How much of JEE Main comes from Complex Numbers and Quadratic Equations?

JEEGenie estimates about 4% of JEE Main marks come from Complex Numbers and Quadratic Equations, based on recent papers. The share varies from year to year.

Which formulas matter most in Complex Numbers and Quadratic Equations?

The Complex Numbers and Quadratic Equations sheet on this page has 24 formula entries, starting with Complex Number Definition, Powers of i, Conjugate of a Complex Number, Modulus of a Complex Number, Argument of a Complex Number. Each one comes with when to use it and the trap to avoid.

What topics does Complex Numbers and Quadratic Equations cover for JEE?

Complex Numbers and Quadratic Equations covers Algebra of Complex Numbers, Modulus and Argument, Argand Plane, De Moivre's Theorem, Quadratic Equations, Nature of Roots.

How should I revise Complex Numbers and Quadratic Equations before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Complex Numbers and Quadratic Equations questions on JEEGenie.

How many previous-year questions are there on Complex Numbers and Quadratic Equations?

JEEGenie has 686 previous-year questions on Complex Numbers and Quadratic Equations, and this page shows a few with worked explanations.

Is practice on Complex Numbers and Quadratic Equations free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Complex Numbers and Quadratic Equations for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 4% of JEE Main marks come from Complex Numbers and Quadratic Equations. This page covers 24 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Confusing $i$ with a real number; $i^{2} = - 1$ is fundamental.

Key points to know

Complex Number Definition. A complex number is of the form $a + i b$ , where $a, b \in R$ and $i = \sqrt{- 1}$ . $a$ is the real part (Re) and $b$ is the imaginary part (Im).
Powers of i. Powers of $i$ cycle every 4: $i^{1} = i$ , $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ . For $i^{n}$ , find $n m o d 4$ .
Conjugate of a Complex Number. The conjugate of $z = a + i b$ is $\overset{ˉ}{z} = a - i b$ . It reflects the number across the real axis.
Modulus of a Complex Number. The modulus $∣ z ∣$ is the distance from origin to point $(a, b)$ in the complex plane: $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ .
nth Roots of Unity. The $n$ th roots of unity are solutions to $z^{n} = 1$ , given by $z = e^{2 π i k / n}$ for $k = 0, 1, . . ., n - 1$ . They form vertices of a regular $n$ -gon on unit circle.
Cube Roots of Unity. Cube roots of unity: $1, ω, ω^{2}$ , where $ω = e^{2 π i / 3} = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ . Key properties: $ω^{3} = 1$ , $1 + ω + ω^{2} = 0$ .
Discriminant and Nature of Roots. For $a x^{2} + b x + c = 0$ , $D = b^{2} - 4 a c$ : $D > 0$ real distinct, $D = 0$ real equal, $D < 0$ complex conjugate (non-real).
Complex Roots Occur in Conjugate Pairs. If a polynomial with real coefficients has a complex root $a + i b$ , then its conjugate $a - i b$ is also a root.
Argand Plane and Geometry. Complex numbers correspond to points $(a, b)$ in plane. $∣ z_{1} - z_{2} ∣$ is distance between points. $\arg (z_{1} / z_{2})$ is angle between vectors.
Key Properties Summary. Essential: $z \overset{ˉ}{z} = ∣ z ∣^{2}$ , $\overline{z_{1} + z_{2}} = \overset{ˉ}{z_{1}} + \overset{ˉ}{z_{2}}$ , $\overline{z_{1} z_{2}} = \overset{ˉ}{z_{1}} \overset{ˉ}{z_{2}}$ , $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣ ∣ z_{2} ∣$ , $\arg (z_{1} z_{2}) \equiv \arg (z_{1}) + \arg (z_{2}) (m o d 2 π)$ , $\arg (z_{1} / z_{2}) \equiv \arg (z_{1}) - \arg (z_{2}) (m o d 2 π)$ .
Quadratic with Real Coefficients: Complex Roots. If discriminant $D < 0$ , roots are complex conjugates $α \pm i β$ . Sum and product formulas still hold.

