24 Solved Complex Numbers Problems for JEE Main & WBJEE

24 Solved Complex Numbers Problems for JEE Main & WBJEE (Year-Wise Master Bank)

This master resource compiles 24 high-yielding, year-wise solved problems on Complex Numbers. Structured explicitly for competitive examinations like JEE Main and WBJEE, these questions emphasize rapid structural reductions, advanced modulus-argument applications, and core properties of the cube roots of unity.

Category I: WBJEE Master Set

Problem 1 (WBJEE)

If ω is an imaginary cube root of unity, evaluate the value of the determinant expression: Δ = | 1 ω ω² | / | ω ω² 1 | / | ω² 1 ω |.

Core Logic

Utilize column transformations combined with the cyclic algebraic identity 1 + ω + ω² = 0.

1. Execute the operational transform column-wise: C₁ → C₁ + C₂ + C₃.
2. This changes the first column matrix elements uniformly into (1 + ω + ω²).
3. Apply the identity 1 + ω + ω² = 0 directly to this column.
4. Since an entire structural column reduces identically to 0, the determinant evaluates to zero.

Correct Answer: 0

Problem 2 (WBJEE)

Determine the principal argument value of the complex fraction: z = (1 + i√3) / (1 - i√3).

Core Logic

Use the difference quotient rule of amplitudes: arg(z₁/z₂) = arg(z₁) - arg(z₂).

1. Identify the numerator position: z₁ = 1 + i√3. It falls within Quadrant I, so arg(z₁) = π/3.
2. Identify the denominator position: z₂ = 1 - i√3. It falls within Quadrant IV, so arg(z₂) = -π/3.
3. Subtract arguments linearly: arg(z) = π/3 - (-π/3) = 2π/3.
4. Check boundary conditions: 2π/3 falls safely inside the range -π < θ ≤ π.

Correct Answer: 2π/3

Problem 3 (WBJEE)

Find the value of the real parameter x such that the complex product (x - i)(3 + 2i) yields a purely imaginary number.

Core Logic

Expand the product fully and isolate the absolute Real Part Re(z) to evaluate to 0.

1. Expand the expression algebraically: (x - i)(3 + 2i) = 3x + 2xi - 3i - 2i².
2. Substitute i² = -1 to collect terms: (3x + 2) + i(2x - 3).
3. For a number to be purely imaginary, enforce Re(z) = 0 ⇒ 3x + 2 = 0.
4. Isolate the target variable: x = -2/3.

Correct Answer: -2/3

Problem 4 (WBJEE)

If (1+i)(1+2i)(1+3i)...(1+ni) = x + iy, find the value of the product sequence 2 · 5 · 10 ... (1+n²).

Core Logic

Apply the multiplicative property of modulus tracking states: |z₁ z₂ ... z_n| = |z₁||z₂|...|z_n|.

1. Take the modulus on both sides of the product statement.
2. Compute intermediate magnitudes: √(1+1²)√(1+2²)...√(1+n²) = √(x²+y²).
3. Square both sides to eliminate radicals: (1+1)(1+4)(1+9)...(1+n²) = x² + y².
4. Simplify to obtain the targeted sequence: 2 · 5 · 10 ... (1+n²) = x² + y².

Correct Answer: x² + y²

Problem 5 (WBJEE)

If the complex number expression evaluates to a singular unity modulus, namely |(z - 5i) / (z + 5i)| = 1, determine the locus of the point z = x + iy.

Core Logic

Interpret the relation geometrically as an equidistant locus between two designated coordinate centers on the plane.

1. Split the quotient modulus across fractions: |z - 5i| / |z + 5i| = 1 ⇒ |z - 5i| = |z + 5i|.
2. This represents the set of points equidistant from (0, 5) and (0, -5).
3. The perpendicular bisector of the line segment joining these points is the horizontal real coordinate axis.
4. Algebraically: x² + (y-5)² = x² + (y+5)² ⇒ -10y = 10y ⇒ y = 0.

Correct Answer: The Real Axis (X-axis, y = 0)

Problem 6 (WBJEE)

Evaluate the value of the finite sum sequence given by: ∑ (from n=1 to 100) of i^n.

Core Logic

Apply the property that the sum of any four consecutive powers of i equals zero.

1. The total number of terms in the sum is exactly 100.
2. Divide the total number of terms by 4: 100 / 4 = 25 complete cycles with zero remainder.
3. Since each cycle of four terms i^(4k+1) + i^(4k+2) + i^(4k+3) + i^(4k+4) = 0, all 25 blocks evaluate to zero.
4. The final total sum reduces to exactly zero.

