Find the range of |z| when $|z−\frac{4}{z}|=2$ I have tried using the triangle inequality but got two answers.
$z≤1+\sqrt5$ and one imaginary quadratic. I think here is one more equation which I am not getting. Please help.
Extended hint:  

$$
\begin{align}
4=\left|z−\frac{4}{z}\right|^2=\left(z−\frac{4}{z}\right)\left(\bar z−\frac{4}{\bar z}\right) &amp; = |z|^2+\frac{16}{|z|^2}-4\left(\frac{z}{\bar z}+\frac{\bar z}{z}\right) \\ &amp; = \left(|z|+\frac{4}{|z|}\right)^2 - 4\left( 2+ 2 \operatorname{Re}\frac{z}{\bar z}\right)
\end{align}
$$

Since $\left|\frac{z}{\bar z}\right| = 1$ it follows that $-1 \le \operatorname{Re}\frac{z}{\bar z} \le 1$ so:

$$
\left(|z|+\frac{4}{|z|}\right)^2 = 4\left( 3+ 2 \operatorname{Re}\frac{z}{\bar z}\right) \in [\,4,\,20\,]
$$

$$
|z|+\frac{4}{|z|} \in [\,2,\,2\sqrt{5}\,]
$$

From there on it's left to determine the range of the positive real roots of the quadratics in $|z|\,$:

$$
\begin{cases}
\begin{align}
|z|+\frac{4}{|z|} &amp; \ge 2 &amp;&amp;\iff |z|^2 - 2|z| + 4 \ge 0 \tag{1}\\
|z|+\frac{4}{|z|} &amp; \le 2 \sqrt{5} &amp;&amp; \iff |z|^2 - 2 \sqrt{5} \,|z| + 4 \le 0 \tag{2}\\
\end{align}
\end{cases}
$$

$(1)$ is satisfied for $\forall |z| \in \mathbb{R^+}$ while $(2)$ gives the solution $|z| \in [\sqrt{5}-1,\sqrt{5}+1]$.
As suggested using  Inequalities Among Complex Numbers,

$$\left|z-\dfrac4z\right|\ge|z|-\dfrac4{|z|}\implies|z|-\dfrac4{|z|}\le2\ \  \  \ (1)$$

and   $$\left|z-\dfrac4z\right|=\left|z+\left(-\dfrac4z\right)\right|\le|z|+\left|\dfrac{-4}z\right|=|z|+\dfrac4{|z|}\implies|z|+\dfrac4{|z|}\ge2\ \  \  \ (2)$$

Now use the followings on $(1),(2)$

$(x-a)(x-b)\le0;a\le b\implies a\le x\le b$

and $(x-a)(x-b)\ge0;a\le b\implies$ either $x\le a$ or $x\ge b$