<s>[INST] The greatest value of $|z|$ if $\Big|z+\frac{1}{z}\Big|=3$ where $z\in\mathbb{C}$ 
$\bigg|z+\dfrac{1}{z}\bigg|=3$ then the greatest value of $|z|$ is ___________

My Attempt
$$
\bigg|z+\frac{1}{z}\bigg|=\bigg|\dfrac{z^2+1}{z}\bigg|=\frac{|z^2+1|}{|z|}=3\\
\bigg|z+\frac{1}{z}\bigg|=3\leq|z|+\frac{1}{|z|}\implies |z|^2-3|z|+1\ge0\\
|z|=\frac{3\pm\sqrt{5}}{2}\implies\color{red}{|z|\in(-\infty,\frac{3-\sqrt{5}}{2}]\cup[\frac{3+\sqrt{5}}{2},+\infty)}
$$
$$
\bigg|z+\frac{1}{z}\bigg|=3\geq\bigg||z|-\frac{1}{|z|}\bigg|\\
|z|^2-3|z|-1\leq0\text{ or }|z|^2+3|z|-1\geq0\\
|z|=\frac{3\pm\sqrt{13}}{2}\text{ or }|z|=\frac{-3\pm\sqrt{13}}{2}\\
\color{red}{|z|\in[\frac{3-\sqrt{13}}{2},\frac{3+\sqrt{13}}{2}]}\text{ or }\color{red}{|z|\in(-\infty,\frac{-3-\sqrt{13}}{2}]\cup[\frac{-3+\sqrt{13}}{2},+\infty)}
$$
The solution given in my reference is $\dfrac{3+\sqrt{13}}{2}$, why am I not able to find it in my attempt ?
Note: A similar question has been asked before, If $∣z+\frac{1}{z}∣=a$ then find the minimum and maximum value of $|z|$, but it does not have any answer which address why we choose a specific version of the triangle inequality.
 [/INST] A possible approach is using polar representation


*

*$z = |z|e^{i\phi}$

*$\Big|z+\frac{1}{z}\Big|=3 \Leftrightarrow (z+\frac{1}{z})(\bar z + \frac{1}{\bar z}) = 9 \Leftrightarrow \boxed{\left(|z|e^{i\phi} + \frac{1}{|z|} e^{-i\phi}\right)\left(|z|e^{-i\phi} + \frac{1}{|z|} e^{i\phi}\right) = 9}$
Expanding gives
\begin{eqnarray*}
|z|^2 + e^{-2i\phi} + e^{2i\phi} + \frac{1}{|z|^2} & = & 9 \\
& \Leftrightarrow & \\
|z|^2 + 2\cos 2\phi + \frac{1}{|z|^2} & = & 9 \\
\end{eqnarray*}
So, the maximum $|z|$ is reached for $\cos 2\phi = -1$, since the maximum solution of $q+\frac{1}{q}= C$ is $q= \frac{1}{2}(C+\sqrt{C^2-4})$ . Hence, solve
\begin{eqnarray*}
|z|^2 + \frac{1}{|z|^2} & = & 11 \\
& \stackrel{|z|\: max.}{\Rightarrow} & \\
|z| & = & \sqrt{\frac{11+\sqrt{117}}{2}}\\
\end{eqnarray*}
Here I show that this is the same as the given solution:
\begin{eqnarray*}
|z| & = & \sqrt{\frac{11+\sqrt{117}}{2}}\\
 & = & \sqrt{\frac{22+2\cdot 3 \sqrt{13}}{4}}\\
  & = & \sqrt{\frac{9 + 13+2\cdot 3 \sqrt{13}}{4}}\\
    & = & \sqrt{\frac{(3 +\sqrt{13})^2}{4}}\\
    & = & \frac{3 +\sqrt{13}}{2}\\
\end{eqnarray*}
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