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

Q18 jee_main_2026_28_january_evening Geometry of Complex Regions

Let

A = z in C : |z - 2| \leq 4

and

B = z in C : |z - 2| + |z + 2| = 5

. Then the max

|z₁ - z₂| : z₁ in A and z₂ in B

is

A. $(15)/(2)$ $\frac{15}{2}$
B. $8$ $8$
C. $(17)/(2)$ $\frac{17}{2}$
D. $9$ $9$

Solution

Core Logic

Analyze region $A$ $A$: $|z - 2| \leq 4$ $|z - 2| \le 4$ represents a solid disk with center $(2, 0)$ $(2, 0)$ and radius $R=4$ $R=4$. Its rightmost point on the real axis is $(6, 0)$ $(6, 0)$ and leftmost is $(-2, 0)$ $(-2, 0)$.

Analyze locus $B$ $B$: $|z - 2| + |z + 2| = 5$ $|z - 2| + |z + 2| = 5$ represents an ellipse with foci at $(2, 0)$ $(2, 0)$ and $(-2, 0)$ $(-2, 0)$, and major axis length $2a = 5 \Rightarrow a = (5)/(2)$ $2a = 5 \Rightarrow a = \frac{5}{2}$. The equation of the ellipse is $(x²)/((5/2)²) + (y²)/(b²) = 1$ . Its leftmost vertex is at $(-(5)/(2), 0)$ $(-\frac{5}{2}, 0)$.

Execution

To maximize $|z₁ - z₂|$ $|z_1 - z_2|$, we need the maximum geometric distance between any point in the disk $A$ $A$ and any point on the ellipse $B$ $B$. By visualizing the placement on the coordinate plane: The rightmost point of the circle $A$ $A$ is $z₁ = 6$ $z_1 = 6$. The leftmost point of the ellipse $B$ $B$ is $z₂ = -(5)/(2)$ $z_2 = -\frac{5}{2}$. The distance between them is the maximum horizontal span since both shapes are symmetric around the real axis and their extremes occur on the real axis.

|z₁ - z₂| = 6 - (-(5)/(2)) = 6 + (5)/(2) = (17)/(2)

Pattern Recognition

Maximum distance problems between loci in the complex plane almost always resolve to finding the extreme diametric points along their shared axis of symmetry (usually the real axis).

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Conic Sections

Q75 jee_main_2025_02_april_evening Quadratic Equations

If the set of all

a in R - 1

, for which the roots of the equation

(1 - a)x² + 2(a - 3)x + 9 = 0

are positive is

(-\infty, -α] [β, γ)

, then

2α + β + γ

is equal to ____________.

Numerical Answer. Answer: 7 to 7

Solution

Related Formula

For a quadratic equation A x² + B x + C = 0 to have two positive real roots:

1. Real roots: D = B² - 4AC \geq 0

2. Sum of roots: -(B)/(A) > 0

3. Product of roots: (C)/(A) > 0

Core Logic

We write down three systems of inequalities based on real and positive root conditions, find their intersection, and map the boundaries to solve for the parameters.

Step 1: Apply the discriminant condition (Real roots)

For real roots, the discriminant $D \geq 0$ $D \ge 0$:

D = [ 2(a - 3) ]² - 4(1 - a)(9) \geq 0

4(a² - 6a + 9) - 36(1 - a) \geq 0

(a² - 6a + 9) - 9(1 - a) \geq 0

a² - 6a + 9 - 9 + 9 a \geq 0

a² + 3a \geq 0 a(a + 3) \geq 0

Thus, the interval is:

a in (-\infty, -3] [0, \infty) --- (1)

Step 2: Apply the sum of roots condition (Positive sum)

For positive roots, the sum of roots must be positive:

-(B)/(A) = (-2(a - 3))/(1 - a) = (2(a - 3))/(a - 1) > 0

Using the wavy curve method for $(a-3)/(a-1) > 0$ $\frac{a-3}{a-1} > 0$:

a in (-\infty, 1) (3, \infty) --- (2)

Step 3: Apply the product of roots condition (Positive product)

For positive roots, the product of roots must be positive:

(C)/(A) = (9)/(1 - a) > 0 1 - a > 0 a < 1

Thus, the interval is:

a in (-\infty, 1) --- (3)

Step 4: Find the intersection of all conditions

Intersecting equations (1), (2), and (3):

First, intersect (2) and (3):

( (-\infty, 1) (3, \infty) ) (-\infty, 1) = (-\infty, 1)

Next, intersect with (1):

( (-\infty, -3] [0, \infty) ) (-\infty, 1) = (-\infty, -3] [0, 1)

Comparing this with $(-\infty, -α] [β, γ)$ :

$α = 3$ $\alpha = 3$
$β = 0$ $\beta = 0$
$γ = 1$ $\gamma = 1$

Now calculate the target sum:

2α + β + γ = 2(3) + 0 + 1 = 7

Pattern Recognition

Location of roots: When both roots are positive, checking sum and product signs along with $D \geq 0$ $D \ge 0$ is the standard and fastest set of inequalities, avoiding complex vertex projections.

