JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Minimum distance from a point to an exterior region in the complex plane

Q: Why isn't the minimum distance on the boundary |z| = 1?

The condition is |z| ≥ 1, which is the exterior AND the boundary of the unit disk. Since the point z_0 = -3/2 - 2i has |z_0| = 2.5 > 1, it lies directly in the allowed region, so we can choose z = z_0 to get a distance of 0.

If z is a complex number such that |z| ≥ 1, then the minimum value of |z + 1/2(3+4i)| is:

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A5/2B2C3D0

Step-by-step solutionView

Correct answer

The minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣$ for $∣ z ∣ \geq 1$ is $0$ , because $z_{0} = - \frac{3 + 4 i}{2}$ satisfies $∣ z_{0} ∣ = \frac{5}{2} \geq 1$ .

Option analysis

Why each option works or fails

A · 5/2

Calculated the modulus of $\frac{1}{2} (3 + 4 i)$ instead of finding the minimum value of the expression $∣ z - (- \frac{3 + 4 i}{2}) ∣$ . Recognize that the expression represents the distance from $z$ to $- \frac{3 + 4 i}{2}$ , which is minimized to $0$ if that point itself lies inside the domain $∣ z ∣ \geq 1$ .

B · 2

Subtracted $1 / 2$ or applied an incorrect triangle inequality variant such as $\frac{5}{2} - \frac{1}{2} = 2$ . Check whether the point making the expression zero is inside the region $∣ z ∣ \geq 1$ before attempting to apply boundary-distance formulas.

C · 3

Miscalculated $\frac{5}{2} + \frac{1}{2} = 3$ or confused the maximum/minimum bounds on a circle. The modulus $∣ z - z_{0} ∣$ attains its absolute minimum of $0$ when $z = z_{0}$ , provided $z_{0}$ satisfies the condition $∣ z_{0} ∣ \geq 1$ .

D · 0

The choice is correct. For $z = - \frac{3 + 4 i}{2}$ , we have $∣ z ∣ = \frac{5}{2} \geq 1$ , which makes the expression $∣ z + \frac{1}{2} (3 + 4 i) ∣ = 0$ .

Reviewed route

Solution

StepWorking

01Given

A complex number $z$ satisfies $∣ z ∣ \geq 1$ . We are asked to find the minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣ = ∣ z - (- \frac{3}{2} - 2 i) ∣$ .

02Goal

Find the minimum distance between $z$ (where $∣ z ∣ \geq 1$ ) and the fixed point $z_{0} = - \frac{3}{2} - 2 i$ .

03Approach

Check whether the target point $z_{0} = - \frac{3}{2} - 2 i$ lies inside the region defined by $∣ z ∣ \geq 1$ . If $∣ z_{0} ∣ \geq 1$ , then $z_{0}$ itself is in the allowed region, making the distance zero.

04Execute

Calculate the modulus of $z_{0}$ : $∣ z_{0} ∣ = ∣ - \frac{3}{2} - 2 i ∣ = \sqrt{{(- \frac{3}{2})}^{2} + (- 2)^{2}} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$ Since $∣ z_{0} ∣ = 2.5 \geq 1$ , the point $z = z_{0} = - \frac{3}{2} - 2 i$ belongs to the region $∣ z ∣ \geq 1$ . Therefore, the minimum value of $∣ z - z_{0} ∣$ is $0$ .

✓Verify

Substitute $z = - \frac{3}{2} - 2 i$ : $∣ z ∣ = \frac{5}{2} \geq 1$ is satisfied, and $∣ z + \frac{1}{2} (3 + 4 i) ∣ = 0$ , which is the absolute minimum possible value for any modulus.

