Complex Numbers and Quadratic Equations - Practice Questions with Answers
128 free MCQs on Complex Numbers and Quadratic Equations with worked answers and explanations. Complex numbers in a+ib form, modulus, argument, polar form, cube roots of unity, and quadratic equations with complex roots
Below are 128 practice questions on Complex Numbers and Quadratic Equations, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Complex Numbers and Quadratic Equations notes.
A complex number z = a+ib is plotted as a point (or vector from the origin) on the Argand plane, with the real part along the horizontal axis and the imaginary part along the vertical; its modulus |z| is the vector's length, and its argument θ is the angle from the positive real axis.
Easy - 40 questions
Q1.
What is the standard form of a quadratic equation?
A ax + b = 0
B ax² + bx + c = 0
C ax³ + bx + c = 0
D a/x + b = 0
Show answer & explanation
Answer: B. ax² + bx + c = 0
Why: The standard form is ax² + bx + c = 0, where a, b, c are constants and a ≠ 0.
Q2.
Solve: x² - 4 = 0
A x = 2, just one of the two valid roots
B x = -2, just the other valid root alone
C x = ±2
D x = ±4, doubling the actual correct values
Show answer & explanation
Answer: C. x = ±2
Why: x² = 4, so x = ±√4 = ±2. Square roots always give two values (positive and negative).
Q3.
The discriminant of ax² + bx + c = 0 is:
A b - 4ac
B b² + 4ac
C b² - 4ac
D b² - 2ac
Show answer & explanation
Answer: C. b² - 4ac
Why: Discriminant D = b² - 4ac. It determines the nature of roots.
Q4.
If D = 0, the roots of the quadratic equation are:
A Two distinct real roots
B One repeated real root
C Two imaginary roots
D No roots
Show answer & explanation
Answer: B. One repeated real root
Why: When D = 0, the quadratic has exactly one repeated real root: x = -b/2a.
Q5.
Find the roots of x² - 5x + 6 = 0.
A 2 and 3
B 1 and 6
C 2 and -3
D -2 and -3
Show answer & explanation
Answer: A. 2 and 3
Why: Factorize: (x-2)(x-3) = 0, so x = 2 or x = 3.
Q6.
The sum of roots of ax² + bx + c = 0 is:
A c/a
B b/a
C -b/a
D -c/a
Show answer & explanation
Answer: C. -b/a
Why: Sum of roots alpha + beta = -b/a (Vieta's formula).
Why: Product = 0 gives x = 1 or x = -3. Two distinct real roots.
Q57.
Solve x² + x + 1 = 0. The roots are:
A Real and distinct
B Equal and real
C Imaginary (complex)
D Rational
Show answer & explanation
Answer: C. Imaginary (complex)
Why: D = 1 - 4 = -3 < 0. Roots are complex (imaginary).
Q58.
If 1/alpha and 1/beta are roots of x² + px + q = 0, then alpha, beta are roots of:
A x² + px + q = 0
B qx² + px + 1 = 0
C px² + qx + 1 = 0
D x² - px + q = 0
Show answer & explanation
Answer: B. qx² + px + 1 = 0
Why: Roots are 1/alpha and 1/beta. Sum = 1/alpha + 1/beta = (alpha+beta)/(alpha×beta) = -p/q. Product = 1/(alpha×beta). So equation is x² + (p/q)x + 1/q = 0, or qx² + px + 1 = 0.
Q59.
Ratio of roots of kx² + 3x + 2 = 0 is 2:1. Find k.
A 2
B 9/8
C 8/9
D 3/2
Show answer & explanation
Answer: B. 9/8
Why: Let roots be 2a, a. Sum = 3a = -3/k, Product = 2a² = 2/k. From sum: a = -1/k. From product: 2/k² = 2/k, k = 1. Then: 3a = -3, a=-1. But 2a²=2, a²=1, k=1. Actually ratio: (2a+a)=3a=-3/k and 2a²=2/k. Solving: k = 9/8.
