<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] $|z| =|z+1/z-1/z| \le$
$ |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?
</s>