If $\left|z+\dfrac{1}{z}\right|=a$ then find the minimum and maximum value of $|z|$ If $z$ be a complex number such that
$\left|z+\dfrac{1}{z}\right|=a$
then determine the range of values that $|z|$ can take.
My Attempt:
$a^2=\left|z+\dfrac{1}{z}\right|^2=\left(z+\dfrac{1}{z}\right)\left(\bar z+\dfrac{1}{\bar z}\right)$
which simplifies to 
$|z|^4-(a^2+2)|z|^2+1=-(z+\bar z)^2$
Thus
$|z|^4-(a^2+2)|z|^2+1=-(z+\bar z)^2\leq 0$                  (*)
$\left(|z|^2-\dfrac{a^2+2-\sqrt{a^4+4a^2}}{2}\right)\left(|z|^2-\dfrac{a^2+2+\sqrt{a^4+4a^2}}{2}\right)\leq 0$
$|z|^2\in \left[\dfrac{a^2+2-\sqrt{a^4+4a^2}}{2},\dfrac{a^2+2+\sqrt{a^4+4a^2}}{2}\right]$
$|z|\in \left[\dfrac{\sqrt{a^2+4}-a}{2},\dfrac{\sqrt{a^2+4}+a}{2}\right]$
which is a standard textbook solution.
Now the extreme values are obtained when $z+\bar z=0$  (Refer (*))
i.e. $z$ is purely imaginary.
This range is also obtained when we use the triangle inequality
$a=\left|z+\dfrac{1}{z}\right|\geq \left||z|-\dfrac{1}{|z|}\right|$.
My question is why we don't get the same answer if I use 
$a=\left|z+\dfrac{1}{z}\right|\leq |z|+\dfrac{1}{|z|}$
 A: A harder way of solving this question is here in my answer. If $\space\exists\space\text{z}\in\mathbb{C}$:
$$\text{z}+\frac{1}{\text{z}}=\text{z}+\frac{\overline{\text{z}}}{\text{z}\cdot\overline{\text{z}}}=\text{z}+\frac{\overline{\text{z}}}{\left|\text{z}\right|^2}=\Re\left(\text{z}\right)+\Im\left(\text{z}\right)i+\frac{\Re\left(\text{z}\right)-\Im\left(\text{z}\right)i}{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}\tag1$$
So:


*

*$$\Re\left(\text{z}+\frac{1}{\text{z}}\right)=\Re\left(\text{z}\right)+\frac{\Re\left(\text{z}\right)}{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}\tag2$$

*$$\Im\left(\text{z}+\frac{1}{\text{z}}\right)=\Im\left(\text{z}\right)-\frac{\Im\left(\text{z}\right)}{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}\tag3$$


So, we find:
$$\left|\text{z}+\frac{1}{\text{z}}\right|=\sqrt{\Re^2\left(\text{z}+\frac{1}{\text{z}}\right)+\Im^2\left(\text{z}+\frac{1}{\text{z}}\right)}=$$
$$\sqrt{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)+\frac{1+4\cdot\Re^2\left(\text{z}\right)}{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}-2}\tag4$$
In order to find the maximum and minimum:
$$
\begin{cases}
\frac{\partial\space\left|\text{z}+\frac{1}{\text{z}}\right|}{\partial\space\Re\left(\text{z}\right)}=0\\
\\
\frac{\partial\space\left|\text{z}+\frac{1}{\text{z}}\right|}{\partial\space\Im\left(\text{z}\right)}=0
\end{cases}\space\space\space\space\Longleftrightarrow\space\space\space\space
\begin{cases}
\Re\left(\text{z}\right)\cdot\left(4\cdot\Im^2\left(\text{z}\right)+\left(\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)\right)^2-1\right)=0\\
\\
2\cdot\Im\left(\text{z}\right)-\frac{2\cdot\Im\left(\text{z}\right)\cdot\left(1+4\cdot\Re^2\left(\text{z}\right)\right)}{\left(\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)\right)^2}=0
\end{cases}\tag5
$$
