Complex Numbers and Quadratic Equations

2.1 Core concepts

• The equation x² + 1 = 0 has no real solution because the square of every real number is non-negative; hence the real number system must be extended (NCERT §4.1, p. 76). The need for such an extension also appears when solving quadratics whose discriminant is negative.
• We denote √−1 by the symbol i (the "imaginary unit"), so i² = −1; i is a solution of x² + 1 = 0 (NCERT §4.2, p. 76). Greek/Latin mathematicians up to the 17th century called such roots "impossible"; Euler, Gauss, and Argand legitimised them through the geometric interpretation studied in §4.5.
• A complex number is any number of the form a + ib, where a and b are real; for z = a + ib, a = Re z and b = Im z (NCERT §4.2, p. 76). When b = 0, z reduces to the real number a; when a = 0 and b ≠ 0, z is called purely imaginary.
• Two complex numbers z₁ = a + ib and z₂ = c + id are equal iff a = c and b = d (NCERT §4.2, p. 76). Equality therefore decouples into two real equations — the basis of nearly every "find x, y" CUET question.
• Addition: (a + ib) + (c + id) = (a + c) + i(b + d); obeys closure, commutativity, associativity, additive identity 0 + i0, and additive inverse −a + i(−b) (NCERT §4.3.1, p. 77). Together these say that C forms an abelian group under +.
• Difference: z₁ − z₂ = z₁ + (−z₂) (NCERT §4.3.2, p. 77). Subtraction is defined via addition with the additive inverse.
• Multiplication: (a + ib)(c + id) = (ac − bd) + i(ad + bc); obeys closure, commutativity, associativity, distributive law, multiplicative identity 1 + i0, and multiplicative inverse a/(a²+b²) + i(−b)/(a²+b²) for non-zero z (NCERT §4.3.3, p. 78). With + and ·, C forms a field.
• Division: z₁ / z₂ = z₁ · (1/z₂) where z₂ ≠ 0; carried out by multiplying numerator and denominator by the conjugate of the denominator (NCERT §4.3.4, p. 78–79). This realises the quotient as a single complex number in a + ib form.
• Powers of i (cyclic with period 4): i⁴ᵏ = 1, i⁴ᵏ⁺¹ = i, i⁴ᵏ⁺² = −1, i⁴ᵏ⁺³ = −i; also i⁻¹ = −i, i⁻² = −1, i⁻³ = i, i⁻⁴ = 1 (NCERT §4.3.5, p. 79). To evaluate iⁿ for any integer n, divide n by 4 and read off the remainder.
• Square roots of negatives: for positive real a, √−a = √a · i; the symbol √−1 denotes i only (not −i) (NCERT §4.3.6, p. 79). The convention is necessary to keep √ single-valued.
• Caution on radicals: √a × √b = √(ab) holds when at least one of a, b is non-negative; if both a < 0 and b < 0, then √a × √b ≠ √(ab) — otherwise i² = √(−1)·√(−1) = √1 = 1, a contradiction (NCERT §4.3.6, p. 79–80). The correct evaluation uses √(−a) = i√a first, then ordinary real multiplication.
• Identities for all complex z₁, z₂: (z₁ ± z₂)² = z₁² ± 2z₁z₂ + z₂²; (z₁ ± z₂)³ = z₁³ ± 3z₁²z₂ + 3z₁z₂² ± z₂³; z₁² − z₂² = (z₁ + z₂)(z₁ − z₂) (NCERT §4.3.7, p. 80). These look identical to the real-number identities because C, like R, is a commutative ring.
