Question

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

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

Answer 1

The correct option is B

$\frac{3}{2}$

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

$\left|z\right|=2$ is a circle with centre as origin and radius 2. $\left|z\right|\ge 2$ $\Rightarrow$ all the points outside and on the circle

$\left|z\right|$ =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 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-$\left|\frac{1}{2}\right|$
= $\frac{3}{2}$