Correct Answer: 0

Problem 7 (WBJEE)

If α and β are imaginary roots of the quadratic equation x² + x + 1 = 0, calculate the value of α^(2026) + β^(2026).

Core Logic

Recognize the roots as the standard cube roots of unity, ω and ω², and reduce their higher powers using ω³ = 1.

1. Assign roots: Let α = ω and β = ω².
2. Substitute these into the target expression: ω^(2026) + (ω²)^(2026) = ω^(2026) + ω^(4052).
3. Divide the exponents by 3 to find the remainders: 2026 = 3(675) + 1 and 4052 = 3(1350) + 2.
4. Simplify based on remainders: ω¹ + ω². Since 1 + ω + ω² = 0, it follows that ω + ω² = -1.

Correct Answer: -1

Problem 8 (WBJEE)

Find the minimum positive integer value of n such that the complex ratio satisfies the condition: ((1+i) / (1-i))^n = 1.

Core Logic

Rationalize the internal complex fraction before applying the exponential power bounds.

1. Multiply the numerator and denominator by the conjugate of the denominator: ((1+i)/(1-i)) · ((1+i)/(1+i)) = (1+i)² / (1 - i²).
2. Expand the terms: (1 + 2i - 1) / 2 = 2i / 2 = i.
3. Rewrite the original equation with this simplified base: i^n = 1.
4. The smallest positive integer that satisfies this condition is n = 4.

Correct Answer: 4

Problem 9 (WBJEE)

If the amplitude condition satisfies arg(z) < 0, find the principal argument of the conjugate tracking position arg(-z).

Core Logic

Analyze the positions of z and -z geometrically across the origin within the bounded range of principal arguments.

1. Let arg(z) = θ. Given the condition arg(z) < 0, θ lies in the interval (-π, 0).
2. Multiplying by -1 reflects the point through the origin, which adds π to the argument.
3. To keep the final value within the principal range (-π, π], add π to the negative angle: arg(-z) = θ + π.
4. Thus, the expression evaluates cleanly to arg(z) + π.

Correct Answer: arg(z) + π

Problem 10 (WBJEE)

Determine the complete geometric locus specified by the complex constraint equation: Re(1/z) = 1/4, where z ≠ 0.

Core Logic

Substitute the algebraic form z = x + iy and isolate the real component of the reciprocal fraction.

1. Express the reciprocal in standard form: 1/z = 1 / (x+iy) = (x-iy) / (x²+y²).
2. Isolate the real part: Re(1/z) = x / (x²+y²).
3. Set this equal to the given value: x / (x²+y²) = 1/4 ⇒ x² + y² = 4x.
4. Rearrange into standard form: x² - 4x + y² = 0, which represents a circle centered at (2,0) with radius 2.

Correct Answer: A circle given by x² + y² - 4x = 0

Problem 11 (WBJEE)

Find the value of the square root expression: √(-7 + 24i).

Core Logic

Express the radicand as a perfect square of a binomial matching the form (a + ib)².

1. Let -7 + 24i = a² - b² + 2abi. This implies 2ab = 24 ⇒ ab = 12.
2. Find two factors of 12 whose squares have a difference of -7: choose a = 3 and b = 4.
3. Verify: a² - b² = 3² - 4² = 9 - 16 = -7. This matches the real part.
4. Rewrite as a perfect square: -7 + 24i = (3 + 4i)². Take the square root: ±(3 + 4i).

Correct Answer: ±(3 + 4i)

Problem 12 (WBJEE)

If |z₁| = |z₂| = 1, compute the value of the expression: (z₁ + z₂)(1/z₁ + 1/z₂).

Core Logic

Use the modulus identity z · z̄ = |z|² = 1, which allows the reciprocals to be rewritten as conjugates: 1/z = z̄.

1. Since |z₁| = 1, substitute 1/z₁ = z̄₁. Similarly, substitute 1/z₂ = z̄₂.
2. Rewrite the target expression: (z₁ + z₂)(z̄₁ + z̄₂).
3. Apply the sum property of conjugates: (z₁ + z₂)(overline{z₁ + z₂}).
4. Simplify using the modulus identity: This matches the form Z · Z̄ = |Z|² = |z₁ + z₂|².

Correct Answer: |z₁ + z₂|²

Category II: JEE Main Master Set

Problem 13 (JEE Main)

If (1 + ω²)^n = (1 + ω^4)^n, where ω represents the imaginary cube root of unity, determine the least positive non-zero integer value for n.

