Question

If $z$ is a complex number such that $\left|z\right|$ greater than or equal to $2,$ then the $\text{minimum value of}|z+\frac{1}{2}|.$

• A. $\frac{3}{2}$
• B. Less than $\frac{2}{3}$
• C. $\frac{5}{2}$
• D. $\frac{2}{5}$

Answer 1

The correct option is A

$\frac{3}{2}$

$|z+\frac{1}{2}|\text{is the distance of}z\text{from}-\frac{1}{2}$

$\left|z\right|=2$ is a circle with centre as origin and radius $2.$ $\left|z\right|\ge 2\text{i.e.}$ all the points outside and on the circle $|z|=2,$ it is the shaded region in the figure.

The closest point in that region on the negative $x-$axis, which is $(-2,0).$ So the $\text{least distance is}(\frac{-1}{2}-(-2)=\frac{3}{2})$

Or

Another way of solving is by applying triangle inequality.
$|z+\frac{1}{2}|\le \left|z\right|-\left|\frac{1}{2}\right|=2-\left|\frac{1}{2}\right|=\frac{3}{2}$