Formula sheet

Name	Formula	When to use
Complex Number Definition	$z = a + i b$	When representing numbers that involve square roots of negatives or solving equations with no real solutions.
Powers of i	$i^{4 k} = 1$ , $i^{4 k + 1} = i$ , $i^{4 k + 2} = - 1$ , $i^{4 k + 3} = - i$	Simplifying expressions with high powers of $i$ .
Conjugate of a Complex Number	$\overset{ˉ}{z} = a - i b$	Rationalizing denominators, finding modulus, proving properties.
Modulus of a Complex Number	$∣ z ∣ = \sqrt{a^{2} + b^{2}}$	Finding magnitude, distance between points, or in inequalities.
Argument of a Complex Number	$\arg (z) = \tan^{- 1} (\frac{b}{a})$ (with quadrant adjustment)	When converting to polar form or solving equations involving angles.
Polar and Euler Forms	$z = r (\cos θ + i \sin θ) = r e^{i θ}$	Multiplying/dividing complex numbers, finding powers/roots via De Moivre.
De Moivre's Theorem	$(\cos θ + i \sin θ)^{n} = \cos (n θ) + i \sin (n θ)$	Finding powers, roots, or simplifying trigonometric expressions.
nth Roots of Unity	$z_{k} = e^{2 π i k / n} = \cos \frac{2 π k}{n} + i \sin \frac{2 π k}{n}$	Solving $z^{n} = 1$ , factorizing $z^{n} - 1$ , or in geometry problems.
Cube Roots of Unity	$ω = \frac{- 1 + i \sqrt{3}}{2}$ , $ω^{2} = \frac{- 1 - i \sqrt{3}}{2}$	Simplifying expressions like $a^{3} + b^{3}$ factorization, or problems with $x^{2} + x + 1 = 0$ .
Quadratic Formula	$x = \frac{- b \pm \sqrt{D}}{2 a}$ , $D = b^{2} - 4 a c$	Solving any quadratic equation with real coefficients.
Discriminant and Nature of Roots	$D = b^{2} - 4 a c$	Quickly determining root type without solving.
Sum and Product of Roots	$α + β = - \frac{b}{a}$ , $α β = \frac{c}{a}$	Forming quadratic equations from roots, or finding symmetric functions of roots.
Forming Quadratic from Roots	$x^{2} - S x + P = 0$ , where $S = α + β$ , $P = α β$	Given roots, find the quadratic equation.
Complex Roots Occur in Conjugate Pairs	If $P (a + i b) = 0$ , then $P (a - i b) = 0$ for real coefficients.	When given one complex root of a real polynomial, find others.
Solving Quadratic with Complex Coefficients	$z = \frac{- b \pm \sqrt{D}}{2 a}$ where $D$ may be complex.	When coefficients are complex numbers.
Square Root of a Complex Number	$\sqrt{a + i b} = \pm (\sqrt{\frac{∣ z ∣ + a}{2}} + i sgn (b) \sqrt{\frac{∣ z ∣ - a}{2}})$	Solving quadratic with complex discriminant, or finding roots of complex numbers.
Condition for Common Roots	$(a_{1} c_{2} - a_{2} c_{1})^{2} = (a_{1} b_{2} - a_{2} b_{1}) (b_{1} c_{2} - b_{2} c_{1})$	When given two quadratics and asked if they share a root.
Location of Roots (Quadratic Inequalities)	For both roots $> k$ : $D \geq 0$ , $a f (k) > 0$ , $- b / (2 a) > k$ .	Problems asking 'find $k$ such that both roots are greater than 2' etc.
Argand Plane and Geometry	$∣ z_{1} - z_{2} ∣ = \sqrt{(a_{1} - a_{2})^{2} + (b_{1} - b_{2})^{2}}$	Geometry problems involving circles, lines, or triangles in complex plane.
Locus Problems	Circle: $∣ z - z_{0} ∣ = r$ ; Line: $∣ z - z_{1} ∣ = ∣ z - z_{2} ∣$ ; Ray: $\arg (z - z_{0}) = θ$ .	When asked to find the geometric figure described by a complex equation.
Triangle Inequality	$∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣$	Proving bounds or inequalities involving moduli.
De Moivre for Trigonometric Identities	$\cos n θ + i \sin n θ = \sum_{k = 0}^{n} (\binom{n}{k}) \cos^{n - k} θ (i \sin θ)^{k}$	Deriving formulas like $\cos 3 θ = 4 \cos^{3} θ - 3 \cos θ$ .
Key Properties Summary	$z \overset{ˉ}{z} = ∣ z ∣^{2}$	In almost every complex number problem for simplification.
Quadratic with Real Coefficients: Complex Roots	$α \pm i β = \frac{- b \pm i \sqrt{∣ D ∣}}{2 a}$	When solving quadratics with negative discriminant.