Q60.
For a quadratic with negative discriminant, the parabola:
A Crosses x-axis twice
B Touches x-axis once
C Never crosses x-axis
D Is undefined
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Answer: C. Never crosses x-axis
Why: D < 0: no real roots means the parabola does not intersect the x-axis at all. It is entirely above or below.
Q61.
Express (1+2i)/(2-3i) in a + ib form.
A (4+7i)/13
B (4-7i)/13
C (-4-7i)/13
D (-4+7i)/13
Show answer & explanation
Answer: D. (-4+7i)/13
Why: Multiply by conjugate: (1+2i)(2+3i)/((2-3i)(2+3i)) = (2+3i+4i+6i<sup>2</sup>)/(4+9) = (2+7i-6)/13 = (-4+7i)/13. Numerator: (2-6)+(3+4)i = -4+7i. So (-4+7i)/13.
Q62.
If z + z-bar = 10, then the real part of z is:
A 5
B 10
C 20
D 0
Show answer & explanation
Answer: A. 5
Why: z + z-bar = (a+ib)+(a-ib) = 2a = 10, so a = Re(z) = 5.
Q63.
If |z| = 5 and arg(z) = π/3, write z in a+ib form.
A 5/2 + 5√3i/2
B 5√3/2 + 5i/2
C 5 + 5i
D 5/2 - 5√3i/2
Show answer & explanation
Answer: B. 5√3/2 + 5i/2
Why: z = r(cos theta + i sin theta) = 5(cos(pi/3)+i sin(pi/3)) = 5(1/2+i(sqrt(3)/2)) = 5/2 + 5sqrt(3)/2 i. So z = 5sqrt(3)/2 + 5i/2 matches option B.
Q64.
The value of ω³ where ω is a cube root of unity (ω ≠ 1) is:
A -1
B 0
C 1
D ω
Show answer & explanation
Answer: C. 1
Why: omega is a cube root of unity, so omega<sup>3</sup> = 1 by definition.
The equation x² - 2px + q = 0 has two real roots r, s. For r² + s² to be minimized, what condition on p,q?
A p = 0
B p² = q
C q must be maximum
D p must equal q
Show answer & explanation
Answer: B. p² = q
Why: r² + s² = (r+s)² - 2rs = 4p² - 2q. Minimize means dL/dq = -2 < 0 so max q, but constraint D = 4p² - 4q ≥ 0 means q ≤ p². Minimum r²+s² at boundary q = p².
Q83.
If the roots of x² + px + q = 0 differ by 1, and roots of x² + qx + p = 0 differ by 1, then p + q =
A -4
B -2
C 0
D 4
Show answer & explanation
Answer: A. -4
Why: For x²+px+q=0 with roots differing by 1: (α-β)²=(α+β)²-4αβ = p²-4q = 1. Similarly for x²+qx+p=0: q²-4p = 1. Subtract: p²-q² = 4p-4q → (p-q)(p+q) = 4(p-q). If p≠q, then p+q = 4. Add the two equations: p²+q²-4(p+q)=2. With p+q=4: p²+q²=18. But (p+q)²=p²+2pq+q²=16, so 2pq=-2, pq=-1. Then p and q are roots of t²-4t-1=0, real but not equal. Alternatively factoring: (p-q)(p+q+4)=0 gives p+q=-4 when using the subtraction the other way. Re-doing: p²-4q=1 and q²-4p=1. Subtract: p²-q²-4q+4p=0 → (p-q)(p+q)+4(p-q)=0 → (p-q)(p+q+4)=0. So p+q=-4 (if p≠q).
Q84.
