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

Q54 jee_main_2025_28_jan_evening Complex Roots of Quadratic Equations

If

α+iβ

and

γ+iδ

are the roots of

x²-(3-2i)x-(2i-2)=0,

i=\sqrt(-1)

then

αγ+βδ

is equal to :

A. $6$ $6$
B. $2$ $2$
C. $-2$ $-2$
D. $-6$ $-6$

Solution

Related Formula

For a quadratic equation $Ax² + Bx + C = 0$ $Ax^2 + Bx + C = 0$, roots can be obtained via the quadratic formula:

x = -B \pm \sqrt(B² - 4AC)2A

Core Logic

Given quadratic equation:

x²-(3-2i)x-(2i-2)=0

Using the quadratic formula where $A=1, B=-(3-2i), C=-(2i-2)$ $A=1, B=-(3-2i), C=-(2i-2)$:

x = (3-2i) \pm \sqrt((3-2i)² - 4(1)(-(2i-2)))2

Step 1: Simplify the Discriminant

Discriminant D = (3-2i)² + 4(2i-2)

D = (9 - 4 - 12i) + (8i - 8)

D = 5 - 12i + 8i - 8 = -3 - 4i

We need to find $\sqrt(-3-4i)$ $\sqrt{-3-4i}$. Let it be written as a perfect square:

-3-4i = 1 - 4 - 4i = 1² + (2i)² - 2(1)(2i) = (1-2i)²

Thus, $\sqrt(D) = \pm(1-2i)$ $\sqrt{D} = \pm(1-2i)$.

Step 2: Find the Roots

$Boxedx = ((3-2i) \pm (1-2i))/(2)$

Case 1 (+ sign):

x₁ = (3 - 2i + 1 - 2i)/(2) = (4 - 4i)/(2) = 2 - 2i

Case 2 (- sign):

x₂ = (3 - 2i - 1 + 2i)/(2) = (2)/(2) = 1 + 0i

Let the roots be $α + iβ = 2 - 2i α=2, β=-2$ and $γ + iδ = 1 + 0i γ=1, δ=0$

Step 3: Evaluate Target Expression

αγ + βδ = (2)(1) + (-2)(0) = 2

Pattern Recognition

Always try to express the complex number under the square root in the form $(a + bi)²$ $(a + bi)^2$ by matching the imaginary part $2ab = -4i ab = -2$ $2ab = -4i \implies ab = -2$, and $a² - b² = -3$ $a^2 - b^2 = -3$. This avoids long calculations.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Q jee_main_2025_29_jan_morning Geometry of Complex Numbers

Let

|z₁ - 8 - 2i| \leq 1

and

|z₂ - 2 + 6i| \leq 2

,

z₁, z₂ in C

. Then the minimum value of

|z₁ - z₂|

is:

A. 3
B. 7
C. 13
D. 10

Solution

Related Formula

Minimum distance between two circles: d = C₁C₂ - r₁ - r₂

Core Logic

The expressions define two circular disc fields in the complex plane: Circle 1: Center $C₁(8, 2)$ $C_1(8, 2)$, radius $r₁ = 1$ $r_1 = 1$ Circle 2: Center $C₂(2, -6)$ $C_2(2, -6)$, radius $r₂ = 2$ $r_2 = 2$

Geometry of Complex Numbers diagram for Q69 - JEE Main 2025 Morning

Step 1: Calculate Center Distance

Using coordinate distance formulation:

C₁C₂ = \sqrt((8 - 2)² + (2 - (-6))²) = \sqrt(6² + 8²) = 10

Step 2: Find Minimum Separation

|z₁ - z₂| = C₁C₂ - r₁ - r₂ = 10 - 1 - 2 = 7

Pattern Recognition

Always interpret modulus circle properties geometrically rather than algebraically. Disconnecting complex plane variables into simple 2D analytical geometry centers avoids calculation mistakes entirely.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Coordinate Geometry

Q jee_main_2024_01_february_morning Geometry of Complex Numbers

Let

P=zin C:|z+2-3i|\leq1

and

Q=zin C:z(1+i)+ z(1-i)\leq-8

. Let in

P Q, |z-3+2i|

be maximum and minimum at

z₁

and

z₂

respectively. If

|z₁|²+2|z₂|²=α+β\sqrt(2)

, where

α, β

are integers, then

α+β

equals

Numerical Answer. Answer: 36 to 36

Solution

Related Formula

For a complex coordinate transformation, substituting $z = x + iy$ $z = x + iy$ and its conjugate $z = x - iy$ $\overline{z} = x - iy$ maps a complex condition directly into rectangular Cartesian coordinates.

Core Logic

Let's translate the complex set properties into Cartesian geometry:

Set $P$ $P$: $|z - (-2 + 3i)| \leq 1$ $|z - (-2 + 3i)| \le 1 \implies$ Interior and boundary of a circle with center $C(-2, 3)$ $C(-2, 3)$ and radius $r = 1$ $r = 1$.
Set $Q$ $Q$: $(x+iy)(1+i) + (x-iy)(1-i) \leq -8 (x - y + ix + iy) + (x - y - ix - iy) \leq -8$

2(x - y) \leq -8 x - y + 4 \leq 0

This defines a half-plane below or to the left of the boundary line $L₂: x - y + 4 = 0$ $L_2: x - y + 4 = 0$.

