If S=z C:|z-i|=|z+i|=|z-1|, then n(S) is: — jee_main_2024_27_jan

Q: If S=z C:|z-i|=|z+i|=|z-1|, then n(S) is:

### Related Formula This represents the perpendicular bisector of the line segment joining the points z1 and z2 in the complex plane. ### Core Logic The given set defines a complex number z that is equidistant from three fixed points: A equiv (0, 1) corresponding to i B equiv (0, -1) corresponding to -i C equiv (1, 0) corresponding to 1 The condition |z-i|=|z+i|=|z-1| implies that z is the point of intersection of the perpendicular bisectors of the sides of the triangle formed by A, B, and C. ##

Year	2026	2025	2024	Total
Questions	11	16	14	41

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

|z - z₁| = |z - z₂|

This represents the perpendicular bisector of the line segment joining the points $z₁$ $z_1$ and $z₂$ $z_2$ in the complex plane.

Core Logic

The given set defines a complex number $z$ $z$ that is equidistant from three fixed points: $A \equiv (0, 1)$ $A \equiv (0, 1)$ corresponding to $i$ $i$ $B \equiv (0, -1)$ $B \equiv (0, -1)$ corresponding to $-i$ $-i$ $C \equiv (1, 0)$ $C \equiv (1, 0)$ corresponding to $1$ $1$

The condition $|z-i|=|z+i|=|z-1|$ $|z-i|=|z+i|=|z-1|$ implies that $z$ $z$ is the point of intersection of the perpendicular bisectors of the sides of the triangle formed by $A$ $A$, $B$ $B$, and $C$ $C$.

Step 1: Finding the Circumcenter

The point of intersection of the perpendicular bisectors of a triangle is its circumcenter. Since $A(0,1)$ $A(0,1)$, $B(0,-1)$ $B(0,-1)$, and $C(1,0)$ $C(1,0)$ form a unique, non-degenerate triangle, they have exactly one unique circumcenter.

Step 2: Final Conclusion

Therefore, there is only one such complex number $z$ $z$ that satisfies the condition. $n(S) = 1$ $n(S) = 1$

Pattern Recognition

Recognize that $|z - z₁| = |z - z₂| = |z - z₃|$ $|z - z_1| = |z - z_2| = |z - z_3|$ is geometrically identical to finding the circumcenter of a triangle with vertices at $z₁$ $z_1$, $z₂$ $z_2$, and $z₃$ $z_3$. A non-collinear set of three points always yields exactly 1 circumcenter.

Chapter Mix

Class 11 Maths: Complex Numbers Class 11 Maths: Straight Lines

More Complex Numbers Previous-Year Questions — Page 9

Q26 jee_main_2024_31_jan_morning Properties of Modulus and Argument

α

denotes the number of solutions of

|1 - i|^x = 2^x

and

β = ((|z|)/( (z)))

, where

z = (π)/(4) (1 + i)⁴ ( 1 - \sqrt(π) i\sqrt(π) + i + \sqrt(π) - i1 + \sqrt(π) i), i = \sqrt(-1)

, then the distance of the point

(α, β)

from the line

4x - 3y = 7

Numerical Answer. Answer: 3 to 3

Solution

Core Logic

|1 - i|^x = 2^x (\sqrt(2))^x = 2^x 2x/2 = 2^x

This implies $(x)/(2) = x x = 0$ $\frac{x}{2} = x \implies x = 0$. There is exactly 1 solution, so $α = 1$ $\alpha = 1$.

Step 1: Simplify complex number z

(1+i)⁴ = ((1+i)²)² = (1 + i² + 2i)² = (2i)² = -4

Thus, $z = -π ( (1-\sqrt(π)i)(\sqrt(π)-i)π + 1 + (\sqrt(π)-i)(1-\sqrt(π)i)1 + π )$

Step 2: Simplify Bracket

Let's expand the terms directly: $z = (π)/(4)(-4) [ \sqrt(π) - π i - i - \sqrt(π)π + 1 + \sqrt(π) - i - π i - \sqrt(π)1 + π ]$

= -π [ (-i(π+1))/(π+1) + (-i(π+1))/(π+1) ]

= -π [ -i - i ] = 2π i

Step 3: Find beta

For $z = 2π i$ $z = 2\pi i$: $|z| = 2π$ $|z| = 2\pi$ and $(z) = (π)/(2)$ $\arg(z) = \frac{\pi}{2}$.

β = (|z|)/( (z)) = (2π)/(π/2) = 4

Step 4: Distance from Line

Distance of point $(α, β) = (1, 4)$ $(\alpha, \beta) = (1, 4)$ from the line $4x - 3y - 7 = 0$ $4x - 3y - 7 = 0$:

D = |4(1) - 3(4) - 7|\sqrt(4² + (-3)²) = (|4 - 12 - 7|)/(5) = (|-15|)/(5) = 3

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding the Circumcenter

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

More Complex Numbers Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Simplify complex number z

Step 2: Simplify Bracket

Step 3: Find beta

Step 4: Distance from Line

Chapter Mix

More Complex Numbers Questions — jee_main_2024_27_jan_morning

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding the Circumcenter

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

More Complex Numbers Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Simplify complex number z

Step 2: Simplify Bracket

Step 3: Find beta

Step 4: Distance from Line

Chapter Mix

More Complex Numbers Questions — jee_main_2024_27_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

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Exam Questions

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