Finding the minimum value of a complex number

Question

If $z$ is a complex number satisfying $|z^2+1| = 4|z|$ . Then prove that the minimum value of $|z|$ is $4$

This is how I attempted the problem ,

$\frac{|z^2+1|}{|z|} = 4$

Therefore ,

$|z + \frac{1}{z}| = 4$

How do I proceed from here ?

According to the solution of the above problem the next step would be as follows

$||z| - |\frac{1}{z}|| ≤ 4$

We can then form a quadratic in $z$ to find out the least value of $z$. However , I’m stuck at the first step of the solution . How does $|z + \frac{1}{z}| = 4$ imply $||z| - |\frac{1}{z}|| ≤ 4$ ? Please help .

Answer 1

The implication$$\left|z+\frac1z\right|=4\implies\left||z|-\left|\frac1z\right|\right|\leqslant4$$comes from the inequality$$(\forall z,w\in\mathbb{C}):|z-w|\geqslant\bigl||z|-|w|\bigr|.$$

Answer 2

This follows from a version of the reverse triangle inequality, in your case $$\lvert\lvert z\rvert-\lvert\frac{1}{z}\rvert\rvert\leq\lvert z+\frac{1}{z}\rvert.$$

Answer 3

To tell it in another way, since: $$ \left| {z + 1/z} \right| = \left| {\left| z \right|e^{\;i\,\varphi } + 1/\left| z \right|e^{\; - i\,\varphi } } \right| = \left| {e^{\;i\,\varphi } \left( {\left| z \right| + 1/\left| z \right|e^{\; - i2\,\varphi } } \right)} \right| = \left| {\left| z \right| + 1/\left| z \right|e^{\; - i2\,\varphi } } \right| $$ then $$ \left| {\left| z \right| - 1/\left| z \right|} \right| \le \left| {z + 1/z} \right| \le \left| {\left| z \right| + 1/\left| z \right|} \right| $$