, $$r^2+\dfrac 1{r^2} +2\cos2\theta=a^2$$ $$\left(r-\dfrac 1r\right)^2=a^2-4\cos^2\theta\le a^2$$ $$$$ Thus we want to maximize $a^2-4\cos^2\theta$ in order to find the maximum (and minimum) value of $\left(r-\dfrac1r\right)^2$. Now, $0\le|\cos\theta|\le 1\Rightarrow 0\le\cos^2\theta\le 1$. Hence, the maximum value of $\left(r-\dfrac1r\right)^2=a^2-4\cos^2\theta$ is $\left(r-\dfrac1r\right)^2=a^2$. Note that to simultaneously maximize (and minimize) $\left(r+\dfrac1r\right)$, we choose the maximum value of $a^2-4\cos^2\theta$ ie we let $$\left(r+\dfrac1r\right)^2=a^2$$
$$$$ $$$$ For $\theta=\dfrac\pi2$, $\left(r-\dfrac 1r\right)^2= a^2\Rightarrow \left(r-\dfrac 1r\right)= a$ $$\Rightarrow r^2-ar-1=0$$ which has roots $r=\dfrac{a\pm\sqrt{a^2+4}}{2}$ $$$$ $\textbf{To maximize r:}$ $$$$We ignore the negative root and assign $r$ the value $r=\dfrac{a+\sqrt{a^2+4}}{2}$
$$$$ Thus:$$r=\dfrac{a+\sqrt{a^2+4}}2$$ It follows that the greatest value of $|z|$ is attained at $$z=\left(\dfrac{a+\sqrt{a^2+4}}2\right)e^{i\pi/2}$$ Hence, $$|z|\le \dfrac{a+\sqrt{a^2+4}}2$$ $$$$ $\textbf{To minimize r:}$ $$$$We ignore the positive root and assign $r$ the value $r=\left|\dfrac{a-\sqrt{a^2+4}}{2}\right|=\dfrac{-a+\sqrt{a^2+4}}{2}$
Note that $ a\ge0\Rightarrow \sqrt{a^2+4}>a\Rightarrow a-\sqrt{a^2+4}<0 $. But $r\ge0$, hence we take the modulus. $$$$