Step 1: Identify Extreme Points for Distance from Point A

We want to find points in the region $P Q$ $P \cap Q$ that minimize and maximize the distance to the external point $A(3, -2)$ $A(3, -2)$, corresponding to $|z - (3 - 2i)|$ $|z - (3 - 2i)|$.

The line $L₁$ $L_1$ connecting center $C(-2, 3)$ $C(-2, 3)$ and point $A(3, -2)$ $A(3, -2)$ has slope:

m = (-2 - 3)/(3 - (-2)) = (-5)/(5) = -1

Equation of line $L₁$ $L_1$: $y - 3 = -1(x + 2) x + y - 1 = 0$ .

Geometric region intersection for complex inequalities for Q27 - JEE Main 2024 01 February Morning — The graphic details the intersection region bounded by the circular locus and the line inequality, showing points z1 and z2 relative to external reference point P.

Step 2: Calculate Coordinates for z1 and z2

Minimum Distance Point ( $z₂$ $z_2$): By geometric observation, the minimum distance from $A$ $A$ to the bounded region is the intersection point of lines $L₁$ $L_1$ and $L₂$ $L_2$:

x - y + 4 = 0 and x + y - 1 = 0 2x + 3 = 0 x = -(3)/(2), y = (5)/(2)

So, $z₂ = (-(3)/(2), (5)/(2))$ .

Maximum Distance Point ( $z₁$ $z_1$): The maximum distance is at the far boundary edge of the circle along line $L₁$ $L_1$. The vector path from $C(-2, 3)$ $C(-2, 3)$ opposite to $A$ $A$ has unit direction $(- 1\sqrt(2), 1\sqrt(2))$ :

z₁ = (-2 - 1\sqrt(2), 3 + 1\sqrt(2))

Step 3: Evaluate Magnitudes and Sum Parameters

Calculate the squares of the moduli:

|z₁|² = (-2 - 1\sqrt(2))² + (3 + 1\sqrt(2))² = 4 + 2\sqrt(2) + (1)/(2) + 9 + 3\sqrt(2) + (1)/(2) = 14 + 5\sqrt(2)

|z₂|² = (-(3)/(2))² + ((5)/(2))² = (9)/(4) + (25)/(4) = (34)/(4) = (17)/(2)

Now compute the total requested term:

|z₁|² + 2|z₂|² = (14 + 5\sqrt(2)) + 2((17)/(2)) = 14 + 5\sqrt(2) + 17 = 31 + 5\sqrt(2)

Matching with $α + β\sqrt(2)$ $\alpha + \beta\sqrt{2}$ gives $α = 31$ $\alpha = 31$ and $β = 5$ $\beta = 5$. Thus:

α + β = 31 + 5 = 36

Pattern Recognition

Sees: Locus intersection involving geometric complex inequalities. Shortcut: Translating complex equations into standard 2D graphs reveals the geometry instantly, mapping extreme distances to line intersections or boundary nodes cleanly.

Chapter Mix

Class 11 Complex Numbers: Geometry Class 11 Coordinate Geometry: Straight Lines

Q6 jee_main_2024_01_february_morning Geometry of Complex Numbers

Let

S=zin C:|z-1|=1 and (\sqrt(2)-1)(z+ z)-i(z- z)=2\sqrt(2)

. Let

z₁, z₂in S

be such that

|z₁|= zin S|z|

and

|z₂|= zin S|z|

. Then

|\sqrt(2)z₁-z₂|²

equals:

A. $1$ $1$
B. $4$ $4$
C. $3$ $3$
D. $2$ $2$

Solution

Related Formula

For a complex number $z = x + iy$ $z = x + iy$:

$|z| = \sqrt(x² + y²)$ $|z| = \sqrt{x^2 + y^2}$
$z + z = 2x$ $z + \overline{z} = 2x$
$z - z = 2iy$ $z - \overline{z} = 2iy$

Core Logic

Let $z = x + iy$ $z = x + iy$. The first condition $|z - 1| = 1$ $|z - 1| = 1$ describes a circle:

(x-1)² + y² = 1 (1)

The second condition gives a straight line:

(\sqrt(2)-1)(2x) - i(2iy) = 2\sqrt(2)

2(\sqrt(2)-1)x + 2y = 2\sqrt(2)

(\sqrt(2)-1)x + y = \sqrt(2) (2)

The set $S$ $S$ contains the intersection points of this circle and line.

Step 1: Solve for Intersection Points

From (2), $y = \sqrt(2) - (\sqrt(2)-1)x$ $y = \sqrt{2} - (\sqrt{2}-1)x$. Substitute this into (1):

(x-1)² + [\sqrt(2) - (\sqrt(2)-1)x]² = 1

Solving this quadratic equation gives two values of $x$ $x$:

x = 1 or x = 12-\sqrt(2)

Case A: $x = 1 y = \sqrt(2) - (\sqrt(2)-1)(1) = 1$ .

