Question

If z is a complex number such that $\left|z\right|\ge 2$, then the minimum value of $|z+\frac{1}{2}|$

• A. is equal to $\frac{5}{2}$
• B. lies in the interval $(1,2)$
• C. is strictly greater than $\frac{5}{2}$
• D. is strictly greater than $\frac{3}{2}$ but less than $\frac{5}{2}$

Answer 1

The correct option is B lies in the interval $(1,2)$

Solution -

$\Rightarrow \left|z\right|\ge 2$ $z=x+iy$

$\Rightarrow \sqrt{x^2+y^2}=2$

$x^2+y^2=4$

$\Rightarrow |z+\frac{1}{2}|=\sqrt{(x+\frac{1}{2}{)}^2+y^2}$

$=\sqrt{x^2+y^2+x+\frac{1}{4}}$

$=\sqrt{4+x+\frac{1}{4}}$

Range of x is $[2,-2]$

Range of $\sqrt{4+x+\frac{1}{4}}$ is from $\frac{3}{2}$ to $\frac{5}{2}$

Minimum value is $\frac{3}{2}$

lies in the interval $(1,2)$

$B$ is correct.