Complex Numbers and Quadratic Equations — Class 11 MCQs with Answers

Complex Numbers and Quadratic Equations — important questions & MCQs with answers (Class 11 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Value of i² is:

   AiB1C0D−1
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   Answer: −1

   Explanation: i is defined so that i² = −1 — the whole point of introducing imaginary numbers. Mixing it up with i⁴ = 1 (two squarings later) is the common slip.

2. Q2 of 6Easy

   Roots of x²−5x+6=0:

   A1, 6B−2, −3C2 and 3D5, 1
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   Answer: 2 and 3

   Explanation: Factor x²−5x+6 as (x−2)(x−3), so the roots are 2 and 3 — check: sum 5 matches −(−5)/1 and product 6 matches 6/1. Picking the sign-flipped pair, −2 and −3, is the usual mix-up.

3. Q3 of 6Medium

   Conjugate of (z₁ + z₂) equals:

   Az̄₁ · z̄₂Bz̄₁ − z̄₂Cz̄₁ + z̄₂Dz₁ + z₂
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   Answer: z̄₁ + z̄₂

   Explanation: Conjugation distributes over addition: the conjugate of z₁+z₂ is z̄₁+z̄₂ — flip each imaginary part separately. Leaving the sum un-conjugated, as plain z₁+z₂, misses that both parts flip.

4. Q4 of 6Medium

   Assertion (A): (2 + 3i)(1 – i) = 5 + i. Reason (R): For z₁ = a + ib and z₂ = c + id, z₁z₂ = (ac – bd) + i(ad + bc).

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: A is true: with a = 2, b = 3, c = 1, d = –1, ac – bd = 2 + 3 = 5 and ad + bc = –2 + 3 = 1, giving 5 + i. R is the product rule, and applying it gives A directly, so R explains A.

5. Q5 of 6Hard

   For complex z₁, z₂: |z₁ + z₂| ≤ ?

   A|z₁| − |z₂|B|z₁| + |z₂|C|z₁| · |z₂|D|z₁|
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   Answer: |z₁| + |z₂|

   Explanation: The triangle inequality for complex numbers states |z₁+z₂| ≤ |z₁|+|z₂| — the direct path is never longer than the sum of two legs. Reversing it into a lower bound like |z₁|−|z₂| gets the direction wrong.

6. Q6 of 6Hard

   Assertion (A): The least positive integer m for which ((1 + i)/(1 – i))ᵐ = 1 is 4. Reason (R): (1 + i)/(1 – i) = –i.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is true but R is false

   Explanation: A is true: (1 + i)/(1 – i) = (1 + i)²/(1 – i²) = 2i/2 = i, and iᵐ = 1 first happens at m = 4. R is false: the quotient equals i, not –i.

Complex Numbers and Quadratic Equations notes: key concepts (Class 11 Mathematics)

When x²+1=0 has no real root, the imaginary unit i=√(−1) extends the reals to the complex numbers z=a+ib. This chapter develops their algebra and conjugates, the modulus–argument (polar) form on the Argand plane, and returns to quadratics, showing that a negative discriminant delivers a pair of complex-conjugate roots.

Imaginary unit

i=√(−1) with i²=−1. Its powers cycle with period four — i, −1, −i, 1 — so iⁿ depends only on n taken modulo 4.

Complex number

z=a+ib with a,b real; a=Re(z) is the real part and b=Im(z) the imaginary part. z is purely imaginary when a=0.

Algebra of complex numbers

Add and subtract componentwise; multiply using i²=−1. Complex numbers form a field, obeying the usual commutative and distributive laws.

Conjugate

The conjugate of z=a+ib is z̄=a−ib. Then z·z̄ = a²+b² = |z|² is real, which is the key to dividing complex numbers.

Division by rationalising

To divide, multiply numerator and denominator by the conjugate of the denominator, turning the denominator into the real number |z|².

Modulus

|z|=√(a²+b²) is the distance of z from the origin on the Argand plane. It is always non-negative and equals 0 only when z=0.

Argument

arg(z)=θ is the angle the line to z makes with the positive real axis, with tanθ=b/a. The principal value lies in (−π, π].

Argand plane

The plane in which z=a+ib is plotted as the point (a,b). Addition of complex numbers behaves like vector addition of position vectors.

Polar form

z=r(cosθ+i sinθ) where r=|z| and θ=arg(z). Multiplication multiplies the moduli and adds the arguments, easing powers and roots.

Modulus properties

|z₁z₂|=|z₁||z₂| and |z₁/z₂|=|z₁|/|z₂|. Also |z|=|z̄| and |z|²=z·z̄, all handy for simplifying expressions.

Quadratic with negative discriminant

When b²−4ac<0, the square root of a negative introduces i, so the two roots are complex conjugates a±ib, symmetric about the real axis.

Nature of roots

The discriminant D=b²−4ac decides root type: D>0 real distinct, D=0 real equal, D<0 a conjugate pair of complex roots.