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

Q17 jee_main_2026_21_jan_morning Cube Roots of Unity

If

x² + x + 1 = 0

, then the value of

(x+ 1x)⁴+(x²+ 1x²)⁴+(x³+ 1x³)⁴+\dots+(x²⁵+ 1x²⁵)⁴

is :

A. 128
B. 162
C. 175
D. 145

Solution

Related Formula

x² + x + 1 = 0 \Rightarrow x = ω, ω²

Properties of cube roots of unity: $ω³ = 1$ $\omega^3 = 1$ and $1 + ω + ω² = 0$ $1 + \omega + \omega^2 = 0$.

Core Logic

Let $α = ω$ $\alpha = \omega$. Then $(1)/(x) = (1)/(ω) = ω²$ . The series is $Σk=1²⁵ (ω^k + ω2k)⁴$ . Evaluate the term $Tk = (ω^k + ω2k)⁴$ $T_k = (\omega^k + \omega^{2k})^4$ based on the modulo of $k$ $k$ with 3.

Step 1: Cyclic Evaluation

Case 1: $k = 3m$ $k = 3m$ (multiples of 3)

T3m = (ω3m + ω6m)⁴ = (1 + 1)⁴ = 2⁴ = 16

There are $8$ $8$ such multiples up to 25 ( $3, 6, 9, , 24$ $3, 6, 9, \dots, 24$).

Case 2: $k \neq 3m$ $k \neq 3m$ (non-multiples of 3) For $k = 1, 2, 4, 5,$ $k = 1, 2, 4, 5, \dots$ $ω^k + ω2k$ $\omega^k + \omega^{2k}$ will always be $ω + ω²$ $\omega + \omega^2$ or $ω² + ω$ $\omega^2 + \omega$. Since $1 + ω + ω² = 0$ $1 + \omega + \omega^2 = 0$, we have $ω + ω² = -1$ $\omega + \omega^2 = -1$.

Tk \neq 3m = (-1)⁴ = 1

There are $25 - 8 = 17$ $25 - 8 = 17$ such non-multiples up to 25.

Step 2: Total Sum

Sum = 17 \times 1 + 8 \times 16

= 17 + 128 = 145

Pattern Recognition

Powers of $x+1/x$ $x+1/x$ when $x$ $x$ solves $x² \pm x + 1 = 0$ $x^2 \pm x + 1 = 0$ perfectly orbit around periods of 3 or 6. Isolate the $3m$ $3m$ resonant beats (which hit pure scalars like $1+1=2$ $1+1=2$) versus the out-of-phase beats (which collapse to $-1$ $-1$ or $1$ $1$ via basic $ω$ $\omega$ identities).

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations

Q19 jee_main_2026_21_jan_evening Geometry of Complex Numbers

Let

z

be the complex number satisfying

|z-5|\leq 3

and having maximum positive principal argument. Then

34|(5z-12)/(5iz+16)|²

is equal to:

A. $16$ $16$
B. $12$ $12$
C. $26$ $26$
D. $20$ $20$

Solution

Related Formula

For maximum argument of z on a circle, the ray from origin is tangent to the circle.

If |z-a| \leq r, maximum argument implies θ = (r)/(|a|) and coordinates are x=a ²θ, y=a θ θ

Core Logic

Complex geometry maximum argument diagram for Q19 - JEE Main 2026 Evening

The condition

|z-5|\leq 3

represents a disk centered at

(5,0)

with radius

3

. To maximize the principal argument

θ

, the ray from origin must touch the circle in the first quadrant. The tangent, origin, and center form a right-angled triangle.

Step 1: Locate the Point z

From geometry, hypotenuse $c = 5$ $c = 5$, opposite (radius) $r = 3$ $r = 3$. The adjacent side (length of tangent) is $\sqrt(5² - 3²) = 4$ $\sqrt{5^2 - 3^2} = 4$. The angle of tangency $θ$ $\theta$ satisfies $θ = (3)/(5)$ $\sin\theta = \frac{3}{5}$ and $θ = (4)/(5)$ $\cos\theta = \frac{4}{5}$. The point $P(z)$ $P(z)$ lies on the circle and the tangent ray:

z \equiv (4 θ, 4 θ) = (4((4)/(5)), 4((3)/(5))) = ((16)/(5), (12)/(5))

z = (16)/(5) + (12)/(5)i

Step 2: Evaluate the Target Expression

We need $34|(5z - 12)/(5iz + 16)|²$ . Substitute $5z = 16 + 12i$ $5z = 16 + 12i$: Numerator: $5z - 12 = 16 + 12i - 12 = 4 + 12i$ $5z - 12 = 16 + 12i - 12 = 4 + 12i$ Denominator: $5iz + 16 = i(16 + 12i) + 16 = 16i - 12 + 16 = 4 + 16i$ Expression:

