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.

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 $z = \pm \frac{1+\sqrt{5}}{2} i$.)
