RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 11 questions · 26 marks
Some quadratics, such as x² + 9 = 0, simply have no real root, and rather than declaring them unsolvable mathematicians invented a new number i with i² = -1. This chapter shows how a + ib behaves under the usual arithmetic, how conjugates clear an imaginary part out of a denominator, and how modulus and argument give every complex number a position on a plane. By the end, every quadratic with real coefficients has exactly two roots.
The value of i²⁴⁷ is:
Answer
i²⁴⁷ = -i. The powers of i repeat every four steps, so divide the exponent by 4 and keep the remainder. 247 = 4 × 61 + 3, so i²⁴⁷ = (i⁴)⁶¹ × i³ = 1⁶¹ × i³. Since i³ = i² × i = -1 × i = -i, the value is -i.
The modulus of the complex number z = 5 - 12i is:
Answer
|z| = 13. By definition |a + ib| = √(a² + b²), with a = 5 and b = -12. So |z| = √(5² + (-12)²) = √(25 + 144) = √169 = 13. The answer 7 comes from subtracting the parts and 17 from adding them, neither of which is the modulus.
The conjugate of the complex number (3 + 2i)/(1 - i) is:
Answer
The conjugate is (1 - 5i)/2. First simplify the number by multiplying numerator and denominator by the conjugate 1 + i: (3 + 2i)(1 + i)/[(1 - i)(1 + i)] = (3 + 3i + 2i + 2i²)/(1 + 1) = (3 + 5i - 2)/2 = (1 + 5i)/2. The conjugate is obtained by changing the sign of the imaginary part, giving (1 - 5i)/2.
The roots of the equation x² + 4x + 13 = 0 are:
Answer
The roots are -2 ± 3i. Here a = 1, b = 4, c = 13, so D = b² - 4ac = 16 - 52 = -36, which is negative. Using the quadratic formula, x = (-4 ± √(-36))/2 = (-4 ± 6i)/2. Dividing both terms by 2 gives x = -2 ± 3i, a conjugate pair as expected for real coefficients. The option -4 ± 6i forgets to divide by 2a.
Assertion (A): The argument of the complex number z = -1 + i is 3π/4. Reason (R): If a complex number a + ib has a < 0 and b > 0, it lies in the second quadrant and its argument is π - tan⁻¹|b/a|.
Answer
Both A and R are true and R is the correct explanation of A. Reason R correctly describes how the principal argument is found for a point in the second quadrant, where the raw value of tan⁻¹(b/a) would point into the wrong quadrant. For z = -1 + i we have a = -1 < 0 and b = 1 > 0, so |b/a| = 1 and tan⁻¹1 = π/4. Hence arg z = π - π/4 = 3π/4, confirming Assertion A, and the rule in R is precisely the step used.
Express (2 + 3i)(4 - i) in the form a + ib.
Answer
Multiply out term by term: (2 + 3i)(4 - i) = 2(4) + 2(-i) + 3i(4) + 3i(-i) = 8 - 2i + 12i - 3i². Since i² = -1, the last term -3i² becomes +3. Collecting, the real part is 8 + 3 = 11 and the imaginary part is -2 + 12 = 10. Hence the product is 11 + 10i.
Find the multiplicative inverse of 4 - 3i and express it in the form a + ib.
Answer
For z = 4 - 3i, the inverse is 1/z = z̄/|z|². The conjugate is z̄ = 4 + 3i and |z|² = 4² + (-3)² = 16 + 9 = 25. So 1/z = (4 + 3i)/25 = 4/25 + (3/25)i. Check: (4 - 3i)(4 + 3i)/25 = (16 + 9)/25 = 25/25 = 1, confirming the inverse.
If z = 3 + 4i, verify that z z̄ = |z|².
Answer
The conjugate is z̄ = 3 - 4i. z z̄ = (3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i² = 9 + 16 = 25. The modulus is |z| = √(3² + 4²) = √25 = 5, so |z|² = 25. Both sides equal 25, so z z̄ = |z|² is verified.
Express z = -√3 + i in polar form. State its modulus and argument clearly, and explain how you fixed the quadrant.
Answer
Step 1 — Identify a = -√3 and b = 1. Step 2 — Modulus: r = |z| = √((-√3)² + 1²) = √(3 + 1) = √4 = 2. Step 3 — Reference angle: tan α = |b/a| = 1/√3, so α = π/6. Step 4 — Fix the quadrant. Since a < 0 and b > 0, the point (-√3, 1) lies in the second quadrant, where the principal argument is π - α. So arg z = π - π/6 = 5π/6. Step 5 — Write the polar form: z = 2(cos 5π/6 + i sin 5π/6). Step 6 — Verify: cos 5π/6 = -√3/2 and sin 5π/6 = 1/2, so 2(-√3/2 + i(1/2)) = -√3 + i, which is the original number.
(a) Solve the quadratic equation 3x² - 4x + 5 = 0 in the set of complex numbers. (b) If z = (1 + 2i)/(2 - i), express z in the form a + ib and find |z|.
Answer
(a) Here a = 3, b = -4, c = 5, so D = (-4)² - 4(3)(5) = 16 - 60 = -44. Since D < 0 the roots are complex: x = (4 ± √(-44))/6. Now √(-44) = √44 × i = 2√11 i. So x = (4 ± 2√11 i)/6, and dividing numerator and denominator by 2 gives x = (2 ± √11 i)/3. The roots are (2 + i√11)/3 and (2 - i√11)/3, a conjugate pair. (b) Multiply numerator and denominator by the conjugate of the denominator, 2 + i: z = (1 + 2i)(2 + i)/[(2 - i)(2 + i)]. Numerator = 2 + i + 4i + 2i² = 2 + 5i - 2 = 5i. Denominator = 2² + (-1)² = 4 + 1 = 5. So z = 5i/5 = i, i.e. a = 0 and b = 1. Hence |z| = √(0² + 1²) = 1.
Read the following and answer the questions that follow: In electrical engineering the opposition a component offers to alternating current is called impedance and is written as a complex number Z = R + iX, where R is resistance and X is reactance, both in ohms. A student wires two components in series, with impedances Z₁ = 4 + 3i ohms and Z₂ = 2 - 5i ohms. For components in series the impedances add. (a) Find the total impedance Z of the circuit. (b) Find the magnitude of the total impedance, in exact surd form and correct to two decimal places. (c) Compute the product Z₁Z₂ in the form a + ib. (d) Find Z₁ times its conjugate, and state what this value represents.
Answer
(a) Z = Z₁ + Z₂ = (4 + 3i) + (2 - 5i) = (4 + 2) + (3 - 5)i = 6 - 2i ohms. (b) |Z| = √(6² + (-2)²) = √(36 + 4) = √40 = 2√10 ohms. Numerically 2 × 3.1623 = 6.32 ohms correct to two decimal places. (c) Z₁Z₂ = (4 + 3i)(2 - 5i) = 8 - 20i + 6i - 15i². Since i² = -1, the term -15i² becomes +15. So Z₁Z₂ = (8 + 15) + (-20 + 6)i = 23 - 14i. (d) The conjugate of Z₁ is 4 - 3i, and Z₁ × conjugate = (4 + 3i)(4 - 3i) = 16 + 9 = 25. This is |Z₁|², the square of the magnitude of the first impedance — a real number, which is why multiplying by the conjugate is the standard trick for removing the imaginary part from a denominator.
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