If z = sqrt(3)2 + (i/2), i = sqrt(-1), then (z201

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More Complex Numbers Previous-Year Questions — Page 2

Q9 jee_main_2026_24_january_morning Locus in Complex Plane

Let

S = z in C : | (z - 6i)/(z - 2i) | = 1 and | (z - 8 + 2i)/(z + 2i) | = (3)/(5)

. Then

Σz in S |z|²

is equal to

A. $398$ $398$
B. $413$ $413$
C. $423$ $423$
D. $385$ $385$

Solution

Related Formula

|z - z₁| = |z - z₂| represents the perpendicular bisector of the segment joining z₁ and z₂

|x+iy|² = x² + y²

Core Logic

First condition: $|z - 6i| = |z - 2i|$ $|z - 6i| = |z - 2i|$. This means $z$ $z$ lies on the perpendicular bisector of $(0,6)$ $(0,6)$ and $(0,2)$ $(0,2)$. Let $z = x + iy$ $z = x + iy$. Thus, $y = 4$ $y = 4$.

Step 1: Circle Equation

Second condition: $5|z - 8 + 2i| = 3|z + 2i|$ $5|z - 8 + 2i| = 3|z + 2i|$. Substitute $y = 4$ $y = 4$ into $z$ $z$: $z = x + 4i$ $z = x + 4i$.

5|x + 4i - 8 + 2i| = 3|x + 4i + 2i|

5|x - 8 + 6i| = 3|x + 6i|

Squaring both sides:

25((x - 8)² + 36) = 9(x² + 36)

25(x² - 16x + 64 + 36) = 9x² + 324

25x² - 400x + 2500 = 9x² + 324

16x² - 400x + 2176 = 0

Divide by 16:

x² - 25x + 136 = 0

Step 2: Roots and Modulus

Roots of $x² - 25x + 136 = 0$ $x^2 - 25x + 136 = 0$:

(x - 17)(x - 8) = 0 \Rightarrow x = 17 or x = 8

Thus, the points in $S$ $S$ are $z₁ = 17 + 4i$ $z_1 = 17 + 4i$ and $z₂ = 8 + 4i$ $z_2 = 8 + 4i$.

Σz in S |z|² = |17 + 4i|² + |8 + 4i|²

= (17² + 4²) + (8² + 4²)

= (289 + 16) + (64 + 16) = 305 + 80 = 385

Pattern Recognition

Whenever an absolute value ratio equals 1, immediately map it to a line (perpendicular bisector) and substitute its constraint directly into the second curve equation to reduce dimensionality.

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations

Q23 jee_main_2026_24_january_evening Properties of Moduli

Let

z = (1 + i)(1 + 2i)(1 + 3i) (1 + ni)

, where

i = \sqrt(-1)

. If

|z|² = 44200

, then

n

is equal to

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

|z₁ \cdot z₂ zₙ| = |z₁| \cdot |z₂| |zₙ|

|1 + ri|² = 1² + r² = 1 + r²

Core Logic

Given $z = Πr=1ⁿ (1 + ri)$ $z = \prod_{r=1}^{n} (1 + ri)$. Take the modulus of both sides:

|z| = Πr=1ⁿ |1 + ri|

Square both sides:

|z|² = Πr=1ⁿ |1 + ri|² = Πr=1ⁿ (1 + r²)

Step 1: Factoring the Target Number

We are given $|z|² = 44200$ $|z|^2 = 44200$. Factorizing $44200$ $44200$:

44200 = 442 \times 100 = (2 \times 221) \times (10²)

= 2 \times (13 \times 17) \times (2² \times 5²)

= 2³ \cdot 5² \cdot 13 \cdot 17

Step 2: Expanding the Product

Calculate the product series step-by-step for small values of $n$ $n$: For $r=1$ $r=1$: $1 + 1² = 2$ $1 + 1^2 = 2$ For $r=2$ $r=2$: $1 + 2² = 5$ $1 + 2^2 = 5$ For $r=3$ $r=3$: $1 + 3² = 10 = 2 \times 5$ $1 + 3^2 = 10 = 2 \times 5$ For $r=4$ $r=4$: $1 + 4² = 17$ $1 + 4^2 = 17$ For $r=5$ $r=5$: $1 + 5² = 26 = 2 \times 13$ $1 + 5^2 = 26 = 2 \times 13$

Now, multiply these first 5 terms together:

P₅ = 2 \times 5 \times 10 \times 17 \times 26

= 2 \times 5 \times (2 \times 5) \times 17 \times (2 \times 13)

= 2³ \times 5² \times 13 \times 17

Step 3: Comparing and Concluding

The calculated product for $n=5$ $n=5$ exactly matches the prime factorization of $44200$ $44200$. Therefore, $n = 5$ $n = 5$.

Pattern Recognition

Modulus is multiplicative. In problems featuring chains of complex multiplications set equal to a huge real magnitude, instantly switch to magnitudes and map to integer factorization.

Chapter Mix

Class 11 Maths: Complex Numbers

Q5 jee_main_2026_28_january_morning Geometry of Complex Numbers

Let

z

be a complex number such that

|z - 6| = 5

and

|z + 2 - 6i| = 5

. Then the value of

z³ + 3z² - 15z + 141

is equal to

A. $42$ $42$
B. $37$ $37$
C. $50$ $50$
D. $61$ $61$

Solution

Core Logic

The given equations represent two circles in the complex plane: Circle 1: Center

C₁(6, 0)

, radius

r₁ = 5

Circle 2: Center

C₂(-2, 6)

, radius

r₂ = 5

Distance between the centers $C₁$ $C_1$ and $C₂$ $C_2$:

C₁C₂ = \sqrt((-2 - 6)² + (6 - 0)²) = \sqrt(64 + 36) = 10

Notice that $C₁C₂ = r₁ + r₂ = 5 + 5 = 10$ $C_1C_2 = r_1 + r_2 = 5 + 5 = 10$. This means the two circles touch each other externally at exactly one point.

