JEE Main 2024 — 1 February, Evening Shift. Previous Year Question.
Problem
If $z$ is a complex number such that $|z|\leq1$, then the minimum value of $\left|z+\dfrac{1}{2}(3+4i)\right|$ is:
Key insight. $\left|z-w\right|$ is just the distance between the points $z$ and $w$ in the complex plane. Since $|z|\leq1$ restricts $z$ to the closed unit disk, the expression asks: what’s the shortest distance from any point in that disk to the fixed point $w=-\dfrac{1}{2}(3+4i)$? That’s a pure geometry question, not an algebra one.
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Approach
Rewrite the expression as the distance from $z$ to the fixed point $w = -\dfrac{1}{2}(3+4i)$. Find the distance from the origin (the disk’s centre) to $w$, and compare it to the disk’s radius $1$. Since $w$ lies outside the disk, the minimum distance from any point in the disk to $w$ is simply that distance minus the radius.
Solution
Step 1 — Identify the fixed point
$$\left|z+\frac{1}{2}(3+4i)\right| = \left|z-\left(-\frac{3}{2}-2i\right)\right|$$
So this is the distance from $z$ to the fixed point $w=\left(-\dfrac{3}{2},-2\right)$.
Step 2 — Find the distance from the origin to w
$$|w| = \sqrt{\left(\frac{3}{2}\right)^2+2^2} = \sqrt{\frac{9}{4}+4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$
Step 3 — Compare with the disk’s radius
The set $|z|\leq1$ is the closed disk of radius $1$ centred at the origin. Since $|w|=\dfrac{5}{2}>1$, the point $w$ lies outside this disk.
Step 4 — Apply the minimum-distance rule
When a fixed point lies outside a disk, the closest point inside the disk to it lies along the straight line joining the centre to that fixed point — at the boundary of the disk. So the minimum distance from any $z$ in the disk to $w$ is:
$$|w| – \text{radius} = \frac{5}{2} – 1 = \frac{3}{2}$$
Answer
$$\frac{3}{2}$$
Common mistakes
- Computing $|w|$ but forgetting to subtract the radius of the disk. $|w|=\dfrac{5}{2}$ itself is only the distance to the centre, not the minimum distance to the disk — the closest point of the disk is $1$ unit closer, along the line to the centre.
- Assuming the minimum occurs at $z=0$. Since $w$ is outside the disk and the disk is centred at the origin, the closest point of the disk to $w$ is on the disk’s boundary, in the direction of $w$ — not at the centre itself.
Practise next
- If $|z|\leq2$, find the minimum value of $|z-(6+8i)|$, using the same distance-to-a-disk reasoning.
Show answer
$8$. The points $z$ form a closed disk of radius $2$ about the origin, and $|z-(6+8i)|$ is the distance from $z$ to the fixed point $6+8i$.
That point is at distance $|6+8i|=\sqrt{36+64}=10$ from the centre, so it lies outside the disk. The nearest point of the disk is on the segment joining the centre to it.
So the minimum is $10-2=8$, attained at $z=\tfrac{2}{10}(6+8i)=1.2+1.6i$.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.