<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] 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 $a = 0$.
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