Let alpha, beta be the roots of the equation x2-x+2=0 with Im(alpha) Im(beta). Then alpha6+alpha4+beta4-5alpha2 is…

Q61 jee_main_2025_07_april_morning Geometry of Complex Numbers

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.

A. $both are incorrect$ $\text{both are incorrect}$
B. $only (S1) is correct$ $\text{only (S1) is correct}$
C. $only (S2) is correct$ $\text{only (S2) is correct}$
D. $both are correct$ $\text{both are correct}$

Solution

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

Q64 jee_main_2025_08_april_evening Purely Real/Imaginary Conditions

Let

A = θ in [0,2π ]:1 + 10Re( 2 θ + i θ θ - 3i θ) = 0

. Then

Σθ in Aθ²

is equal to

A. $(21)/(4)π²$ $\frac{21}{4}\pi^2$
B. $8π²$ $8\pi^{2}$
C. $(27)/(4)π²$ $\frac{27}{4}\pi^2$
D. $6π²$ $6\pi^{2}$

Solution

Related Formula

z + z = 2Re(z)

Core Logic

Isolate the real fractional component block by conjugating the complex quotient matrix expression, then resolve the structural wave equations across bounds boundaries.

Step 1: Expand Complex Real Operator

(2 ²θ - 3 ²θ)/( ²θ + 9 ²θ) = -(1)/(10)

20 ²θ - 30 ²θ = - ²θ - 9 ²θ

Step 2: Factor Trigonometric Expressions

21 ²θ - 21 ²θ = 0 (2θ) = 0

Step 3: Collect Domain Solutions and Evaluate Squares

Since angular coordinate parameters scan $[0, 2π]$ $[0, 2\pi]$, multi frequency vectors trace out:

2θ = (π)/(2), (3π)/(2), (5π)/(2), (7π)/(2)

Σ θ² = (π²)/(16) + (9π²)/(16) + (25π²)/(16) + (49π²)/(16) = (84π²)/(16) = (21)/(4)π²

Pattern Recognition

Transforming algebraic equations to clean forms like $(2θ) = 0$ $\\cos(2\\theta) = 0$ guarantees evenly distributed coordinate solutions across standard periodicity ranges.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Trigonometric Functions

Q74 jee_main_2025_29_jan_evening Determinants and Roots of Unity

Let integers

a, b in [-3, 3]

be such that

a + b \neq 0

. Then the number of all possible ordered pairs

(a, b)

, for which

| (z - a)/(z + b) | = 1

and

| arraycccz + 1 & ω & ω² ω & z + ω² & 1 ω² & 1 & z + ω array | = 1, z in C

, where

ω

and

ω²

are the roots of

x² + x + 1 = 0

, is equal to

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

Properties of cube roots of unity:

1 + ω + ω² = 0, ω³ = 1

Core Logic

Simplify the determinant by performing row operation $R₁ \to R₁ + R₂ + R₃$ $R_1 \to R_1 + R_2 + R_3$:

Δ = vmatrix z + 1 + ω + ω² & z + 1 + ω + ω² & z + 1 + ω + ω² ω & z + ω² & 1 ω² & 1 & z + ω vmatrix

Using $1 + ω + ω² = 0$ $1 + \omega + \omega^2 = 0$, the top row simplifies to vector $[z, z, z]$ $[z, z, z]$. Factoring out $z$ $z$:

Δ = z \cdot (z²) = z³

Given modulus constraint $|z³| = 1 |z| = 1$ $|z^3| = 1 \implies |z| = 1$. The root solutions are:

z in 1, ω, ω²

Step 1: Evaluate Geometric Magnitude Metric

The condition $|(z - a)/(z + b)| = 1 |z - a| = |z + b|$ . This equation represents the perpendicular bisector of the segment connecting real coordinate points $a$ $a$ and $-b$ $-b$ on the complex plane.

Since $a$ $a$ and $b$ $b$ are integers, the bisector is a vertical line: $x = (a - b)/(2)$ $x = \frac{a - b}{2}$.

Step 2: Match Root Solutions and Count Pairs

For $z=1$ $z=1$, it must lie on the line: $(a-b)/(2) = 1 a - b = 2$ $\frac{a-b}{2} = 1 \implies a - b = 2$. For $z = ω, ω²$ $z = \omega, \omega^2$, their real part is $-(1)/(2)$ $-\frac{1}{2}$, so the line must be: $(a-b)/(2) = -(1)/(2) a - b = -1$ .

Counting integer pairs $(a,b) in [-3, 3]²$ $(a,b) \in [-3, 3]^2$ with $a+b \neq 0$ $a+b \neq 0$: From $a - b = 2$ $a - b = 2$: valid pairs are $(3,1), (1,-1), (0,-2), (-1,-3)$ $(3,1), (1,-1), (0,-2), (-1,-3)$. Note: $(2,0)$ $(2,0)$ is valid, but $a+b=2 \neq 0$ $a+b=2 \neq 0$. Total = 5 pairs. From $a - b = -1$ $a - b = -1$: valid pairs match another 5 configurations.

Combining both groups gives a final count of 10 pairs.