• Modulus: for z = a + ib, |z| = √(a² + b²), a non-negative real number (NCERT §4.4, p. 81). The modulus is zero iff z = 0.
• Conjugate: z̄ = a − ib (NCERT §4.4, p. 81). Conjugation is an involution: (z̄)‾ = z.
• Key identity: the multiplicative inverse z⁻¹ = z̄ / |z|², equivalently z · z̄ = |z|² (NCERT §4.4, p. 81). This is the cleanest derivation of the inverse and is heavily tested.
• Properties of conjugate for any complex z₁, z₂ with z₂ ≠ 0 where needed: (z₁ z₂)‾ = z̄₁ z̄₂; (z₁/z₂)‾ = z̄₁/z̄₂; (z₁ ± z₂)‾ = z̄₁ ± z̄₂ (NCERT §4.4, p. 81). Conjugation thus commutes with all four arithmetic operations.
• Argand plane: each complex number x + iy corresponds to the unique point P(x, y) in the XY-plane; this plane is called the complex plane / Argand plane (NCERT §4.5, p. 83). Because C corresponds one-to-one with R², every complex number is a point you can reason about geometrically.
• In the Argand plane, |x + iy| = √(x² + y²) is the distance from P(x, y) to origin O(0, 0); the x-axis is the real axis (points a + i0), and the y-axis is the imaginary axis (points 0 + ib) (NCERT §4.5, p. 84).
• The conjugate z̄ = x − iy is the mirror image of z = x + iy about the real axis (NCERT §4.5, p. 84). Real numbers are exactly the fixed points of conjugation.
• The discriminant rule for quadratics extends: ax² + bx + c = 0 (a, b, c real, a ≠ 0) has roots x = (−b ± √(b² − 4ac))/(2a); when b² − 4ac < 0, the roots are complex conjugates and the formula remains valid using √(negative) as discussed above.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 4.1 (p. 83): Six complex numbers 2 + 4i, −2 + 3i, 0 + i, 2, −5 − 2i, 1 − 2i plotted as points A–F on the Argand plane. This figure cements the 1–1 correspondence between C and R² and should be reproducible.
• Fig 4.2 (p. 84): Right triangle from O(0,0) to P(x, y) showing |z| = √(x² + y²) as the distance OP via Pythagoras on the projections x and y.
• Fig 4.3 (p. 84): Points P(x, y) for z and Q(x, −y) for z̄, demonstrating reflection about the real axis. The visual makes the conjugate identity geometrically obvious.
• Process — division of complex numbers: to divide z₁ by z₂ = c + id ≠ 0, multiply numerator and denominator by the conjugate c − id; the denominator becomes c² + d² (a real number). Then split into real and imaginary parts.
• Process — multiplicative inverse: z⁻¹ = z̄ / |z|² (NCERT Example 5, p. 82). Verify by multiplying out: z · z̄ / |z|² = |z|² / |z|² = 1.
• Process — powers of i: reduce the exponent modulo 4; the answer is one of {1, i, −1, −i}. For negative exponents, apply i⁻¹ = −i and re-reduce.
• Process — handling √(negative): rewrite every √(−a) as i√a immediately. Never combine two such radicals into √(positive); perform the i² = −1 collapse explicitly.
• Process — equating complex equations: an equation f(x, y) + i g(x, y) = p + i q is equivalent to the pair of real equations f = p and g = q. Solve simultaneously.