Open the Complex Numbers and Quadratic Equations formula page

What JEE asks

Complex Number Definition. JEE often tests basic identification of real and imaginary parts, and equality of complex numbers.
Powers of i. Direct simplification problems or as part of larger expressions.
Conjugate of a Complex Number. Used in division, finding real/imaginary parts, and solving equations.
Modulus of a Complex Number. JEE often asks to find modulus after operations or use properties like $∣ z ∣^{2} = z \overset{ˉ}{z}$ .
Argument of a Complex Number. JEE tests quadrant-specific argument calculation and properties like $\arg (z_{1} z_{2}) = \arg (z_{1}) + \arg (z_{2})$ .
Polar and Euler Forms. JEE uses Euler form for quick multiplication and De Moivre's theorem.
De Moivre's Theorem. JEE often asks to find $n$ th roots of unity or simplify $\cos 3 θ$ etc.
nth Roots of Unity. JEE tests properties like sum=0, product= $(- 1)^{n - 1}$ , and primitive roots.
Cube Roots of Unity. JEE often uses $1 + ω + ω^{2} = 0$ to evaluate sums or products.
Quadratic Formula. JEE tests direct application and conditions for equal/real/complex roots.
Discriminant and Nature of Roots. JEE often asks conditions for real roots or equal roots.
Sum and Product of Roots. JEE tests forming equation given roots, or evaluating $α^{2} + β^{2}$ etc.
Forming Quadratic from Roots. Direct JEE problem: 'Find quadratic with roots $2 + 3 i$ and $2 - 3 i$ '.
Complex Roots Occur in Conjugate Pairs. JEE uses this to find all roots of a quadratic or higher degree polynomial.
Solving Quadratic with Complex Coefficients. JEE may ask to solve $z^{2} + (1 + i) z + i = 0$ .
Square Root of a Complex Number. JEE often asks to find $\sqrt{3 + 4 i}$ or similar.
Condition for Common Roots. JEE tests this condition in parameter-based problems.
Location of Roots (Quadratic Inequalities). JEE often combines with quadratic inequalities and parameter ranges.
Argand Plane and Geometry. JEE uses to find locus: e.g., $∣ z - z_{0} ∣ = r$ is circle.
Locus Problems. JEE often asks to identify locus as circle, line, ellipse, etc.
Triangle Inequality. JEE uses to find maximum/minimum of $∣ z_{1} + z_{2} ∣$ given constraints.
De Moivre for Trigonometric Identities. JEE may ask to prove such identities or use them in problems.
Key Properties Summary. JEE tests these properties in multi-step problems.
Quadratic with Real Coefficients: Complex Roots. JEE often asks to find roots or verify conjugate property.