Core Logic

Reduce the higher power using ω^4 = ω, and apply the cyclic identity 1 + ω + ω² = 0.

1. Simplify the right side: since ω^4 = ω, the equation becomes (1 + ω²)^n = (1 + ω)^n.
2. Substitute the identities 1 + ω² = -ω and 1 + ω = -ω².
3. Equate the simplified terms: (-ω)^n = (-ω²)^n.
4. Divide both sides to collect terms: (-ω² / -ω)^n = 1 → ω^n = 1.
5. The smallest positive integer that satisfies ω^n = 1 is n = 3.

Correct Answer: 3

Problem 14 (JEE Main)

Find the common roots of the equations z³ + 2z² + 2z + 1 = 0 and z^(2026) + z^(2025) + 1 = 0.

Core Logic

Factor the cubic equation to find its roots, then test those roots in the higher-degree equation.

1. Factor the cubic equation by grouping: (z³ + 1) + 2z(z + 1) = 0 → (z + 1)(z² + z + 1) = 0. The roots are z = -1, z = ω, and z = ω².
2. Test z = -1 in the second equation: (-1)^(2026) + (-1)^(2025) + 1 = 1 - 1 + 1 = 1 ≠ 0. Not a root.
3. Test z = ω: ω^(2026) + ω^(2025) + 1 = ω¹ + 1 + 1 = ω + 2 ≠ 0.
4. Thus, these equations share no common roots.

Correct Answer: No common roots exist

Problem 15 (JEE Main)

Let z be a complex number such that the modulus equation satisfies |z - 1| = |z + 1| = |z - i|. Find the total number of distinct solutions for z.

Core Logic

Interpret the equations geometrically as the intersection of perpendicular bisectors on the coordinate plane.

1. The equation condition |z - 1| = |z + 1| defines the perpendicular bisector between (1,0) and (-1,0), which is the Y-axis (x = 0).
2. The equation condition |z - 1| = |z - i| defines the perpendicular bisector between (1,0) and (0,1), which is the line y = x.
3. Find the unique intersection point of these two lines: x = 0 and y = x ⇒ (0,0).
4. The only point that satisfies these conditions is the origin, z = 0. This yields exactly one distinct solution.

Correct Answer: 1

Problem 16 (JEE Main)

If z = x + iy and satisfies the inequality constraint equation log₀.₅((|z|² + |z| + 4) / (3|z| - 1)) > 0, determine the valid range for |z|.

Core Logic

When solving inequalities where the logarithmic base is less than 1 (0.5 < 1), reverse the inequality sign when removing the logarithm.

1. Set up the inequality: (|z|² + |z| + 4) / (3|z| - 1) < (0.5)⁰ ⇒ (|z|² + |z| + 4) / (3|z| - 1) < 1.
2. Ensure the denominator is positive for the log domain: 3|z| - 1 > 0 ⇒ |z| > 1/3.
3. Cross-multiply safely: |z|² + |z| + 4 < 3|z| - 1 ⇒ |z|² - 2|z| + 5 < 0.
4. The discriminant of this quadratic is (-2)² - 4(1)(5) = -16 (negative). Since the leading coefficient is positive, the expression is always positive. It can never be less than zero.

Correct Answer: No real solution for |z|

Problem 17 (JEE Main)

Determine the real part Re(z) of the complex number solution that satisfies the system relation: z + 2z̄ = (1 + 5i)².

Core Logic

Substitute the algebraic form z = x + iy, expand the terms, and equate the real and imaginary components.

1. Expand the right side of the equation: (1 + 5i)² = 1 + 10i + 25i² = -24 + 10i.
2. Substitute z = x + iy and z̄ = x - iy into the left side: (x + iy) + 2(x - iy) = 3x - iy.
3. Equate the expressions: 3x - iy = -24 + 10i.
4. Match components: 3x = -24 ⇒ x = -8, and -y = 10 ⇒ y = -10. The real part is -8.

Correct Answer: -8

Problem 18 (JEE Main)

Calculate the value of the complex expression: ((-1 + i√3) / 2)³⁰ + ((-1 - i√3) / 2)³⁰.

Core Logic

Recognize the base terms as the standard cube roots of unity, ω and ω², and apply the property ω³ = 1.

1. Rewrite the expression using ω and ω²: ω³⁰ + (ω²)³⁰ = ω³⁰ + ω⁶⁰.
2. Since both exponents are multiples of 3, rewrite them in terms of ω³: (ω³)¹⁰ + (ω³)²⁰.
3. Substitute ω³ = 1: (1)¹⁰ + (1)²⁰ = 1 + 1 = 2.