34 |(4 + 12i)/(4 + 16i)|² = 34 (|4 + 12i|²)/(|4 + 16i|²) = 34 (4² + 12²)/(4² + 16²)

= 34 ((16 + 144)/(16 + 256)) = 34 ((160)/(272))

= (34 \times 160)/(272) = (5440)/(272) = 20

Pattern Recognition

For maximum $(z)$ $\arg(z)$ on $|z-c| = r$ $|z-c| = r$ where $c$ $c$ is real, $z$ $z$ coordinates are given by geometric projection: $z = \sqrt(c²-r²)( θ + i θ)$ where $θ = r/c$ $\sin\theta = r/c$.

Chapter Mix

Class 11 Maths: Complex Numbers

Q21 jee_main_2026_22_january_morning Cube Roots of Unity

Let

α = -1 + i\sqrt(3)2

and

β = -1 - i\sqrt(3)2,i = \sqrt(-1)

. If

(7 - 7α +9β)²⁰ + (9 + 7α -7β)²⁰ + (-7 + 9α +7β)²⁰ + (14 + 7α +7β)²⁰ = m¹⁰

, then m is ____.

Numerical Answer. Answer: 49 to 49

Solution

Related Formula

α = ω, β = ω²

1 + ω + ω² = 0 ω³ = 1

Core Logic

Substitute $ω$ $\omega$ and $ω²$ $\omega^2$ into the terms:

Let $T₁ = 7 - 7ω + 9ω²$ $T_1 = 7 - 7\omega + 9\omega^2$ Let $T₂ = 9 + 7ω - 7ω²$ $T_2 = 9 + 7\omega - 7\omega^2$ Let $T₃ = -7 + 9ω + 7ω²$ $T_3 = -7 + 9\omega + 7\omega^2$ Let $T₄ = 14 + 7ω + 7ω²$ $T_4 = 14 + 7\omega + 7\omega^2$

Step 1: Simplify Each Term Using Properties of Omega

Notice relationships between the terms by factoring powers of $ω$ $\omega$:

T₁ = ω \cdot ω²(7 - 7ω + 9ω²) wait, an easier path is factoring

Observe that:

ω \cdot T₂ = ω(9 + 7ω - 7ω²) = 9ω + 7ω² - 7 = T₃

ω² \cdot T₂ = ω²(9 + 7ω - 7ω²) = 9ω² + 7 - 7ω = T₁

So, $T₁²⁰ + T₂²⁰ + T₃²⁰ = (ω² T₂)²⁰ + T₂²⁰ + (ω T₂)²⁰$

= ω⁴⁰ T₂²⁰ + T₂²⁰ + ω²⁰ T₂²⁰ = T₂²⁰ (1 + ω⁴⁰ + ω²⁰)

Since $ω³ = 1$ $\omega^3 = 1$, we have $ω⁴⁰ = ω$ $\omega^{40} = \omega$ and $ω²⁰ = ω²$ $\omega^{20} = \omega^2$.

Thus, $T₂²⁰ (1 + ω + ω²) = T₂²⁰ (0) = 0$ .

Step 2: Evaluating the Non-Zero Term

The only remaining term is $T₄²⁰$ $T_4^{20}$:

T₄ = 14 + 7(ω + ω²)

Since $ω + ω² = -1$ $\omega + \omega^2 = -1$,

T₄ = 14 + 7(-1) = 14 - 7 = 7

So the entire expression sum is $0 + 7²⁰ = 7²⁰$ $0 + 7^{20} = 7^{20}$.

Step 3: Finding m

Given that the sum equals $m¹⁰$ $m^{10}$:

m¹⁰ = 7²⁰ = (7²)¹⁰ = 49¹⁰

Hence, $m = 49$ $m = 49$.

Cube Roots of Unity diagram for Q21 - JEE Main 2026 Morning

Pattern Recognition

Complex expressions with $ω$ $\omega$ raised to large powers almost always exploit rotational symmetry. If coefficients are permuted circularly $(A, B, C) \to (C, A, B) \to (B, C, A)$ $(A, B, C) \to (C, A, B) \to (B, C, A)$, multiplying by $ω$ $\omega$ maps them to one another, meaning their sum of $3n$ $3n$ or symmetric powers identically nullifies.

Chapter Mix

Class 11 Maths: Complex Numbers

Q4 jee_main_2026_22_january_evening Complex Equations and Modulus

Let

S = z in C : 4z² + z = 0

. Then

Σz in S |z|²

is equal to:

A. $(3)/(16)$ $\frac{3}{16}$
B. $(7)/(64)$ $\frac{7}{64}$
C. $(1)/(16)$ $\frac{1}{16}$
D. $(5)/(64)$ $\frac{5}{64}$

Solution

Related Formula

For $z = x + iy$ $z = x + iy$, $z = x - iy$ $\overline{z} = x - iy$ and $|z|² = x² + y²$ $|z|^2 = x^2 + y^2$.

