If S = {z ∈ C : |z − i| = |z + i| = |z − 1|}, Then n(S) Is

JEE Main 2024 — 27 January, Morning Shift. Previous Year Question.

Problem

If $S = \{z\in\mathbb{C} : |z-i|=|z+i|=|z-1|\}$, then $n(S)$ is:

Key insight. A point equidistant from two fixed points lies on their perpendicular bisector — a standard fact. Since the condition here requires equidistance from three points at once ($i$, $-i$, and $1$), $z$ must lie on the perpendicular bisector of every pair — meaning $z$ is the circumcenter of the triangle formed by these three points, and there’s exactly one such point (unless the three points are collinear).

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Approach

Substitute $z=x+iy$ and translate each pair of equal moduli into an equation. Use $|z-i|=|z+i|$ to get one equation, and $|z-i|=|z-1|$ to get another. Solve the resulting system to find the unique point (or points) satisfying both simultaneously.

Solution

Step 1 — Substitute z = x + iy and set up |z − i| = |z + i|

$$|x+i(y-1)| = |x+i(y+1)|$$

Squaring both sides:

$$x^2+(y-1)^2 = x^2+(y+1)^2$$

$$y^2-2y+1 = y^2+2y+1 \implies -4y=0 \implies y=0$$

Step 2 — Set up |z − i| = |z − 1|

$$x^2+(y-1)^2 = (x-1)^2+y^2$$

Expanding:

$$x^2+y^2-2y+1 = x^2-2x+1+y^2 \implies -2y = -2x \implies y=x$$

Step 3 — Combine both equations

$$y=0 \quad \text{and} \quad y=x \implies x=0,\ y=0$$

Step 4 — Verify z = 0 satisfies all three original conditions

$$|0-i|=1, \qquad |0+i|=1, \qquad |0-1|=1$$

All three are equal, confirming $z=0$ works.

Step 5 — Conclude there is exactly one solution

Since the system $y=0$ and $y=x$ has exactly one solution, $S=\{0\}$, so:

$$n(S)=1$$

Answer

Common mistakes

• Only using one of the two possible equal-modulus pairings. The condition $|z-i|=|z+i|=|z-1|$ is really two independent equations (e.g. $|z-i|=|z+i|$ and $|z-i|=|z-1|$) — using just one leaves an entire line of solutions instead of the single intersection point.
• Assuming the answer is always $1$ for “equidistant from three points.” This only holds when the three given points are not collinear — if they were, no single point could be equidistant from all three (or infinitely many could, if two of the points coincided), giving $n(S)=0$ or a different count entirely.

Practise next

• Find $n(S)$ where $S=\{z\in\mathbb{C} : |z-2i|=|z+2i|=|z-2|\}$, using the same perpendicular-bisector approach.