<s>[INST] With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$ 
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.
 [/INST] 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.
</s>