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

z ≥ 2 i . e ., z - 0 ≥ 2 ∴ The distance of ' z ' from origin is greater than two. We have to find minimum of z + 1 2 z + 1 2 = z - - 1 2 = distance of z from - 1 2 , 0 . On the argand plane, z lies in the dotted regions I , II , III , IV . ∴ The minimum distance is from the point B = 2 - 1 2 = 3 2 ⇒ 3 2 ∈ 1 , 2

Mathematics - Complex Number Question with Solution | TestHub

MathematicsComplex NumberGeometric ApplicationMedium2 minPYQ_2014

MathematicsMediumsingle choice

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

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Answer:

Solution:

$|z| \geq 2 i . e ., |z - 0| \geq 2$

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

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

On the argand plane,

$z$ lies in the dotted regions $I, II, III, IV .$

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

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

Stream:JEESubject:MathematicsTopic:Complex NumberSubtopic:Geometric Application

⏱ 2mℹ️ Source: PYQ_2014

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