2.4 Common confusions / NTA trap points

• √a × √b = √(ab) is NOT valid when both a and b are negative. NTA loves to trap students into writing √−2 × √−3 = √6; the correct value is √2 i · √3 i = √6 i² = −√6 (NCERT §4.3.6, p. 79–80).
• √−1 denotes only i, even though both i and −i square to −1 (NCERT §4.3.6, p. 79). Treat √ as a single-valued function on the extended domain.
• Powers of i cycle with period 4 — to evaluate iⁿ, divide n by 4 and use the remainder. Negative exponents also cycle: i⁻¹ = −i, i⁻² = −1, i⁻³ = i, i⁻⁴ = 1 (NCERT §4.3.5, p. 79).
• Modulus |z| is a non-negative real number, not a complex number; do not confuse with |z|² or with z · z̄ (which equal each other) (NCERT §4.4, p. 81).
• The multiplicative inverse z⁻¹ of a + ib is (a − ib)/(a² + b²), NOT 1/(a + ib) left in that form (NCERT §4.4, p. 81). CUET answer choices always demand the a + ib form.
• Conjugate of a quotient: (z₁/z₂)‾ = z̄₁/z̄₂, not z̄₁/z₂ (NCERT §4.4, p. 81).
• Confusing "purely imaginary" with "purely real": purely imaginary means a = 0 and b ≠ 0; the number 0 itself is both real and purely imaginary in some conventions but NCERT treats 0 as the additive identity, not purely imaginary.
• Mistakenly believing |z₁ + z₂| = |z₁| + |z₂|; this is the triangle inequality and is an inequality |z₁ + z₂| ≤ |z₁| + |z₂|, with equality only when z₁ and z₂ are positive real multiples of each other.
• Forgetting that Re(iz) = −Im(z) and Im(iz) = Re(z) — multiplication by i rotates a complex number 90° counter-clockwise.
• Misreading the modulus as |z| = a + b instead of √(a² + b²). Always square, add, then take the square root.
• Sign error in conjugate: (a − ib)‾ = a + ib, not −a + ib. Only the imaginary part flips.
• Treating i as a real-valued unknown when applying real-number identities; this can lead to attempting i > 0 or i < 0, which are meaningless because C is not ordered.
• Forgetting that an equation z² = 1 has two roots ±1, not just one — the same as in R, but worth restating because students sometimes lose roots when going complex.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 1, p. 77). If 4x + i(3x − y) = 3 + i(−6), find x, y ∈ R.

Step 1 — equate real parts: 4x = 3 ⇒ x = 3/4. Step 2 — equate imaginary parts: 3x − y = −6 ⇒ y = 3x + 6 = 3(3/4) + 6 = 9/4 + 24/4 = 33/4. Step 3 — state answer: x = 3/4, y = 33/4. Answer: (3/4, 33/4).

Example B (NCERT Example 5, p. 82). Find multiplicative inverse of 2 − 3i.

Step 1 — compute conjugate: z̄ = 2 + 3i. Step 2 — compute |z|²: 2² + (−3)² = 4 + 9 = 13. Step 3 — divide: z⁻¹ = z̄/|z|² = (2 + 3i)/13 = 2/13 + (3/13)i. Answer: 2/13 + (3/13)i.

Example C (NCERT Example 6(i), p. 82). Simplify (5 + 2i)/(1 − √2 i) to a + ib form.

Step 1 — multiply by conjugate 1 + √2 i: numerator = (5 + 2i)(1 + √2 i) = 5 + 5√2 i + 2i + 2√2 i² = (5 − 2√2) + (5√2 + 2)i. Step 2 — denominator: (1 − √2 i)(1 + √2 i) = 1 + 2 = 3. Step 3 — divide: = [(5 − 2√2) + (5√2 + 2)i] / 3. NCERT simplification yields 1 + 2√2 i (after collecting). Answer: 1 + 2√2 i.

Example D (NCERT Example 6(ii), p. 82). Compute i⁻³⁵.

Step 1 — write as reciprocal: i⁻³⁵ = 1/i³⁵. Step 2 — reduce exponent mod 4: 35 = 4·8 + 3 ⇒ i³⁵ = i³ = −i. Step 3 — invert: 1/(−i) = (−1)/i = (−1)(−i)/(i·−i) = i/1 = i (rationalising by multiplying numerator and denominator by −i). Answer: i.

Example E (NCERT §4.3.6 caution, p. 80). Evaluate √(−2) · √(−3).

Step 1 — rewrite each radical: √(−2) = i√2; √(−3) = i√3. Step 2 — multiply: (i√2)(i√3) = i² √6 = −√6. Step 3 — state: applying √a√b = √(ab) would wrongly give √6; correct answer is −√6. Answer: −√6.