Common traps

Confusing $i$ with a real number; $i^{2} = - 1$ is fundamental. Complex Number Definition
Forgetting that $i^{0} = 1$ . Powers of i
Forgetting that $\overline{z_{1} + z_{2}} = \overset{ˉ}{z_{1}} + \overset{ˉ}{z_{2}}$ but $\overline{z_{1} z_{2}} = \overset{ˉ}{z_{1}} \overset{ˉ}{z_{2}}$ . Conjugate of a Complex Number
Confusing modulus with absolute value; $∣ z ∣$ is always non-negative real. Modulus of a Complex Number
Using $\tan^{- 1} (b / a)$ directly without considering quadrant of $(a, b)$ . Argument of a Complex Number
Forgetting $\cos θ + i \sin θ$ is not $\cos θ - i \sin θ$ . Polar and Euler Forms
Applying to $(\cos θ - i \sin θ)$ without converting to $\cos (- θ) + i \sin (- θ)$ . De Moivre's Theorem
Forgetting that sum of all $n$ th roots of unity is 0. nth Roots of Unity
Forgetting $ω^{2} = \overset{ˉ}{ω}$ . Cube Roots of Unity
Forgetting the $\pm$ sign; both roots must be considered. Quadratic Formula
Thinking $D = 0$ gives one root; it gives two equal roots. Discriminant and Nature of Roots
Forgetting the negative sign in sum. Sum and Product of Roots
Forgetting the minus sign before $S$ . Forming Quadratic from Roots
Applying this to polynomials with complex coefficients; it fails. Complex Roots Occur in Conjugate Pairs

Topics in Complex Numbers and Quadratic Equations

Algebra of Complex Numbers
Modulus and Argument
Argand Plane
De Moivre's Theorem
Quadratic Equations
Nature of Roots

Previous-year questions on Complex Numbers and Quadratic Equations

JEEGenie has 686 previous-year questions on Complex Numbers and Quadratic Equations. A few worked examples follow.

Year	Questions
2026	36
2025	38
2024	44
2023	48
2022	53
2021	55
2020	36
2019	63
2018	15
2017	13
2016	12
2015	12
2014	17
2013	15
2012	5
2011	8
2010	12
2009	7
2008	11
2007	9
2006	9
2005	11
2004	12
2003	11
2002	17
2001	7
2000	5
1999	3
1998	7
1997	6
1996	4
1995	6
1994	1
1993	2
1992	2
1991	2
1990	2
1989	4
1988	3
1987	7
1986	6
1985	5
1984	6
1983	10
1982	7
1981	5
1980	5
1979	7
1978	5

Question 1 medium

Let

α, α + 2, α \in Z

, be the roots of the quadratic equation

x (x + 2) + (x + 1) (x + 3) + (x + 2) (x + 4) + \dots . + (x + n - 1) (x + n + 1) = 4 n

for some

n \in N

. Then

n + α

is equal to :

Answer: C

Explanation. The given sum is

\sum_{k = 0}^{n - 1} (x + k) (x + k + 2) = 4 n .

Expand the general term:

(x + k) (x + k + 2) = x^{2} + (2 k + 2) x + k (k + 2) .

Summing from

k = 0

n - 1

\sum_{k = 0}^{n - 1} x^{2} = n x^{2},

\sum_{k = 0}^{n - 1} (2 k + 2) x = 2 x \sum_{k = 0}^{n - 1} (k + 1) = 2 x \cdot \frac{n (n + 1)}{2} = n (n + 1) x,

\sum_{k = 0}^{n - 1} k (k + 2) = \sum_{k = 0}^{n - 1} (k^{2} + 2 k) = \frac{(n - 1) n (2 n - 1)}{6} + n (n - 1) .

Simplify the constant term:

\frac{n (n - 1) (2 n - 1)}{6} + n (n - 1) = \frac{n (n - 1) (2 n - 1 + 6)}{6} = \frac{n (n - 1) (2 n + 5)}{6} .

Thus the equation becomes

n x^{2} + n (n + 1) x + \frac{n (n - 1) (2 n + 5)}{6} = 4 n .

Divide by

n

(since

n \in N

x^{2} + (n + 1) x + \frac{(n - 1) (2 n + 5)}{6} = 4.

x^{2} + (n + 1) x + \frac{2 n^{2} + 3 n - 5}{6} - 4 = 0,

x^{2} + (n + 1) x + \frac{2 n^{2} + 3 n - 29}{6} = 0.

Given roots are

α

and

α + 2

. By Vieta: Sum of roots:

2 α + 2 = - (n + 1) ⟹ α = - \frac{n + 3}{2} .

Product of roots:

α (α + 2) = \frac{2 n^{2} + 3 n - 29}{6} .