Chapter Mix

Class 11 Chemistry: Practical Chemistry

Q jee_main_2025_02_april_morning Geometry of Complex Numbers

Let

z

be a complex number such that

|z| = 1

. If

2 + k²zk + z = kz, k in R

, then the maximum distance of

k + ik²

from the circle

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

is:

A. $\sqrt(5) + 1$ $\sqrt{5} + 1$
B. $2$ $2$
C. $3$ $3$
D. $\sqrt(3) + 1$ $\sqrt{3} + 1$

Solution

Related Formula

For a complex number lying on the unit circle:

|z| = 1 z z = 1 z = (1)/(z)

Maximum distance from a point $P$ $P$ to a circle with center $C$ $C$ and radius $r$ $r$ is:

d = PC + r

Core Logic

Simplify the algebraic condition using $z = 1/z$ $\overline{z} = 1/z$ to uniquely determine the value of the parameter $k$ $k$, then compute geometric distances.

Step 1: Solve for k

Cross-multiply the given expression:

2 + k²z = kz(k + z) = k²z + kz z

Since $z z = |z|² = 1$ $z\overline{z} = |z|^2 = 1$:

2 + k²z = k²z + k(1) k = 2

Step 2: Locate Point and Circle Parameters

Substitute $k=2$ $k=2$ into the target point expression $P = k + ik²$ $P = k + ik^2$:

P = 2 + 4i \equiv (2,4)

The circle equation is $|z - (1 + 2i)| = 1$ $|z - (1 + 2i)| = 1$, which represents a circle centered at $C = (1, 2)$ $C = (1, 2)$ with radius $r = 1$ $r = 1$.

Step 3: Compute Geometric Distances

Find the Euclidean distance between $P(2,4)$ $P(2,4)$ and center $C(1,2)$ $C(1,2)$:

PC = \sqrt((2-1)² + (4-2)²) = \sqrt(1 + 4) = \sqrt(5)

The maximum distance from the point to the circle boundary is:

d = PC + r = \sqrt(5) + 1

Pattern Recognition

Notice how $k²z$ $k^2z$ cancels perfectly on both sides during expansion due to the unique property of uni-modular complex numbers ( $z z=1$ $z\overline{z}=1$), rendering the calculation of $k$ $k$ trivial.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Q jee_main_2025_02_april_morning Theory of Equations

Let

Pₙ = αⁿ + βⁿ

,

n in N

. If

P₁₀ = 123

,

P₉ = 76

,

P₈ = 47

and

P₁ = 1

, then the quadratic equation having roots

(1)/(α)

and

(1)/(β)

is:

A. $x² - x + 1 = 0$ $x^2 - x + 1 = 0$
B. $x² + x - 1 = 0$ $x^2 + x - 1 = 0$
C. $x² - x - 1 = 0$ $x^2 - x - 1 = 0$
D. $x² + x + 1 = 0$ $x^2 + x + 1 = 0$

Solution

Related Formula

Newton's Sums for the roots of a quadratic equation $ax² + bx + c = 0$ $ax^2 + bx + c = 0$:

a Pₙ + b Pₙ₋₁ + c Pₙ₋₂ = 0

Core Logic

Observe the recurrence relation from the given numerical values of $Pₙ$ $P_n$ to construct the base quadratic equation satisfied by $α$ $\alpha$ and $β$ $\beta$, then invert the roots.

Step 1: Identify the Linear Recurrence Relation

Compare the provided sequence values:

P₈ + P₉ = 47 + 76 = 123 = P₁₀

This fits the general sequence relation:

Pₙ = Pₙ₋₁ + Pₙ₋₂ Pₙ - Pₙ₋₁ - Pₙ₋₂ = 0

Step 2: Construct the Base Quadratic Equation

The characteristic equation corresponding to this recurrence relation is:

$x² - x - 1 = 0$ $x^2 - x - 1 = 0$

Thus, $α$ $\alpha$ and $\beta$ are roots of $x² - x - 1 = 0$ $x^2 - x - 1 = 0$, giving sum $α+β = 1$ $\alpha+\beta = 1$ and product $αβ = -1$ $\alpha\beta = -1$ (which matches $P₁ = α+β = 1$ $P_1 = \alpha+\beta = 1$).

