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

(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 $\frac{5}{2}$

JEE Main · 2014 · 06 Apr 2014 · Offline · Complex Numbers

If z is a complex number such that

∣ z ∣ \geq 2

, then the minimum value of

z + \frac{1}{2}

Answer: (B)

By the reverse triangle inequality,

z + \frac{1}{2} \geq ∣ z ∣ - \frac{1}{2}

. With

∣ z ∣ \geq 2

this is at least

2 - \frac{1}{2} = \frac{3}{2}

, and the value

\frac{3}{2}

is actually attained, at

z = - 2

, where

- 2 + \frac{1}{2} = \frac{3}{2}

. So the minimum is

\frac{3}{2}

, which lies in the interval

(1, 2)

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option