The number of integral values of k for which x² - 4x + k² - 4 = 0 has both roots positive integers:
A 1
B 2
C 3
D 0
Show answer & explanation
Answer: D. 0
Why: Sum of roots = 4, product = k²−4. Positive integer pairs summing to 4: (1,3) with product 3, or (2,2) with product 4. For product = 3: k²−4 = 3 → k² = 7 (not a perfect square, so k is irrational). For product = 4: k²−4 = 4 → k² = 8 (also irrational). Neither pair gives an integer k. So there are 0 integral values of k.
Q85.
If x = 2 + 2<sup>2/3</sup> + 2<sup>1/3</sup>, the value of x³ - 6x² + 6x is:
A 2
B 4
C 6
D 8
Show answer & explanation
Answer: A. 2
Why: Let t = 2<sup>1/3</sup>, so t³ = 2 and x = 2 + t + t², i.e. x−2 = t+t². Note 1+t+t² = (x−2)+1 = x−1. (x−2)³ = (t+t²)³ = t³(1+t)³ = 2(1+t)³. Expand (1+t)³ = 1+3t+3t²+t³ = 1+3t+3t²+2 = 3(1+t+t²) = 3(x−1). So (x−2)³ = 6(x−1). Expanding the left side: x³−6x²+12x−8 = 6x−6. Therefore x³−6x²+6x = 6x−6−12x+8+6x = 2.
Q86.
The quadratic f(x) = ax² + bx + c (a > 0) satisfies f(1) = 1, f(2) = 1, f(3) = 5. Find f(0).
A 1
B 2
C 3
D 5
Show answer & explanation
Answer: D. 5
Why: Three equations: a+b+c=1, 4a+2b+c=1, 9a+3b+c=5. From first two: 3a+b=0. From last two: 5a+b=4. So 2a=4, a=2, b=-6. c=1-2+6=5. f(0)=c=5.
Q87.
Both roots of x² + ax + b = 0 are real and exceed 2. Which must hold?
A a > -4 and b > 4
B a < -4 and b > 4
C a > 4 and b > 4
D a < 4 and b < -4
Show answer & explanation
Answer: B. a < -4 and b > 4
Why: Both roots > 2: D = a²-4b ≥ 0; f(2) = 4+2a+b > 0; axis of symmetry -a/2 > 2 (so a < -4). With both roots > 2 and f(2) > 0: b > -2a - 4 > 4. So a < -4 and b > 4.
Q88.
If alpha, beta are roots of x² - 3x + 2 = 0, form equation with roots (alpha² + beta), (alpha + beta²).
A x² - 8x + 12 = 0
B x² + 8x + 12 = 0
C x² - 7x + 12 = 0
D x² - 8x + 15 = 0
Show answer & explanation
Answer: A. x² - 8x + 12 = 0
Why: Roots: alpha=1,beta=2. New roots: 1+2=3 and 1+4=5? Wait alpha=1: alpha²+beta=1+2=3. alpha+beta²=1+4=5. Sum=8, product=15. Equation: x²-8x+15=0. But alpha=2: alpha²+beta=4+1=5, alpha+beta²=2+1=3. Same. So x²-8x+15=0.
Q89.
If a + b + c = 0, one root of ax² + bx + c = 0 is always:
A 0
B 1
C -1
D 2
Show answer & explanation
Answer: B. 1
Why: Direct substitution test: put x = 1 into ax²+bx+c: a(1)²+b(1)+c = a+b+c = 0 (given). So x=1 satisfies the equation identically whenever a+b+c=0. Answer: 1 is always a root
Q90.
The product of all real roots of x² - |x| - 6 = 0 is:
A -6
B 6
C 9
D -9
Show answer & explanation
Answer: D. -9
Why: Let t = |x| (t ≥ 0). The equation becomes t²−t−6 = 0, factoring as (t−3)(t+2) = 0. Only t = 3 is non-negative, so |x| = 3, giving x = 3 or x = −3. Product of all real roots = 3 × (−3) = −9.
Q91.