Step 1: Finding z

Since the circles touch externally, the common point $z$ $z$ is the midpoint of the line segment joining the centers $C₁$ $C_1$ and $C₂$ $C_2$.

z = (6 + (-2))/(2) + i (0 + 6)/(2)

$z = 2 + 3i$ $z = 2 + 3i$

Step 2: Simplifying the Polynomial

We have $z = 2 + 3i$ $z = 2 + 3i$. $z - 2 = 3i$ $z - 2 = 3i$ Squaring both sides: $(z - 2)² = -9$ $(z - 2)^2 = -9$

z² - 4z + 4 = -9 z² - 4z + 13 = 0

$z² = 4z - 13$ $z^2 = 4z - 13$

Step 3: Evaluating the Expression

We need to evaluate $z³ + 3z² - 15z + 141$ $z^3 + 3z^2 - 15z + 141$. First, find $z³$ $z^3$:

z³ = z \cdot z² = z(4z - 13) = 4z² - 13z

Substitute $z² = 4z - 13$ $z^2 = 4z - 13$ again:

z³ = 4(4z - 13) - 13z = 16z - 52 - 13z = 3z - 52

Now substitute $z³$ $z^3$ and $z²$ $z^2$ into the target expression:

(3z - 52) + 3(4z - 13) - 15z + 141

= 3z - 52 + 12z - 39 - 15z + 141

= (3z + 12z - 15z) + (-52 - 39 + 141)

$= 0 + 50 = 50$ $= 0 + 50 = 50$

Pattern Recognition

When given two complex distance modulus equations $|z-z₁|=r₁$ $|z-z_1|=r_1$ and $|z-z₂|=r₂$ $|z-z_2|=r_2$, always check the distance between centers $|z₁ - z₂|$ $|z_1 - z_2|$. If it exactly equals $r₁ + r₂$ $r_1 + r_2$, the single unique solution is the section formula midpoint.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Q10 jee_main_2026_28_january_morning Nature of Roots

α, β

, where

α < β

, are the roots of the equation

λ x² - (λ + 3)x + 3 = 0

such that

(1)/(α) - (1)/(β) = (1)/(3)

, then the sum of all possible values of

λ

is:

A. $6$ $6$
B. $2$ $2$
C. $4$ $4$
D. $8$ $8$

Solution

Related Formula

For a quadratic equation $ax² + bx + c = 0$ $ax^2 + bx + c = 0$: Sum of roots: $α + β = -(b)/(a)$ $\alpha + \beta = -\frac{b}{a}$ Product of roots: $αβ = (c)/(a)$ $\alpha\beta = \frac{c}{a}$

Core Logic

From the given equation $λ x² - (λ + 3)x + 3 = 0$ $\lambda x^2 - (\lambda + 3)x + 3 = 0$:

α + β = (λ + 3)/(λ)

αβ = (3)/(λ)

We are given the condition:

(1)/(α) - (1)/(β) = (1)/(3) (β - α)/(αβ) = (1)/(3)

β - α = (αβ)/(3) = (3/λ)/(3) = (1)/(λ)

Step 1: Square Identity

We know that $(α + β)² - 4αβ = (β - α)²$ . Substitute the known values:

(β - α)² = (1)/(λ²)

(α + β)² = ((λ + 3)²)/(λ²)

So, we substitute these into the identity:

((λ + 3)²)/(λ²) - 4((3)/(λ)) = (1)/(λ²)

Step 2: Solving for Lambda

Multiply the entire equation by $λ²$ $\lambda^2$ (assuming $λ \neq 0$ $\lambda \neq 0$ because it's a quadratic leading coefficient):

(λ + 3)² - 12λ = 1

λ² + 6λ + 9 - 12λ - 1 = 0

λ² - 6λ + 8 = 0

Step 3: Finding Roots

(λ - 2)(λ - 4) = 0

So, $λ = 2$ $\lambda = 2$ or $λ = 4$ $\lambda = 4$. (The solution notes $λ=0$ $\lambda=0$ derived from $λ³ - 6λ² + 8λ = 0$ $\lambda^3 - 6\lambda^2 + 8\lambda = 0$ by cross multiplication, but $λ=0$ $\lambda=0$ degrades the quadratic. Thus valid $λ in 2, 4$ $\lambda \in \{2, 4\}$). Sum of possible values of $λ = 2 + 4 = 6$ $\lambda = 2 + 4 = 6$.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

If z = √(3)2 + (i)/(2), i = √(-1)$z = \frac{\sqrt{3}}{2} + \frac{i}{2}, i = \sqrt{-1}$, then (z²⁰¹ - i)⁸$(z^{201} - i)^8$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify Exponent

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Complex Numbers Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Circle Equation

Step 2: Roots and Modulus

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Factoring the Target Number

Step 2: Expanding the Product

Step 3: Comparing and Concluding

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Finding z

Step 2: Simplifying the Polynomial

Step 3: Evaluating the Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Square Identity

Step 2: Solving for Lambda

Step 3: Finding Roots

Chapter Mix

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If $z = \sqrt(3)2 + (i)/(2), i = \sqrt(-1)$ , then $(z²⁰¹ - i)⁸$ $(z^{201} - i)^8$ is equal to