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Complex Numbers and Quadratic Equations: Mathematics | JEE Main

Q: Why isn't the minimum distance on the boundary ?

The condition is , which is the exterior AND the boundary of the unit disk. Since the point has , it lies directly in the allowed region, so we can choose to get a distance of 0.

z

is a complex number such that

|z| \ge 1

, then the minimum value of

\left|z + \frac{1}{2}(3+4i)\right|

is:

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\frac{5}{2}

2

3

Step-by-step solutionView

Correct answer

The minimum value of $\left|z + \frac{1}{2}(3+4i)\right|$ for $|z| \ge 1$ is $0$ , because $z_0 = -\frac{3+4i}{2}$ satisfies $|z_0| = \frac{5}{2} \ge 1$ .

Option analysis

Why each option works or fails

A ·

\frac{5}{2}

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

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 $|z| \ge 1$ 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 $|z_0| \ge 1$ .

D · 0

None. The choice is correct. For $z = -\frac{3+4i}{2}$ , we have $|z| = \frac{5}{2} \ge 1$ , which makes the expression $\left|z + \frac{1}{2}(3+4i)\right| = 0$ .

Reviewed route

Solution

StepWorking

01given

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

02goal

Find the minimum distance between $z$ (where $|z| \ge 1$ ) and the fixed point $z_0 = -\frac{3}{2} - 2i$ .

03approach

Check whether the target point $z_0 = -\frac{3}{2} - 2i$ lies inside the region defined by $|z| \ge 1$ . If $|z_0| \ge 1$ , then $z_0$ itself is in the allowed region, making the distance zero.

04execute

Calculate the modulus of $z_0$ : $|z_0| = \left|-\frac{3}{2} - 2i\right| = \sqrt{\left(-\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$ Since $|z_0| = 2.5 \ge 1$ , the point $z = z_0 = -\frac{3}{2} - 2i$ belongs to the region $|z| \ge 1$ . Therefore, the minimum value of $|z - z_0|$ is $0$ .

✓verify

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

Hints that build this answer step by step

What does the expression $\left|z + \frac{1}{2}(3+4i)\right|$ represent geometrically in the complex plane?
The distance from $z$ to the point $z_0 = -\frac{1}{2}(3+4i)$
What is the modulus of $z_0 = -\frac{1}{2}(3+4i)$ ?
$|z_0| = \frac{5}{2}$
Does the point $z_0 = -\frac{1}{2}(3+4i)$ satisfy the domain restriction $|z| \ge 1$ ?
Yes, because $|z_0| = 2.5 \ge 1$ , 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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why isn't the minimum distance on the boundary

|z| = 1

The condition is $|z| \ge 1$ , which is the exterior AND the boundary of the unit disk. Since the point $z_0 = -\frac{3}{2} - 2i$ 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.

Answer

The minimum value of $\left|z + \frac{1}{2}(3+4i)\right|$ for $|z| \ge 1$ is $|z| \ge 1$ , because $z_0 = -\frac{3+4i}{2}$ satisfies $|z_0| = \frac{5}{2} \ge 1$ .

Why each option works or fails

A: $\frac{5}{2}$ - Calculated the modulus of $\frac{1}{2}(3+4i)$ instead of finding the minimum value of the expression $\left|z - \left(-\frac{3+4i}{2}\right)\right|$ . Recognize that the expression represents the distance from $z$ to $-\frac{3+4i}{2}$ , which is minimized to $0$ if that point itself lies inside the domain $|z| \ge 1$ .
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 $|z| \ge 1$ 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 $|z_0| \ge 1$ .
D · correct: 0 - None. The choice is correct. For $z = -\frac{3+4i}{2}$ , we have $|z| = \frac{5}{2} \ge 1$ , which makes the expression $\left|z + \frac{1}{2}(3+4i)\right| = 0$ .

Step-by-step solution

given: A complex number $z$ satisfies $|z| \ge 1$ . We are asked to find the minimum value of $\left|z + \frac{1}{2}(3+4i)\right| = \left|z - \left(-\frac{3}{2} - 2i\right)\right|$ .
goal: Find the minimum distance between $z$ (where $|z| \ge 1$ ) and the fixed point $z_0 = -\frac{3}{2} - 2i$ .
approach: Check whether the target point $z_0 = -\frac{3}{2} - 2i$ lies inside the region defined by $|z| \ge 1$ . If $|z_0| \ge 1$ , then $z_0$ itself is in the allowed region, making the distance zero.
execute: Calculate the modulus of $z_0$ : $|z_0| = \left|-\frac{3}{2} - 2i\right| = \sqrt{\left(-\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$ Since $|z_0| = 2.5 \ge 1$ , the point $z = z_0 = -\frac{3}{2} - 2i$ belongs to the region $|z| \ge 1$ . Therefore, the minimum value of $|z - z_0|$ is $0$ .
verify: Substitute $z = -\frac{3}{2} - 2i$ : $|z| = \frac{5}{2} \ge 1$ is satisfied, and $\left|z + \frac{1}{2}(3+4i)\right| = 0$ , which is the absolute minimum possible value for any modulus.

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

The minimum value of $\left|z + \frac{1}{2}(3+4i)\right|$ for $|z| \ge 1$ is $0$ , because $z_0 = -\frac{3+4i}{2}$ satisfies $|z_0| = \frac{5}{2} \ge 1$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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