Substitute

α = - \frac{n + 3}{2}

(- \frac{n + 3}{2}) (- \frac{n + 3}{2} + 2) = \frac{2 n^{2} + 3 n - 29}{6} .

Simplify LHS:

\frac{n + 3}{2} \cdot \frac{n - 1}{2} = \frac{(n + 3) (n - 1)}{4} .

\frac{(n + 3) (n - 1)}{4} = \frac{2 n^{2} + 3 n - 29}{6} .

Cross-multiply:

6 (n^{2} + 2 n - 3) = 4 (2 n^{2} + 3 n - 29),

6 n^{2} + 12 n - 18 = 8 n^{2} + 12 n - 116,

0 = 2 n^{2} - 98 ⟹ n^{2} = 49 ⟹ n = 7.

Then

α = - \frac{7 + 3}{2} = - 5.

Hence

n + α = 7 - 5 = 2.

C

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 hard

Let

α

and

β

be the roots of

x^{2} + \sqrt{3} x - 16 = 0

, and

γ

and

δ

be the roots of

x^{2} + 3 x - 1 = 0

. If

P_{n} = α^{n} + β^{n}

and

Q_{n} = γ^{n} + {\hat{o}}^{n}

, then

\frac{P_{25} + \sqrt{3} P_{24}}{2 P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}}

is equal to

Answer: C

Explanation. Let

α, β be roots of x^{2} + \sqrt{3} x - 16 = 0

α + β = - \sqrt{3}, α β = - 16.

Also let

γ, δ be roots of x^{2} + 3 x - 1 = 0

γ + δ = - 3, γ δ = - 1.

Define

P_{n} = α^{n} + β^{n}, Q_{n} = γ^{n} + δ^{n} .

For

P_{n}

, the recurrence is

P_{n} = (α + β) P_{n - 1} - α β P_{n - 2} = - \sqrt{3} P_{n - 1} + 16 P_{n - 2} .

Thus

P_{n} + \sqrt{3} P_{n - 1} = 16 P_{n - 2} .

Putting

n = 25

P_{25} + \sqrt{3} P_{24} = 16 P_{23} .

Hence

\frac{P_{25} + \sqrt{3} P_{24}}{2 P_{23}} = \frac{16 P_{23}}{2 P_{23}} = 8.

For

Q_{n}

, the recurrence is

Q_{n} = (γ + δ) Q_{n - 1} - γ δ Q_{n - 2} = - 3 Q_{n - 1} + Q_{n - 2} .

Q_{n} + 3 Q_{n - 1} - Q_{n - 2} = 0.

We need

\frac{Q_{25} - Q_{23}}{Q_{24}} .

From the recurrence with

n = 25

Q_{25} = - 3 Q_{24} + Q_{23} .

Therefore

Q_{25} - Q_{23} = - 3 Q_{24} .

\frac{Q_{25} - Q_{23}}{Q_{24}} = - 3.

Thus the required expression is

8 + (- 3) = 5.

Answer: (C)

From NTA/IIT papers: JEE Main 2025 · 3 Apr Shift 1

Question 3 medium

If 2 and 6 are the roots of the equation

a x^{2} + b x + 1 = 0

, then the quadratic equation, whose roots are

\frac{1}{2 a + b}

and

\frac{1}{6 a + b}

, is :

$x^{2} + 8 x + 12 = 0$
$2 x^{2} + 11 x + 12 = 0$
$4 x^{2} + 14 x + 12 = 0$
$x^{2} + 10 x + 16 = 0$

Answer: A

Explanation. Since

2

and

6

are roots of

a x^{2} + b x + 1 = 0,

the sum and product of roots give:

2 + 6 = - \frac{b}{a} \Rightarrow 8 = - \frac{b}{a} \Rightarrow b = - 8 a

2 \cdot 6 = \frac{1}{a} \Rightarrow 12 = \frac{1}{a} \Rightarrow a = \frac{1}{12}

Thus,

b = - 8 \cdot \frac{1}{12} = - \frac{2}{3} .