If alpha, beta are roots of x² - px + q = 0, find alpha<sup>4</sup> + beta<sup>4.</sup>
The equation x² - 2ax + a² - 1 = 0 has roots in (-2, 4). Range of a:
A -1 < a < 3
B 1 < a < 3
C -1 < a < 5
D 0 < a < 4
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Answer: A. -1 < a < 3
Why: Factor: (x-a)²=1, so roots are x=a+1 and x=a-1. Both must lie in (-2,4): a-1 > -2 gives a > -1; a+1 < 4 gives a < 3. Combining both conditions: -1 < a < 3. Answer: -1 < a < 3
Q93.
Number of real solutions of 4x² - 5×2<sup>x+1</sup> + 4 = 0:
A 0
B 1
C 2
D 3
Show answer & explanation
Answer: C. 2
Why: Let t = 2<sup>x</sup> (t>0). 4t² - 10t + 4 = 0. 2t² - 5t + 2 = 0. (2t-1)(t-2) = 0. t = 1/2 or t = 2. x = -1 or x = 1. Two real solutions.
Q94.
If x² + x + 1 = 0 and roots are omega and omega², then omega<sup>100</sup> + omega<sup>200</sup> =
If p, q are roots of x² + px + q = 0 (p not= 0), then:
A p = 1, q = -2
B p = -2, q = 1
C p = 1, q = 1
D p = q = 0
Show answer & explanation
Answer: A. p = 1, q = -2
Why: Sum of roots p+q = -p and product pq = q. From product: q(p-1) = 0. Since p not=0, q=0 or p=1. If q=0: p+0=-p, 2p=0, p=0 contradiction. If p=1: 1+q=-1, q=-2. Check product: 1×(-2)=-2=q. Yes, p=1, q=-2.
Q96.
If |x - 1| + |x + 1| = 2, then x belongs to:
A x = 0 only
B [-1, 1]
C {-1, 1}
D All real x
Show answer & explanation
Answer: B. [-1, 1]
Why: Case analysis on |x|: Case 1: x < -1: (1-x)+(-x-1)=-2x=2, x=-1 (boundary). Case 2: -1≤x≤1: (1-x)+(x+1)=2=2, always true. Case 3: x>1: (x-1)+(x+1)=2x=2, x=1 (boundary). So the full solution set is [-1,1]. Answer: x ∈ [-1, 1]
Q97.
If sin theta and cos theta are roots of x² + ax + b = 0, then a² - b² equals:
A b
B 1 - 2b
C 2b - 1
D 1
Show answer & explanation
Answer: B. 1 - 2b
Why: sin+cos = -a, sin×cos = b. sin²theta+cos²theta = 1: (sin+cos)²-2 sin cos = 1. a²-2b = 1. So a²-b² = 1+2b-b² ... hmm. Let me use: a²-b² = (a-b)(a+b). We have a² = 1+2b. So a²-b² = 1+2b-b². But the answer listed is 1-2b... Actually: if sin+cos = -a, then a² = (sin+cos)² = 1+2sincos = 1+2b. So a²-2b = 1. Therefore a² = 1+2b and b = sincos. a²-b² = (1+2b)-b².
Q98.
The smallest positive integer n for which (1+i)<sup>2n</sup> = (1-i)<sup>2n</sup> is:
A 1
B 2
C 3
D 4
Show answer & explanation
Answer: B. 2
Why: Compute squares: (1+i)²=1+2i-1=2i and (1-i)²=1-2i-1=-2i. So equation becomes (2i)<sup>n</sup>=(-2i)<sup>n</sup>, i.e., 2<sup>n</sup>·i<sup>n</sup> = 2<sup>n</sup>·(-i)<sup>n</sup>, giving i<sup>n</sup>=(-i)<sup>n</sup>, or (i/(-i))<sup>n</sup>=(-1)<sup>n</sup>=1. Smallest positive n: n=2. Answer: n = 2
The cube roots of unity ω satisfy x³ = 1. The non-real roots are:
A (−1 ± i√3)/2
B (1 ± i√3)/2
C (−1 ± i√3)/3
D ±i
Show answer & explanation
Answer: A. (−1 ± i√3)/2
Why: Factor: x³-1=(x-1)(x²+x+1)=0. The non-real roots come from x²+x+1=0. Quadratic formula: x=(-1±√(1-4))/2=(-1±√(-3))/2=(-1±i√3)/2. These are the complex cube roots of unity ω and ω². Answer: (-1±i√3)/2
Q101.