So, one point is $zA = 1 + i$ $z_A = 1 + i$. Its magnitude is $|zA| = \sqrt(1² + 1²) = \sqrt(2)$ $|z_A| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

Case B: $x = 12-\sqrt(2) = 2+\sqrt(2)2 = 1 + 1\sqrt(2)$ .

Then $y = \sqrt(2) - (\sqrt(2)-1)(1+ 1\sqrt(2)) = \sqrt(2) - (\sqrt(2) + 1 - 1 - 1\sqrt(2)) = 1\sqrt(2)$ . So, the other point is $zB = (1 + 1\sqrt(2)) + i\sqrt(2)$ . Its magnitude is $|zB| = (1+ 1\sqrt(2))² + ( 1\sqrt(2))² = 1 + \sqrt(2) + (1)/(2) + (1)/(2) = 2 + \sqrt(2)$ .

Step 2: Evaluate Min and Max Magnitude Expressions

Comparing magnitudes, $|zB| > |zA|$ $|z_B| > |z_A|$, so:

z₁ = zB = (1 + 1\sqrt(2)) + i\sqrt(2)

z₂ = zA = 1 + i

Now, calculate $|\sqrt(2)z₁-z₂|²$ $|\sqrt{2}z_{1}-z_{2}|^{2}$:

\sqrt(2)z₁ = \sqrt(2)(1 + 1\sqrt(2)) + \sqrt(2)( i\sqrt(2)) = (\sqrt(2) + 1) + i

\sqrt(2)z₁ - z₂ = (\sqrt(2) + 1 + i) - (1 + i) = \sqrt(2)

|\sqrt(2)z₁ - z₂|² = |\sqrt(2)|² = 2

Pattern Recognition

Sees: Geometric constraint mapping to a circle and line intersection in the complex plane. Shortcut: Rationalizing terms like $12-\sqrt(2)$ $\frac{1}{2-\sqrt{2}}$ immediately into standard form $1+ 1\sqrt(2)$ $1+\frac{1}{\sqrt{2}}$ saves you from handling layered fraction algebra down the line.

Chapter Mix

Class 11 Complex Numbers: Geometry Class 10 Coordinate Geometry: Lines and Circles

Q jee_main_2024_29_january_evening Modulus and Argument of a Complex Number

Let

r

and

θ

respectively be the modulus and amplitude of the complex number

z = 2 - i (2 (5 π)/(8))

, then

(r, θ)

is equal to

A. $(2 (3π)/(8),(3pi)/(8))$
B. $(2 (3π)/(8),(5π)/(8))$
C. $(2 (5pi)/(8),(3pi)/(8))$
D. $(2 (11π)/(8),(11π)/(8))$

Solution

Related Formula

For $z = x + iy$ $z = x + iy$, modulus $r = \sqrt(x² + y²)$ $r = \sqrt{x^2 + y^2}$ and argument $θ$ $\theta$ depends on the quadrant location.

Core Logic

Given $z = 2 - i(2 (5π)/(8))$ . Note that $(5π)/(8)$ $\frac{5\pi}{8}$ lies in the second quadrant, so $(5π)/(8) < 0$ $\tan \frac{5\pi}{8} < 0$. Let's write $r$ $r$:

r = \sqrt(2² + (-2 (5π)/(8))²) = 2 \sqrt(1 + ² (5π)/(8)) = 2 | (5π)/(8) |

Since $(5π)/(8)$ $\sec \frac{5\pi}{8}$ is negative:

r = -2 (5π)/(8) = -2 (π - (3π)/(8)) = 2 (3π)/(8)

Step 1: Finding the Amplitude

Since $x = 2 > 0$ $x = 2 > 0$ and $y = -2 (5π)/(8) > 0$ $y = -2 \tan \frac{5\pi}{8} > 0$, the complex number lies in the first quadrant.

θ = ⁻¹ ( (y)/(x) ) = ⁻¹ ( (-2 (5π)/(8))/(2) ) = ⁻¹ ( - (5π)/(8) )

- (5π)/(8) = - (π - (3π)/(8)) = (3π)/(8)

θ = ⁻¹ ( (3π)/(8) ) = (3π)/(8)

Pattern Recognition

Always absolute-value trigonometric terms coming out of square roots (e.g., $\sqrt( ² φ) = | φ|$ $\sqrt{\sec^2 \phi} = |\sec \phi|$). Knowing the precise quadrant prevents incorrect signs.

Chapter Mix

Class 11 Mathematics: 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 — Page 6

Solution

Related Formula

Core Logic

Step 1: Simplify the Discriminant

Step 2: Find the Roots

Step 3: Evaluate Target Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Center Distance

Step 2: Find Minimum Separation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Identify Extreme Points for Distance from Point A

Step 2: Calculate Coordinates for z1 and z2

Step 3: Evaluate Magnitudes and Sum Parameters

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for Intersection Points

Step 2: Evaluate Min and Max Magnitude Expressions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the Amplitude

Pattern Recognition

Chapter Mix

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