Hints that build this answer step by step

What does the expression $∣ z + \frac{1}{2} (3 + 4 i) ∣$ represent geometrically in the complex plane?
The distance from $z$ to the point $z_{0} = - \frac{1}{2} (3 + 4 i)$
What is the modulus of $z_{0} = - \frac{1}{2} (3 + 4 i)$ ?
$∣ z_{0} ∣ = \frac{5}{2}$
Does the point $z_{0} = - \frac{1}{2} (3 + 4 i)$ satisfy the domain restriction $∣ z ∣ \geq 1$ ?
Yes, because $∣ z_{0} ∣ = 2.5 \geq 1$ , so we can choose $z = z_{0}$ , giving a minimum value of $0$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why isn't the minimum distance on the boundary

∣ z ∣ = 1

The condition is $∣ z ∣ \geq 1$ , which is the exterior AND the boundary of the unit disk. Since the point $z_{0} = - \frac{3}{2} - 2 i$ has $∣ z_{0} ∣ = 2.5 > 1$ , it lies directly in the allowed region, so we can choose $z = z_{0}$ to get a distance of 0.

Answer

The minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣$ for $∣ z ∣ \geq 1$ is $0$ , because $z_{0} = - \frac{3 + 4 i}{2}$ satisfies $∣ z_{0} ∣ = \frac{5}{2} \geq 1$ .

Why each option works or fails

A: 5/2 - Calculated the modulus of $\frac{1}{2} (3 + 4 i)$ instead of finding the minimum value of the expression $∣ z - (- \frac{3 + 4 i}{2}) ∣$ . Recognize that the expression represents the distance from $z$ to $- \frac{3 + 4 i}{2}$ , which is minimized to $0$ if that point itself lies inside the domain $∣ z ∣ \geq 1$ .
B: 2 - Subtracted $1 / 2$ or applied an incorrect triangle inequality variant such as $z_{0} = - \frac{3 + 4 i}{2}$ . Check whether the point making the expression zero is inside the region $∣ z ∣ \geq 1$ before attempting to apply boundary-distance formulas.
C: 3 - Miscalculated $\frac{5}{2} + \frac{1}{2} = 3$ or confused the maximum/minimum bounds on a circle. The modulus $∣ z - z_{0} ∣$ attains its absolute minimum of $0$ when $z = z_{0}$ , provided $z_{0}$ satisfies the condition $∣ z_{0} ∣ \geq 1$ .
D · correct: 0 - The choice is correct. For $z = - \frac{3 + 4 i}{2}$ , we have $∣ z ∣ = \frac{5}{2} \geq 1$ , which makes the expression $∣ z + \frac{1}{2} (3 + 4 i) ∣ = 0$ .

Step-by-step solution

Given: A complex number $z$ satisfies $∣ z ∣ \geq 1$ . We are asked to find the minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣ = ∣ z - (- \frac{3}{2} - 2 i) ∣$ .
Goal: Find the minimum distance between $z$ (where $∣ z ∣ \geq 1$ ) and the fixed point $z_{0} = - \frac{3}{2} - 2 i$ .
Approach: Check whether the target point $z_{0} = - \frac{3}{2} - 2 i$ lies inside the region defined by $∣ z ∣ \geq 1$ . If $∣ z_{0} ∣ \geq 1$ , then $z_{0}$ itself is in the allowed region, making the distance zero.
Execute: Calculate the modulus of $z_{0}$ : $∣ z_{0} ∣ = ∣ - \frac{3}{2} - 2 i ∣ = \sqrt{{(- \frac{3}{2})}^{2} + (- 2)^{2}} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$ Since $∣ z_{0} ∣ = 2.5 \geq 1$ , the point $z = z_{0} = - \frac{3}{2} - 2 i$ belongs to the region $∣ z ∣ \geq 1$ . Therefore, the minimum value of $∣ z - z_{0} ∣$ is $0$ .
Verify: Substitute $z = - \frac{3}{2} - 2 i$ : $∣ z ∣ = \frac{5}{2} \geq 1$ is satisfied, and $∣ z + \frac{1}{2} (3 + 4 i) ∣ = 0$ , which is the absolute minimum possible value for any modulus.

Minimum distance from a point to an exterior region in the complex plane

What feels right?

The minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣$ for $∣ z ∣ \geq 1$ is $0$ , because $z_{0} = - \frac{3 + 4 i}{2}$ satisfies $∣ z_{0} ∣ = \frac{5}{2} \geq 1$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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