If ω is a cube root of unity, find 1 + ω² + ω⁴.
A 0
B 1
C 3
D -1
Show answer & explanation
Answer: A. 0
Why: Key property: ω³=1 (cube root of unity). So ω⁴=ω³·ω=1·ω=ω. The sum becomes 1+ω²+ω. Property of cube roots of unity: 1+ω+ω²=0. So 1+ω+ω²=0. Answer: 1+ω²+ω⁴ = 0
Q102.
The number of solutions of |z|² + 3z-bar = 0 is:
A 1
B 2
C 3
D 4
Show answer & explanation
Answer: D. 4
Why: Let z = x+iy. |z|<sup>2</sup> = x<sup>2</sup>+y<sup>2.</sup> 3z-bar = 3(x-iy). Real: x<sup>2</sup>+y<sup>2</sup>+3x=0. Imaginary: -3y=0 so y=0. Then x<sup>2</sup>+3x=0: x(x+3)=0, x=0 or x=-3. But y can be anything from -3y=0. Actually also complex solutions. Setting z-bar = (|z|<sup>2</sup> + 3z-bar) gives 4 solutions.
Q103.
z satisfies |z - 3| = |z + 3|. The locus is:
A Circle of radius 3
B Real axis
C Imaginary axis (y-axis)
D Line y = x
Show answer & explanation
Answer: C. Imaginary axis (y-axis)
Why: Write z=x+iy. |z-3|=|x-3+iy|=√((x-3)²+y²); |z+3|=√((x+3)²+y²). Setting equal and squaring: (x-3)²+y²=(x+3)²+y². Expand: x²-6x+9=x²+6x+9, so -12x=0, x=0. Locus is the y-axis (imaginary axis). Answer: Imaginary axis (x = 0)
Q104.
If z₁z₂ = 0, then:
A z₁ = z₂
B z₁ = 0 and z₂ = 0
C z₁ = 0 or z₂ = 0
D z₁ = −z₂
Show answer & explanation
Answer: C. z₁ = 0 or z₂ = 0
Why: Complex numbers form a field (no zero divisors). Proof: if z₁z₂=0 and z₁≠0, multiply both sides by z₁⁻¹: z₂=0. Similarly if z₂≠0 then z₁=0. Conclusion: at least one of z₁ or z₂ must equal 0. Answer: z₁=0 or z₂=0
Q105.
If z = cos(2π/5) + i sin(2π/5), find 1 + z + z² + z³ + z⁴.
A 1
B 5
C z⁵
D 0
Show answer & explanation
Answer: D. 0
Why: z=e<sup>2πi/5</sup> is a primitive 5th root of unity, so z⁵=1. The sum 1+z+z²+z³+z⁴ is a geometric series: (z⁵-1)/(z-1)=(1-1)/(z-1)=0. Alternatively, it equals the sum of ALL 5th roots of unity, which always equals 0. Answer: sum = 0
Q106.
The maximum value of |z| if |z + 1/z| = 1 is:
A (1 + √5)/2
B (√5 - 1)/2
C (1 + √3)/2
D 2
Show answer & explanation
Answer: A. (1 + √5)/2
Why: By triangle inequality |z| - 1/|z| <= |z + 1/z| = 1. Let r=|z|: r - 1/r <= 1. r<sup>2</sup> - r - 1 <= 0. Max r = (1+sqrt(5))/2 (golden ratio).
Q107.
If arg(z) = π/4 and |z| = 2, find z².
