JEE Main 2024 — 9 April, Morning Shift. Previous Year Question.
Problem
The sum of the square of the modulus of the elements in the set $\{z=a+ib : a,b\in\mathbb Z,\ z\in\mathbb C,\ |z-1|\leq1,\ |z-5|\leq|z-5i|\}$ is:
Key insight. $|z-1|\le1$ describes a disc, and $|z-5|\le|z-5i|$ — “closer to $5$ than to $5i$” — describes a half-plane, since the boundary of equal distance to two fixed points is always a straight line. Intersecting a disc with finitely many integer points against a half-plane just means checking each integer point in the disc individually.
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Approach
Translate $|z-1|\leq1$ into a disc equation in $(a,b)$, and $|z-5|\leq|z-5i|$ into a linear inequality by squaring and simplifying. List all integer points $(a,b)$ satisfying the disc equation, check each against the linear inequality, and sum $a^2+b^2$ over the surviving points.
Solution
Step 1 — Translate |z − 1| ≤ 1 into a disc equation
With $z=a+ib$:
$$(a-1)^2+b^2 \leq 1$$
This is a closed disc centred at $(1,0)$ with radius $1$.
Step 2 — Translate |z − 5| ≤ |z − 5i| into a linear inequality
$$\sqrt{(a-5)^2+b^2} \leq \sqrt{a^2+(b-5)^2}$$
Squaring both sides:
$$(a-5)^2+b^2 \leq a^2+(b-5)^2$$
$$a^2-10a+25+b^2 \leq a^2+b^2-10b+25$$
$$-10a \leq -10b \implies a\geq b$$
Step 3 — List all integer points inside or on the disc
Since the disc is centred at $(1,0)$ with radius $1$, $a$ ranges over $\{0,1,2\}$ and $b$ over $\{-1,0,1\}$. Checking $(a-1)^2+b^2\leq1$ for each combination:
$$(1,0):\ 0\leq1\ \checkmark \qquad (0,0):\ 1\leq1\ \checkmark \qquad (2,0):\ 1\leq1\ \checkmark$$ $$(1,1):\ 1\leq1\ \checkmark \qquad (1,-1):\ 1\leq1\ \checkmark$$
(all other combinations like $(0,1)$ give $1+1=2>1$, failing)
Step 4 — Apply a ≥ b to filter the list
$$(1,0):\ 1\geq0\ \checkmark \qquad (0,0):\ 0\geq0\ \checkmark \qquad (2,0):\ 2\geq0\ \checkmark$$ $$(1,1):\ 1\geq1\ \checkmark \qquad (1,-1):\ 1\geq-1\ \checkmark$$
All five points survive both conditions.
Step 5 — Sum the squares of the modulus (a² + b²) over all surviving points
$$(1,0):\ 1+0=1 \qquad (0,0):\ 0+0=0 \qquad (2,0):\ 4+0=4$$ $$(1,1):\ 1+1=2 \qquad (1,-1):\ 1+1=2$$
Step 6 — Add them up
$$1+0+4+2+2 = 9$$
Answer
$$9$$
Common mistakes
- Squaring the inequality $|z-5|\leq|z-5i|$ incorrectly, or forgetting the direction flips if both sides were negative. Here both sides are non-negative distances, so squaring preserves the inequality direction safely — but this should always be checked before squaring any inequality.
- Missing integer points near the disc’s boundary. Since the radius is exactly $1$, points on the boundary (like $(0,0)$, $(2,0)$, $(1,1)$, $(1,-1)$) satisfy the condition with equality and must be included — only points strictly outside are excluded.
Practise next
- Find the sum of the squares of the modulus of elements in $\{z=a+ib : a,b\in\mathbb Z,\ |z-2|\leq1,\ |z-6|\leq|z-6i|\}$, using the same disc-and-half-plane approach.
Show answer
$24$. The disc $|z-2|\le1$ contains exactly five points with integer coordinates: $(1,0)$, $(2,0)$, $(3,0)$, $(2,1)$ and $(2,-1)$.
The second condition, $|z-6|\le|z-6i|$, squares to $(a-6)^2+b^2\le a^2+(b-6)^2$, which simplifies to $a\ge b$ — the half-plane on or below $y=x$.
All five points satisfy $a\ge b$, so the sum of $|z|^2$ is $1+4+9+5+5=24$.
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.