Step 3: Construct Equation with Reciprocal Roots

To find the equation with roots $(1)/(α)$ $\frac{1}{\alpha}$ and $(1)/(β)$ $\frac{1}{\beta}$, apply the transformations:

Sum of new roots = (1)/(α) + (1)/(β) = (α + β)/(αβ) = (1)/(-1) = -1

Product of new roots = (1)/(αβ) = (1)/(-1) = -1

The new quadratic equation is:

x² - (Sum)x + (Product) = 0 x² - (-1)x + (-1) = 0 x² + x - 1 = 0

Pattern Recognition

The recurrence pattern $Pₙ = Pₙ₋₁ + Pₙ₋₂$ $P_n = P_{n-1} + P_{n-2}$ is the Fibonacci sequence recurrence line. Its roots generate the Golden Ratio layout from $x²-x-1=0$ $x^2-x-1=0$. Inverting roots swaps the coefficients of $x²$ $x^2$ and the constant term, yielding $x²+x-1=0$ $x^2+x-1=0$ instantly.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 11 Mathematics: Sequences and Series

Q65 jee_main_2025_03_april_evening Geometry of Complex Numbers

If

z₁, z₂, z₃ in C

are the vertices of an equilateral triangle, whose centroid is

z₀

, then

Σk=1³ (zk - z₀)²

is equal to

A. $0$ $0$
B. $1$ $1$
C. $i$ $i$
D. $-i$ $-i$

Solution

Related Formula

For any equilateral triangle with vertices $z₁, z₂, z₃$ $z_1, z_2, z_3$:

z₁² + z₂² + z₃² = z₁ z₂ + z₂ z₃ + z₃ z₁ --- (1)

Centroid equation:

z₀ = (z₁ + z₂ + z₃)/(3) z₁ + z₂ + z₃ = 3z₀

Core Logic

Let's expand the target sum:

Σk=1³ (zk - z₀)² = (z₁ - z₀)² + (z₂ - z₀)² + (z₃ - z₀)²

Step 1: Expansion and Algebraic Grouping

Expanding each quadratic term:

= (z₁² + z₂² + z₃²) - 2z₀(z₁ + z₂ + z₃) + 3z₀²

Substitute $z₁ + z₂ + z₃ = 3z₀$ $z_1 + z_2 + z_3 = 3z_0$:

= (z₁² + z₂² + z₃²) - 2z₀(3z₀) + 3z₀²

= z₁² + z₂² + z₃² - 3z₀²

Step 2: Resolving using Equilateral Condition

Substitute $3z₀² = 3 ((z₁+z₂+z₃)/(3))² = ((z₁+z₂+z₃)²)/(3)$ :

= (z₁² + z₂² + z₃²) - (z₁² + z₂² + z₃² + 2(z₁z₂ + z₂z₃ + z₃z₁))/(3)

= (2(z₁² + z₂² + z₃²) - 2(z₁z₂ + z₂z₃ + z₃z₁))/(3)

= (2)/(3)(z₁² + z₂² + z₃² - (z₁ z₂ + z₂ z₃ + z₃ z₁))

Using condition (1) for equilateral triangles, the terms inside the parentheses equal $0$ $0$. Thus: $= 0$ $= 0$

Pattern Recognition

This is a standard invariant of equilateral triangles. Any translation to center of mass coordinates leaves the shape invariant, making the sum of squares of coordinate vectors relative to the centroid equal to zero.

Chapter Mix

Class 11 Mathematics: Complex Numbers

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 3

Solution

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply the discriminant condition (Real roots)

Step 2: Apply the sum of roots condition (Positive sum)

Step 3: Apply the product of roots condition (Positive product)

Step 4: Find the intersection of all conditions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for k

Step 2: Locate Point and Circle Parameters

Step 3: Compute Geometric Distances

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Identify the Linear Recurrence Relation

Step 2: Construct the Base Quadratic Equation

Step 3: Construct Equation with Reciprocal Roots

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Expansion and Algebraic Grouping

Step 2: Resolving using Equilateral Condition

Pattern Recognition

Chapter Mix

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