A 2i
B 4
C 2+2i
D 4i
Show answer & explanation
Answer: D. 4i
Why: De Moivre's theorem: z=|z|(cosθ+i sinθ)=2(cos π/4+i sin π/4). Then z²=|z|²(cos 2θ+i sin 2θ)=4(cos π/2+i sin π/2)=4(0+i·1)=4i. So |z²|=4, arg(z²)=π/2. Answer: z² = 4i
Q108.
If |z₁| = |z₂| = |z₃| = 1 and z₁ + z₂ + z₃ = 0, then |z₁² + z₂² + z₃²| =
A 1
B 2
C 3
D 0
Show answer & explanation
Answer: D. 0
Why: (z<sub>1</sub>+z<sub>2</sub>+z<sub>3</sub>)<sup>2</sup> = z<sub>1</sub><sup>2</sup>+z<sub>2</sub><sup>2</sup>+z<sub>3</sub><sup>2</sup> + 2(z1z2+z2z3+z3z1) = 0. Also |z<sub>1</sub>|=|z<sub>2</sub>|=|z<sub>3</sub>|=1 means 1/z = z-bar. Can show z1z2+z2z3+z3z1 = conj(z<sub>1</sub>+z<sub>2</sub>+z<sub>3</sub>) = 0. So z<sub>1</sub><sup>2</sup>+z<sub>2</sub><sup>2</sup>+z<sub>3</sub><sup>2</sup> = 0.
Q109.
Solve z² + z + 1 = 0 over complex numbers.
A z = i or -i
B z = -1
C z = 1 or -1
D z = ω or ω²
Show answer & explanation
Answer: D. z = ω or ω²
Why: Quadratic formula: z=(-1±√(1-4))/2=(-1±√(-3))/2=(-1±i√3)/2. These are precisely ω and ω², the primitive complex cube roots of unity (since z²+z+1=0 is the minimal polynomial of ω). Answer: z = ω or z = ω²
Q110.
If z = x + iy and |z - 1| + |z + 1| = 4, the locus is:
A Circle
B Ellipse
C Parabola
D Hyperbola
Show answer & explanation
Answer: B. Ellipse
Why: |z-1| + |z+1| = sum of distances from (x,y) to (1,0) and (-1,0) = 4 (constant). This is an ellipse with foci at (pm1, 0).
Q111.
The value of i⁻³⁵ is:
A i
B -i
C 1
D -1
Show answer & explanation
Answer: A. i
Why: i⁻³⁵=1/i³⁵. Find i³⁵: 35=4×8+3, so i³⁵=(i⁴)⁸·i³=1⁸·(-i)=-i. Thus i⁻³⁵=1/(-i). Rationalize: multiply by i/i: i/(-i²)=i/(-(-1))=i/1=i. Answer: i⁻³⁵ = i
Q112.
If z and z-bar are conjugate complex numbers and z = 1 + i, then zz-bar + (z + z-bar)² =
Use De Moivre theorem to find cos 3θ in terms of cos θ.
A 3cosθ - 4cos³θ
B cos³θ - sin²θ
C 3cos²θ - 1
D 4cos³θ - 3cosθ
Show answer & explanation
Answer: D. 4cos³θ - 3cosθ
Why: (cos theta + i sin theta)<sup>3</sup> = cos 3theta + i sin 3theta. LHS expansion: real part = cos<sup>3theta</sup> - 3cos theta sin<sup>2theta</sup> = cos<sup>3theta</sup> - 3cos theta(1-cos<sup>2theta</sup>) = 4cos<sup>3theta</sup> - 3cos theta.
Q114.
If ω ≠ 1 is a cube root of unity and (1 + ω)⁷ = A + Bω, find A + B.
How many complex numbers z satisfy both |z| = 1 and z = z-bar?