Now compute:

2 a + b = 2 \cdot \frac{1}{12} - \frac{2}{3} = \frac{1}{6} - \frac{2}{3} = - \frac{1}{2}

6 a + b = 6 \cdot \frac{1}{12} - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = - \frac{1}{6}

So the required roots are:

\frac{1}{2 a + b} = \frac{1}{- 1 / 2} = - 2

\frac{1}{6 a + b} = \frac{1}{- 1 / 6} = - 6

The quadratic equation with roots

- 2

and

- 6

is:

x^{2} - (- 2 - 6) x + (- 2) (- 6) = 0

x^{2} + 8 x + 12 = 0

The other options do not have roots

- 2

and

- 6

. Answer: (A)

From NTA/IIT papers: JEE Main 2024 · 4 Apr Shift 1

Question 4 easy

If the center and radius of the circle

∣ \frac{z - 2}{z - 3} ∣ = 2

are respectively

(α, β)

and

γ

, then

3 (α + β + γ)

is equal to :

Answer: A

Explanation. Let

z = x + i y

. The given condition is

∣ \frac{z - 2}{z - 3} ∣ = 2

∣ z - 2 ∣ = 2 ∣ z - 3 ∣ .

Squaring both sides,

(x - 2)^{2} + y^{2} = 4 ((x - 3)^{2} + y^{2}) .

Expanding:

x^{2} - 4 x + 4 + y^{2} = 4 (x^{2} - 6 x + 9 + y^{2})

x^{2} - 4 x + 4 + y^{2} = 4 x^{2} - 24 x + 36 + 4 y^{2} .

Bringing all terms to one side:

0 = 3 x^{2} - 20 x + 32 + 3 y^{2}

3 x^{2} + 3 y^{2} - 20 x + 32 = 0.

Divide by

3

x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{3} = 0.

Complete the square in

x

{(x - \frac{10}{3})}^{2} + y^{2} = \frac{100}{9} - \frac{32}{3}

= \frac{100}{9} - \frac{96}{9} = \frac{4}{9} .

Thus the center is

(α, β) = (\frac{10}{3}, 0)

and the radius is

γ = \frac{2}{3} .

Therefore,

α + β + γ = \frac{10}{3} + 0 + \frac{2}{3} = 4.

3 (α + β + γ) = 3 \cdot 4 = 12.

Answer: (A)

From NTA/IIT papers: JEE Main 2023 · 1 Feb Shift 1

Question 5 hard

Let

z

be a complex number with non-zero imaginary part. If

\frac{2 + 3 z + 4 z^{2}}{2 - 3 z + 4 z^{2}}

is a real number, then the value of

∣ z ∣^{2}

is ______.

Answer: 0.50

Explanation.

\frac{2 + 3 z + 4 z^{2}}{2 - 3 z + 4 z^{2}} \in R

Let the ratio equal its conjugate:

\frac{2 + 3 z + 4 z^{2}}{2 - 3 z + 4 z^{2}} = \frac{2 + 3 \overset{ˉ}{z} + 4 {\overset{ˉ}{z}}^{2}}{2 - 3 \overset{ˉ}{z} + 4 {\overset{ˉ}{z}}^{2}}

Cross-multiply:

(2 + 3 z + 4 z^{2}) (2 - 3 \overset{ˉ}{z} + 4 {\overset{ˉ}{z}}^{2}) = (2 + 3 \overset{ˉ}{z} + 4 {\overset{ˉ}{z}}^{2}) (2 - 3 z + 4 z^{2})

Expand both sides. Left:

4 - 6 \overset{ˉ}{z} + 8 {\overset{ˉ}{z}}^{2} + 6 z - 9 z \overset{ˉ}{z} + 12 z {\overset{ˉ}{z}}^{2} + 8 z^{2} - 12 z^{2} \overset{ˉ}{z} + 16 z^{2} {\overset{ˉ}{z}}^{2}

Right:

4 - 6 z + 8 z^{2} + 6 \overset{ˉ}{z} - 9 \overset{ˉ}{z} z + 12 \overset{ˉ}{z} z^{2} + 8 {\overset{ˉ}{z}}^{2} - 12 {\overset{ˉ}{z}}^{2} z + 16 {\overset{ˉ}{z}}^{2} z^{2}

