With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$

Question

With $z\in \mathbb C$ find the maximum value for |z| such that $$\left\lvert z+\frac{1}{z}\right\rvert=1.$$ Source: List of problems for math-contest training.

My attempt: it is easy to see that the given condition is equivalent $$\lvert z^2+1\rvert=\lvert z\rvert$$ and if $z=a+bi$, \begin{align*} \lvert z\rvert&=\lvert a^2-b^2+1+2ab i\rvert=\sqrt{(a^2-b^2+1)^2+4 a^2b^2}\\ &=\sqrt{(a^2-b^2)^2+2(a^2-b^2)+1+4a^2b^2}\\ &=\sqrt{(a^2+b^2)^2+2(a^2-b^2)+1}\\ &=\sqrt{\lvert z\rvert^4+2(a^2-b^2)+1} \end{align*} I think it is not leading to something useful... the approach I followed is probably not useful.

Hints and answers are welcomed.

Answer 1

Suppose $z + \frac{1}{z} = e^{it}$ for some $t \in \mathbb{R}$. Then solving for $z$ gives $$z = \frac{e^{it} \pm \sqrt{e^{2it} - 4}}{2} = e^{it} \cdot \frac{1 \pm \sqrt{1 - 4 e^{-2it}}}{2}.$$ Therefore, $$ |z| = \frac{1}{2} \left| 1 \pm \sqrt{1 - 4 e^{-2it}} \right| \le \frac{1}{2} \left( |1| + |\sqrt{1 - 4 e^{-2it}}|\right) = \frac{1}{2} \left( 1 + \sqrt{1 - 4 e^{-2it}} \right)\le \\ \frac{1}{2} \left( 1 + \sqrt{|1| + |4 e^{-2it}|}\right) = \frac{1 + \sqrt{5}}{2}.$$ On the other hand, if we set $t := \frac{\pi}{2}$ so that $e^{-2it} = -1$ then we will get equality. Therefore, the maximum value of $|z|$ is $\frac{1+\sqrt{5}}{2}$, which is achieved for $z = \frac{1+\sqrt{5}}{2} i$. (In fact, a slight refinement of the argument will show that equality holds, i.e. the maximum is achieved, if and only if $z = \pm \frac{1+\sqrt{5}}{2} i$.)

Answer 2

Since $$|z| = |z^2-(-1)|\geq |z|^2-|-1|$$

Write $r=|z|$ and we have $$r^2-r-1\leq 0\implies r\in [0,r_2]$$ where $$r_{1,2} ={1\pm \sqrt{5}\over 2} $$

So $r\leq \displaystyle{1+ \sqrt{5}\over 2}$

Answer 3

The reverse triangle inequality implies that $|z^2+1|\geq ||z^2|-1|$ so that if $|z|^2\geq 1$ we see that $$|z|\geq |z|^2-1$$ By solving this inequality using the quadratic formula, we see that $|z|\leq \frac{1}{2}(1+\sqrt{5})$, so the maximum must be less than $\frac{1}{2}(1+\sqrt{5})$. We see that this maximum is attained by setting $z=\frac{1}{2}(1+\sqrt{5})$.

Answer 4

We can maximize $g(a,b) = a^2+b^2$ subject to $(a^2+b^2)^2 =3b^2-1$. Solve for $a^2$ we have: $a^2= \sqrt{3b^2-1} -b^2\implies g(a,b) = \sqrt{3b^2-1}$. Thus the problem now is the find the maximum value of $b^2$. To this end, let $c = b^2 \ge 0$, we have: $a^4+2a^2c+c^2 - 3c+ 1 = 0\implies c^2-3c+1 \le 0\implies c \le \dfrac{3+\sqrt{5}}{2}\implies 3b^2 -1 = 3c-1\le \dfrac{7+3\sqrt{5}}{2}\implies |z|_{\text{max}} = \sqrt{g(a,b)}_{\text{max}} = \sqrt[4]{\dfrac{7+3\sqrt{5}}{2}}= \dfrac{1+\sqrt{5}}{2}$,and this max occurs when $a = 0$.

Answer 5

Here's a solution using the dual problem. Assume that the solution of the problem occurs at $z = re^{i\theta}.$ Consider $$ F_r(t) = |re^{it}+r^{-1}e^{-it}|^2 = r^2 + r^{-2} + e^{2it}+e^{-2it} = r^2+r^{-2}+2\cos 2t. $$ If $z= re^{i\theta}$ is the maximizer, we claim that $F_r'(\theta)=0$. That is, given $|z|=r$, the angle $\theta$ should be chosen so that $F_r(t)$ is either maximized or minimized. To see this, assume to the contrary that $F_r'(\theta)\neq 0$. Then, there is small $u$ (we allow $u$ to have negative sign) such that $$F_r(\theta-u)<1=F(\theta)<F_r(\theta+u).$$ If it is the case that $r\leq 1$, then by choosing slightly large $r'$, we can make $F_{r'}(\theta+u)=1$. This contradicts maximality of $r$. If $r>1$, then by choosing slightly large $r'$, we can make $F_{r'}(\theta-u) = 1$. This also leads to a contradiction. Now it follows that $\cos(2\theta)$ should be either $-1$ or $1$. However, we can easily see that $\cos(2\theta)$ cannot be $1$. Thus, from $\cos(2\theta)=-1$, we have $$ 1=F_r(\theta)=r^2 +r^{-2} -2 = (r-r^{-1})^2, $$or $$ r-1/r = \pm 1. $$ This gives $$ r= \frac{\pm 1+ \sqrt{5}}{2}. $$ Therefore, the larger one $r=\frac{ 1+ \sqrt{5}}{2}$ is the solution.

Answer 6

$|z| =|z+1/z-1/z| \le$

$ |z+1/z| +|1/z| =1+|1/z|;$

$r:=|z|$;

$r-1/r \le 1;$

$r^2-r -1\le 0;$

$(r-1/2)^2 -1/4-1\le 0;$

$r-1/2 \le (1/2)√5;$

$r \le (1/2)(1+√5)$.

Is the maximum attained?