If z is a complex number such that z ≥ 2 , then the minimum value of z + 1 2 :

1. Is strictly greater than $\frac{5}{2}$
2. Is strictly greater than $\frac{3}{2}$ but less than $\frac{5}{2}$
3. Is equal to $\frac{5}{2}$
4. Lies in the interval $(1, 2)$

Solution

$\left|z\right|\ge 2 i.e., \left|z-0\right|\ge 2$

$\therefore$ The distance of '$z$' from origin is greater than two.
We have to find minimum of $\left|z+\frac{1}{2}\right|$

$\left|z+\frac{1}{2}\right|=\left|z-\left(-\frac{1}{2}\right)\right|$ $=$ distance of $z$ from $\left(-\frac{1}{2}, 0\right)$.

On the argand plane,



$z$ lies in the dotted regions $\mathrm{I}, \mathrm{II}, \mathrm{III}, \mathrm{IV}.$

$\therefore$   The minimum distance is from the point $B=2-\left(\frac{1}{2}\right)$$=\frac{3}{2}$

$\Rightarrow \frac{3}{2}\in \left(1, 2\right)$

Asked in: JEE Main 2014 (06 Apr)

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