Cancel common terms

4

- 9 z \overset{ˉ}{z}

16 ∣ z ∣^{4}

. Remaining:

- 6 \overset{ˉ}{z} + 8 {\overset{ˉ}{z}}^{2} + 6 z + 12 z {\overset{ˉ}{z}}^{2} - 12 z^{2} \overset{ˉ}{z} + 8 z^{2} = - 6 z + 8 z^{2} + 6 \overset{ˉ}{z} + 12 \overset{ˉ}{z} z^{2} - 12 {\overset{ˉ}{z}}^{2} z + 8 {\overset{ˉ}{z}}^{2}

Bring all to one side:

(- 6 \overset{ˉ}{z} - 6 \overset{ˉ}{z}) + (6 z + 6 z) + (8 {\overset{ˉ}{z}}^{2} - 8 {\overset{ˉ}{z}}^{2}) + (8 z^{2} - 8 z^{2}) + 12 z {\overset{ˉ}{z}}^{2} - 12 \overset{ˉ}{z} z^{2} - 12 z^{2} \overset{ˉ}{z} + 12 {\overset{ˉ}{z}}^{2} z = 0

Simplify:

- 12 \overset{ˉ}{z} + 12 z + 12 z {\overset{ˉ}{z}}^{2} - 12 \overset{ˉ}{z} z^{2} - 12 z^{2} \overset{ˉ}{z} + 12 {\overset{ˉ}{z}}^{2} z = 0

Divide by 12:

(z - \overset{ˉ}{z}) + z \overset{ˉ}{z} (\overset{ˉ}{z} - z) + {\overset{ˉ}{z}}^{2} z - z^{2} \overset{ˉ}{z} = 0

Factor:

(z - \overset{ˉ}{z}) - z \overset{ˉ}{z} (z - \overset{ˉ}{z}) + z \overset{ˉ}{z} (\overset{ˉ}{z} - z) = 0

Note

{\overset{ˉ}{z}}^{2} z - z^{2} \overset{ˉ}{z} = z \overset{ˉ}{z} (\overset{ˉ}{z} - z) = - z \overset{ˉ}{z} (z - \overset{ˉ}{z})

. So:

(z - \overset{ˉ}{z}) - z \overset{ˉ}{z} (z - \overset{ˉ}{z}) - z \overset{ˉ}{z} (z - \overset{ˉ}{z}) = 0

(z - \overset{ˉ}{z}) (1 - 2 z \overset{ˉ}{z}) = 0

Since

z

has non-zero imaginary part,

z - \overset{ˉ}{z} \neq 0

. Hence:

1 - 2 ∣ z ∣^{2} = 0 ⟹ ∣ z ∣^{2} = \frac{1}{2} = 0.50

0.50

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Complex Numbers and Quadratic Equations?: JEEGenie estimates about 4% of JEE Main marks come from Complex Numbers and Quadratic Equations, based on recent papers. The share varies from year to year.
Which formulas matter most in Complex Numbers and Quadratic Equations?: The Complex Numbers and Quadratic Equations sheet on this page has 24 formula entries, starting with Complex Number Definition, Powers of i, Conjugate of a Complex Number, Modulus of a Complex Number, Argument of a Complex Number. Each one comes with when to use it and the trap to avoid.
What topics does Complex Numbers and Quadratic Equations cover for JEE?: Complex Numbers and Quadratic Equations covers Algebra of Complex Numbers, Modulus and Argument, Argand Plane, De Moivre's Theorem, Quadratic Equations, Nature of Roots.
How should I revise Complex Numbers and Quadratic Equations before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Complex Numbers and Quadratic Equations questions on JEEGenie.
How many previous-year questions are there on Complex Numbers and Quadratic Equations?: JEEGenie has 686 previous-year questions on Complex Numbers and Quadratic Equations, and this page shows a few with worked explanations.
Is practice on Complex Numbers and Quadratic Equations free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.