Question

If z is a complex number satisfying $|z^2+1|=4|z|$, then the minimum value of $|z|$ is

• A. $2\sqrt{5}+4$
• B. $2\sqrt{5}-4$
• C. $\sqrt{5}-2$
• D. None of these

Answer 1

Answer: (C)

$\sqrt{5}-2$

Given:

$|z^2+1|=4|z|$

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

We know,

$||z|-\frac{1}{|z|}|\le |z+\frac{1}{z}|\le |z|+\frac{1}{|z|}$

$\therefore \pm (|z|-\frac{1}{|z|})\le 4\le |z|+\frac{1}{|z|}$

Equations we get:

$1.$ $|z{|}^2-4|z|-1=0$

$2.$ $|z{|}^2+4|z|-1=0$

$3.$ $|z{|}^2-4|z|+1=0$

$[|z|$ always $\ge 0]$

Solving eq (1):

$|z|=\sqrt{5}+2$

Solving eq (2):

$|z|=\sqrt{5}-2$

Solving eq (3):

$|z|=2\pm \sqrt{3}$

$\therefore$ Minimum $|z|=\sqrt{5}-2$