Among the statements (S1): The set z C - -i:|z| = 1 and (z - i/z + i) is purely real contains exactly two elements, and (S2) : The set z C - -1 : |z| = 1 and (z - 1/z + 1) is purely imaginary contains infinitely many elements.

A complex number w is purely real if w = w. A complex number w is purely imaginary if w + w = 0. Let's evaluate statement **(S1)**: If w is purely real, then w = w: Since z + z = 2Re(z) = 0, z must lie on the imaginary axis (y-axis). Given the condition |z| = 1, the only points are z = i and z = -i. However, the domain excludes z = -i. Let's test z = i: For z = i, (i - i/i + i) = 0, which is purely real. So it contains elements on the unit circle. But the condition z + z = 0 alongside |z|=1 explicitly limits it to z=i only, which is one element, not two. Thus, (S1) is incorrect. Let's evaluate statement **(S2)**: If u is purely imaginary, then u + u = 0: This condition holds true for ALL points on the unit circle |z| = 1 except z = -1 (which makes the denominator zero). Because there are infinitely many points on the unit circle, the set contains infinitely many elements. Thus, (S2) is correct. Geometric shortcut: The transformation w = (z-1/z+1) maps the unit circle |z|=1 directly onto the imaginary axis Re(w)=0. Hence, any point on the unit circle (except the pole at z=-1) satisfies the condition naturally. Class 11 Mathematics: Complex Numbers and Quadratic Equations

Among the statements (S1): The set z C - -i:|z| = 1 and (z - i/z + i) is purely real contains exactly two elements, and (S2) : The set z C - -1 : |z| = 1 and (z - 1/z + 1) is purely imaginary contains

A complex number w is purely real if w = w. A complex number w is purely imaginary if w + w = 0. Let's evaluate statement **(S1)**: If w is purely real, then w = w: Since z + z = 2Re(z) = 0, z must lie on the imaginary axis (y-axis). Given the condition |z| = 1, the only points are z = i and z = -i. However, the domain excludes z = -i. Let's test z = i: For z = i, (i - i/i + i) = 0, which is purely real. So it contains elements on the unit circle. But the condition z + z = 0 alongside |z|=1 expl

Among the statements (S1): The set z C - -i:|z| = 1 and (z - i/z + i) is purely real contains exactly two elements, and (S2) : The set z C - -1 : |z| = 1 and (z - 1/z + 1) is purely imaginary contains infinitely many elements.

A complex number w is purely real if w = w. A complex number w is purely imaginary if w + w = 0. Let's evaluate statement **(S1)**: If w is purely real, then w = w: Since z + z = 2Re(z) = 0, z must lie on the imaginary axis (y-axis). Given the condition |z| = 1, the only points are z = i and z = -i. However, the domain excludes z = -i. Let's test z = i: For z = i, (i - i/i + i) = 0, which is purely real. So it contains elements on the unit circle. But the condition z + z = 0 alongside |z|=1 explicitly limits it to z=i only, which is one element, not two. Thus, (S1) is incorrect. Let's evaluate statement **(S2)**: If u is purely imaginary, then u + u = 0: This condition holds true for ALL points on the unit circle |z| = 1 except z = -1 (which makes the denominator zero). Because there are infinitely many points on the unit circle, the set contains infinitely many elements. Thus, (S2) is correct. Geometric shortcut: The transformation w = (z-1/z+1) maps the unit circle |z|=1 directly onto the imaginary axis Re(w)=0. Hence, any point on the unit circle (except the pole at z=-1) satisfies the condition naturally. Class 11 Mathematics: Complex Numbers and Quadratic Equations

Among the statements (S1): The set z C - -i:|z| = 1 and (z - i/z + i) is purely real contains exactly two elements, and (S2) : The set z C - -1 : |z| = 1 and (z - 1/z + 1) is purely imaginary contains

A complex number w is purely real if w = w. A complex number w is purely imaginary if w + w = 0. Let's evaluate statement **(S1)**: If w is purely real, then w = w: Since z + z = 2Re(z) = 0, z must lie on the imaginary axis (y-axis). Given the condition |z| = 1, the only points are z = i and z = -i. However, the domain excludes z = -i. Let's test z = i: For z = i, (i - i/i + i) = 0, which is purely real. So it contains elements on the unit circle. But the condition z + z = 0 alongside |z|=1 expl

Among the statements (S1): The set z C - -i:|z| = 1 and (z - i/z + i) is purely real contains exactly two elements, and…

Year	2026	2025	2024	Total
Questions	11	16	14	41

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

A complex number $w$ $w$ is purely real if $w = w$ $w = \bar{w}$. A complex number $w$ $w$ is purely imaginary if $w + w = 0$ $w + \bar{w} = 0$.

Core Logic

Let's evaluate statement (S1):

w = (z - i)/(z + i)

If $w$ $w$ is purely real, then $w = w$ $w = \bar{w}$:

(z - i)/(z + i) = z + i z - i

(z - i)( z - i) = (z + i)( z + i)

|z|² - iz - i z - 1 = |z|² + iz + i z - 1

-i(z + z) = i(z + z) 2i(z + z) = 0 z + z = 0

Since $z + z = 2Re(z) = 0$ $z + \bar{z} = 2\text{Re}(z) = 0$, $z$ $z$ must lie on the imaginary axis (y-axis). Given the condition $|z| = 1$ $|z| = 1$, the only points are $z = i$ $z = i$ and $z = -i$ $z = -i$. However, the domain excludes $z = -i$ $z = -i$. Let's test $z = i$ $z = i$: For $z = i$ $z = i$, $(i - i)/(i + i) = 0$ $\frac{i - i}{i + i} = 0$, which is purely real. So it contains elements on the unit circle. But the condition $z + z = 0$ $z + \bar{z} = 0$ alongside $|z|=1$ $|z|=1$ explicitly limits it to $z=i$ $z=i$ only, which is one element, not two. Thus, (S1) is incorrect.

Step 1: Evaluate Statement S2

Let's evaluate statement (S2):

u = (z - 1)/(z + 1)

If $u$ $u$ is purely imaginary, then $u + u = 0$ $u + \bar{u} = 0$:

(z - 1)/(z + 1) + z - 1 z + 1 = 0

(z - 1)( z + 1) + (z + 1)( z - 1)(z + 1)( z + 1) = 0

(|z|² + z - z - 1) + (|z|² - z + z - 1) = 0

2|z|² - 2 = 0 |z|² = 1 |z| = 1

This condition holds true for ALL points on the unit circle $|z| = 1$ $|z| = 1$ except $z = -1$ $z = -1$ (which makes the denominator zero). Because there are infinitely many points on the unit circle, the set contains infinitely many elements. Thus, (S2) is correct.

Pattern Recognition

Geometric shortcut: The transformation $w = (z-1)/(z+1)$ $w = \frac{z-1}{z+1}$ maps the unit circle $|z|=1$ $|z|=1$ directly onto the imaginary axis $Re(w)=0$ $\text{Re}(w)=0$. Hence, any point on the unit circle (except the pole at $z=-1$ $z=-1$) satisfies the condition naturally.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

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

Among the statements (S1): The set $zin C - -i:|z| = 1 and (z - i)/(z + i) is purely real$ contains exactly two elements, and (S2) : The set $z in C - -1 : |z| = 1 and (z - 1)/(z + 1) is purely imaginary$ contains infinitely many elements.

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Statement S2

Pattern Recognition

Chapter Mix

Reference Study Guides

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

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Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Statement S2

Pattern Recognition

Chapter Mix

Reference Study Guides

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_2025_07_april_morning

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