Question

If the complex number z satisfies the condition |z| $\ge$ 3, then the least value of $|z+\frac{1}{z}|$ is equal to.

• A. $2$
• B. $\frac{4}{3}$
• C. $1$
• D. $\frac{8}{3}$

Answer 1

The correct option is D $\frac{8}{3}$

By using triangle inequality: $||z_1-|z_2||\le |z_1+z_2|\le |z_1|+|z_2|$

We have $|z+\frac{1}{z}|\le |z|+|\frac{1}{z}|$

Now Given that
$|z|\ge 3$
$\therefore |z+\frac{1}{z}|\le |3|+|\frac{1}{3}|$
$\Rightarrow |z+\frac{1}{z}|\le 3-\frac{1}{3}$
$\Rightarrow |z+\frac{1}{z}|\le \frac{8}{3}$