Core Logic

Substitute $z = x + iy$ $z = x + iy$ into $4z² + z = 0$ $4z^2 + \overline{z} = 0$:

4(x² - y² + 2ixy) + x - iy = 0

Separating real and imaginary parts:

4x² - 4y² + x = 0 and y(8x - 1) = 0

Case 1: $y = 0 4x² + x = 0 x = 0$ or $x = -1/4$ $x = -1/4$. Solutions: $z₁ = 0 |z₁|² = 0$ $z_1 = 0 \implies |z_1|^2 = 0$, $z₂ = -1/4 |z₂|² = 1/16$ $z_2 = -1/4 \implies |z_2|^2 = 1/16$.

Case 2: $x = 1/8 4(1/64) - 4y² + 1/8 = 0 4y² = 3/16 y = \pm \sqrt(3)8$ . Solutions: $z₃, z₄ = (1)/(8) \pm i \sqrt(3)8 |z₃|² = |z₄|² = (1)/(64) + (3)/(64) = (1)/(16)$ .

Step 1: Sum of Modulus Squared

Σz in S |z|² = 0 + (1)/(16) + (1)/(16) + (1)/(16) = (3)/(16)

Pattern Recognition

Separate complex equation into real and imaginary components to systematically find all roots.

Chapter Mix

Class 11 Maths: Complex Numbers

Q17 jee_main_2026_23_january_morning Geometry of Complex Numbers

Let

S = z : 3 \leq |2z - 3(1 + i)| \leq 7

be a set of complex numbers. Then

z in S | ( z + (1)/(2)(5 + 3i) ) |

is equal to:

A. $(1)/(2)$ $\frac{1}{2}$
B. $(3)/(2)$ $\frac{3}{2}$
C. $2$ $2$
D. $(5)/(2)$ $\frac{5}{2}$

Solution

Core Logic

Simplify the condition for set $S$ $S$:

3 \leq |2(z - (3)/(2)(1 + i))| \leq 7

(3)/(2) \leq |z - (3)/(2)(1 + i)| \leq (7)/(2)

This represents an annular region bounded by two concentric circles centered at $C = (3)/(2) + (3)/(2)i$ $C = \frac{3}{2} + \frac{3}{2}i$ with radii $r₁ = (3)/(2)$ $r_1 = \frac{3}{2}$ and $r₂ = (7)/(2)$ $r_2 = \frac{7}{2}$.

Geometry of Complex Numbers diagram for Q17 - JEE Main 2026 Morning

Step 1: Identify the Target Point

We need to minimize $| z - ( -(5)/(2) - (3)/(2)i ) |$ . Let $P$ $P$ be the point $-(5)/(2) - (3)/(2)i$ $-\frac{5}{2} - \frac{3}{2}i$. The expression represents the distance from point $P$ $P$ to a point $z$ $z$ in the set $S$ $S$.

Step 2: Distance from Center to P

Calculate the distance $PC$ $PC$ between the center of the circles $C((3)/(2), (3)/(2))$ and the point $P(-(5)/(2), -(3)/(2))$ :

PC = \sqrt(( (3)/(2) - (-(5)/(2)) )² + ( (3)/(2) - (-(3)/(2)) )²)

PC = \sqrt(( (8)/(2) )² + ( (6)/(2) )²) = \sqrt(4² + 3²) = \sqrt(16 + 9) = \sqrt(25) = 5

Since $5 > (7)/(2)$ $5 > \frac{7}{2}$, the point $P$ $P$ lies outside the outer circle.

Step 3: Minimum Distance Calculation

The shortest distance from an external point to an annular region is the distance to the outer boundary along the line connecting the point to the center.

z in S |z - P| = PC - router

Minimum Distance = 5 - (7)/(2) = (10 - 7)/(2) = (3)/(2)

Pattern Recognition

Transforming $|az - b|$ $|az - b|$ by factoring out $a$ $a$ immediately reveals the true geometric center. Shortest distance to any ring/circle from an external point is always collinear with the center: $d - Router$ $d - R_{outer}$.

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Statement S2

Pattern Recognition

Chapter Mix

Reference Study Guides

More Complex Numbers Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Cyclic Evaluation

Step 2: Total Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Locate the Point z

Step 2: Evaluate the Target Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Each Term Using Properties of Omega

Step 2: Evaluating the Non-Zero Term

Step 3: Finding m

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Sum of Modulus Squared

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Identify the Target Point

Step 2: Distance from Center to P

Step 3: Minimum Distance Calculation

Pattern Recognition

Chapter Mix

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