Pattern Recognition

Using matrix summation properties ( $1+ω+ω²=0$ $1+\omega+\omega^2=0$) helps simplify large complex variable equations quickly.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 12 Mathematics: Matrices and Determinants

Q jee_main_2025_28_jan_morning Geometry of Complex Numbers

Let O be the origin, the point A be

z₁ = \sqrt(3) + 2\sqrt(2)i

, the point

B(z₂)

be such that

\sqrt(3)|z₂| = |z₁|

and

(z₂) = (z₁) + (π)/(6)

. Then

A. area of triangle ABO is $11\sqrt(3)$ $\frac{11}{\sqrt{3}}$
B. ABO is a scalene triangle
C. area of triangle ABO is $(11)/(4)$ $\frac{11}{4}$
D. ABO is an obtuse angled isosceles triangle

Solution

Related Formula

Complex rotation and scaling vector rule:

z₂ = (|z₂|)/(|z₁|) z₁ eiθ

Core Logic

Given structural rotation conditions:

z₂ = 1\sqrt(3) z₁ ei(π)/(6)

Evaluating the vectors yields coordinates showing $|z₁ - z₂| = |z₂|$ $|z_1 - z_2| = |z_2|$.

Step 1: Analyzing Geometry Metrics

Since $|z₁ - z₂| = |z₂|$ $|z_1 - z_2| = |z_2|$, $Δ ABO$ $\Delta ABO$ forms an isosceles triangle with internal vertex angles evaluating explicitly to $(π)/(6), (π)/(6)$ $\frac{\pi}{6}, \frac{\pi}{6}$, and $(2π)/(3)$ $\frac{2\pi}{3}$.

Step 2: Conclusion

Since $(2π)/(3) > (π)/(2)$ $\frac{2\pi}{3} > \frac{\pi}{2}$, the triangle is an obtuse-angled isosceles triangle.

Pattern Recognition

Complex argument shifts represent pure coordinate system rotations on the Argand plane diagram matrix.

Chapter Mix

Class 11 Maths: Complex Numbers

Q64 jee_main_2025_03_april_morning Roots of Quadratic Equations in Complex Fields

Let

zin C

be such that

(z² + 3i)/(z - 2 + i) = 2 + 3i

. Then the sum of all possible values of

z²

is:

A. 19 - 2i
B. -19 - 2i
C. 19 + 2i
D. -19 + 2i

Solution

Related Formula

For a quadratic system equation $az²+bz+c=0$ $az^2+bz+c=0$ with roots $z₁, z₂$ $z_1, z_2$:

$z₁ + z₂ = -b/a$ $z_1 + z_2 = -b/a$
$z₁ z₂ = c/a$ $z_1 z_2 = c/a$
$z₁² + z₂² = (z₁+z₂)² - 2z₁z₂$ $z_1^2 + z_2^2 = (z_1+z_2)^2 - 2z_1z_2$

Core Logic

Cross-multiply the denominators to configure a linear equation layout:

z² + 3i = (z - 2 + i)(2 + 3i)

z² + 3i = z(2 + 3i) + (-2 + i)(2 + 3i)

z² + 3i = z(2 + 3i) - 4 - 6i + 2i - 3 = z(2 + 3i) - 7 - 4i

Formulate the classic quadratic representation layout:

z² - z(2 + 3i) + 7 + 7i = 0

Step 1: Summing the squared roots

Identify coefficients from the structural template:

z₁ + z₂ = 2 + 3i

z₁ z₂ = 7 + 7i

Evaluate sum of possible squared values ( $z₁² + z₂²$ $z_1^2 + z_2^2$):

z₁² + z₂² = (z₁ + z₂)² - 2z₁ z₂

= (2 + 3i)² - 2(7 + 7i)

= (4 - 9 + 12i) - (14 + 14i) = -5 + 12i - 14 - 14i

$= -19 - 2i$ $= -19 - 2i$

Pattern Recognition

The question asks for the sum of values of $z²$ $z^2$, meaning $z₁² + z₂²$ $z_1^2 + z_2^2$. Avoid using complex quadratic formulas to solve for $z$ $z$ explicitly; structural expansions save massive computational effort.

Chapter Mix

Class 11 Mathematics: Complex Numbers

Let α, β$\alpha, \beta$ be the roots of the equation x²-x+2=0$x^2-x+2=0$ with Im(α) gt Im(β)$Im(\alpha) \gt Im(\beta)$. Then α⁶+α⁴+β⁴-5α²$\alpha^6+\alpha^4+\beta^4-5\alpha^2$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify the Full Expression

Step 2: Apply Sum of Roots

Pattern Recognition

Chapter Mix

Reference Study Guides

More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Evaluate Statement S2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Expand Complex Real Operator

Step 2: Factor Trigonometric Expressions

Step 3: Collect Domain Solutions and Evaluate Squares

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate Geometric Magnitude Metric

Step 2: Match Root Solutions and Count Pairs

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Analyzing Geometry Metrics

Step 2: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Summing the squared roots

Pattern Recognition

Chapter Mix

More Complex Numbers and Quadratic Equations Questions — jee_main_2024_29_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Let $α, β$ $\alpha, \beta$ be the roots of the equation $x²-x+2=0$ $x^2-x+2=0$ with $Im(α) gt Im(β)$ $Im(\alpha) \gt Im(\beta)$. Then $α⁶+α⁴+β⁴-5α²$ $\alpha^6+\alpha^4+\beta^4-5\alpha^2$ is equal to