Correct Answer: 2

Problem 19 (JEE Main)

Find the maximum value of |z| given that the complex number satisfies the constraint inequality: |z - 4/z| = 2.

Core Logic

Apply the triangle inequality modulus property ||z₁| - |z₂|| ≤ |z₁ + z₂| to establish a bounding quadratic equation.

1. Set up the bounding inequality using the triangle property: |z| - |4/z| ≤ |z - 4/z|.
2. Substitute the given boundary value: |z| - 4/|z| ≤ 2.
3. Rearrange into a quadratic inequality: |z|² - 2|z| - 4 ≤ 0.
4. Find the positive roots of the corresponding equation: |z| = (2 + √(4 - 4(1)(-4))) / 2 = 1 + √5.

Correct Answer: 1 + √5

Problem 20 (JEE Main)

Find the principal argument arg(z) of the complex number z = (1+i)^5 / (1-i)^3.

Core Logic

Simplify the expression algebraically using standard operational properties of the imaginary unit base powers.

1. Rewrite the expression by grouping terms: z = ((1+i)/(1-i))³ · (1+i)².
2. Simplify the base fraction: (1+i)/(1-i) = i. This reduces the first term to i³ = -i.
3. Expand the second term: (1+i)² = 1 + 2i - 1 = 2i.
4. Multiply together: z = (-i) · (2i) = -2i² = 2. Since z = 2 is a positive real number, its principal argument is 0.

Correct Answer: 0

Problem 21 (JEE Main)

If z = (√3 + i) / 2, calculate the value of the higher-power expression: z^(102) + z^(108).

Core Logic

Convert the complex number to polar form to simplify index metrics with De Moivre's Theorem.

1. Express z in polar form: z = cos(π/6) + i sin(π/6).
2. Evaluate the first term: z^(102) = cos(102π/6) + i sin(102π/6) = cos(17π) + i sin(17π) = -1.
3. Evaluate the second term: z^(108) = cos(108π/6) + i sin(108π/6) = cos(18π) + i sin(18π) = 1.
4. Sum the two values: -1 + 1 = 0.

Correct Answer: 0

Problem 22 (JEE Main)

Determine the geometric locus of points specified by the equation: |z - 3 - 4i| = 5.

Core Logic

Recognize the absolute modulus equation form |z - z⁰| = r as the standard geometric definition of a circle.

1. Rewrite the expression in standard locus form: |z - (3 + 4i)| = 5.
2. This equation represents the locus of all points z that maintain a constant distance of 5 units from the fixed point (3,4).
3. This matches the geometric definition of a circle. The center is located at (3,4) and the radius is 5 units.

Correct Answer: A circle with center (3,4) and radius 5

Problem 23 (JEE Main)

If arg(z₁ z₂) = 0 and |z₁| = |z₂| = 1, determine the true relation connecting z₁ and z₂.

Core Logic

Use the property that the argument of a product is the sum of their individual arguments: arg(z₁ z₂) = arg(z₁) + arg(z₂).

1. Let arg(z₁) = θ₁ and arg(z₂) = θ₂. The given condition implies θ₁ + θ₂ = 0 ⇒ θ₂ = -θ₁.
2. Express the numbers in exponential form: z₁ = e^(iθ₁) and z₂ = e^(-iθ₁).
3. Find the conjugate of the first number: z̄₁ = e^(-iθ₁).
4. Comparing the terms shows that z₂ = z̄₁.

Correct Answer: z₂ = z̄₁

Problem 24 (JEE Main)

Find the value of the polynomial expression evaluated at the given complex point:

P(x) = 2x⁴ + 5x³ + 7x² - x + 41

where x = -2 - i√3.

Core Logic

Construct a localized quadratic equation from the roots of x to eliminate the imaginary unit, then use polynomial reduction to evaluate the remainder.

1. Isolate the real number constant to one side: x + 2 = -i√3.
2. Square both sides of the equation to eliminate the radical component: (x + 2)² = (-i√3)² ⇒ x² + 4x + 4 = -3.
3. Equate the simplified quadratic relation to zero: x² + 4x + 7 = 0.
4. Express the target polynomial P(x) in terms of our quadratic divisor using long division: P(x) = (x² + 4x + 7)(2x² - 3x + 5) + 6.
5. Substitute the known condition where x² + 4x + 7 = 0: P(x) = (0) · (2x² - 3x + 5) + 6 ⇒ P(x) = 6.

Correct Answer: 6