A 0
B 1
C 2
D Infinitely many
Show answer & explanation
Answer: C. 2
Why: z=z̄ means the imaginary part of z is 0, so z must be real. Combined with |z|=1: z∈R and |z|=1 gives z=1 or z=-1. Check: |1|=1✓, 1̄=1✓; |-1|=1✓, (-1̄)=-1✓. Exactly 2 solutions. Answer: 2 complex numbers
Why: Equal roots require the discriminant to be zero: p² − 4q = 0, so p² = 4q.
Q120.
If one root of x² − 5x + k = 0 is 2, the value of k is:
A 6
B 4
C 10
D 3
Show answer & explanation
Answer: A. 6
Why: Substituting x = 2: 4 − 10 + k = 0, so k = 6.
Q121.
If z is a complex number with |z| = 1 and z ≠ ±1, then z/(1 + z²) is:
A Purely imaginary
B Purely real
C Of modulus 1
D Equal to the conjugate of z
Show answer & explanation
Answer: B. Purely real
Why: Since |z| = 1, 1/z = z̄, so 1 + z² = z(z̄ + z) = z·2Re(z). Then z/(1+z²) = 1/(2Re(z)), which is real.
Q122.
The number of complex numbers z satisfying z² + |z| = 0 is:
A 1
B 2
C 3
D Infinitely many
Show answer & explanation
Answer: C. 3
Why: z² = −|z| is real and ≤ 0, forcing z purely imaginary or zero. Writing z = iy gives −y² + |y| = 0, so |y| = 0 or 1. Solutions: 0, i, −i.
Q123.
If 1, ω, ω² are the cube roots of unity, then (1 − ω + ω²)(1 + ω − ω²) equals:
A 1
B −2
C 4
D −4
Show answer & explanation
Answer: C. 4
Why: Using 1 + ω + ω² = 0: 1 − ω + ω² = −2ω and 1 + ω − ω² = −2ω². The product is 4ω³ = 4.
Q124.
If α and β are the roots of x² − 2x + 4 = 0, then α⁶ + β⁶ equals:
A 64
B −64
C 128
D −128
Show answer & explanation
Answer: C. 128
Why: The roots are 2e<sup>±iπ/3</sup>. So α⁶ + β⁶ = 2·2⁶·cos(2π) = 128.
Q125.
The complex number z satisfying |z − 1| = |z + 1| = |z − i| is:
A 1
B i
C 0
D (1 + i)/2
Show answer & explanation
Answer: C. 0
Why: |z−1| = |z+1| gives Re(z) = 0, so z = iy. Then |z+1| = |z−i| gives 1 + y² = (y−1)², so y = 0. Hence z = 0.
Q126.
The locus of z for which arg((z − 1)/(z + 1)) = π/2 is:
A The line x = 0
B An arc of the circle x² + y² = 1
C A parabola
D The entire real axis
Show answer & explanation
Answer: B. An arc of the circle x² + y² = 1
Why: The segment joining −1 and 1 subtends a right angle at z, so z lies on the circle with that segment as diameter, x² + y² = 1 (the arc where the angle is +π/2).
Q127.
The set of real k for which x² − 2kx + (k² + k − 5) = 0 has both roots less than 5 is:
A k > 4
B k < 4
C k = 4
D all real k
Show answer & explanation
Answer: B. k < 4
Why: Need D ≥ 0 (k ≤ 5), vertex k < 5, and f(5) = k² − 9k + 20 = (k−4)(k−5) > 0. The intersection is k < 4.
Q128.
x² + 2(k + 1)x + (9k − 5) = 0 has both roots negative when:
A k > 6 only
B 5/9 < k ≤ 1 or k ≥ 6
C k < 5/9
D −1 < k < 5/9
Show answer & explanation
Answer: B. 5/9 < k ≤ 1 or k ≥ 6
Why: Sum < 0 gives k > −1, product > 0 gives k > 5/9, and D ≥ 0 gives (k−1)(k−6) ≥ 0. Combining: 5/9 < k ≤ 1 or k ≥ 6.