1. Chapter at a Glance
The equation x² + 1 = 0 has no real solution, because the square of a real number is never negative. Mathematicians did not give up: they introduced a new number i with i² = −1. Numbers of the form a + ib — complex numbers — turned out to be enormously useful in electrical engineering, signal processing, quantum physics and of course in algebra, where every quadratic equation now has two roots. This chapter covers the algebra of complex numbers, their modulus and conjugate, the Argand plane and polar form, and solving quadratic equations with negative discriminant.
- The imaginary unit i and powers of i.
- Complex numbers z = a + ib: real part, imaginary part, equality.
- Addition, subtraction, multiplication, division; multiplicative inverse.
- Conjugate and modulus; their properties.
- Argand plane; polar form z = r(cos θ + i sin θ), argument.
- Quadratic equations with real coefficients and D < 0.
2. The Imaginary Unit
i = √−1, so i² = −1. For a > 0, √−a = i√a.
| Power | i¹ | i² | i³ | i⁴ | i⁵ | i⁶ | … |
|---|---|---|---|---|---|---|---|
| Value | i | −1 | −i | 1 | i | −1 | repeats every 4 |
So iⁿ depends on the remainder when n is divided by 4. Example: i³⁷ = i^(36+1) = i; i⁻³⁵ = 1/i³⁵ = 1/i³ = 1/(−i) = i.
√a × √b = √(ab) is not valid when both a and b are negative: √−1 × √−1 = i² = −1, not √1 = 1.
3. Algebra of Complex Numbers
z = a + ib with a, b ∈ ℝ: Re(z) = a, Im(z) = b. Two complex numbers are equal iff their real parts and imaginary parts are equal.
| Operation | Rule |
|---|---|
| Addition | (a + ib) + (c + id) = (a + c) + i(b + d) |
| Subtraction | (a + ib) − (c + id) = (a − c) + i(b − d) |
| Multiplication | (a + ib)(c + id) = (ac − bd) + i(ad + bc) |
| Inverse | 1/(a + ib) = (a − ib)/(a² + b²) |
| Division | z₁/z₂ = z₁ × (1/z₂) |
Complex addition and multiplication are commutative and associative, and multiplication distributes over addition. The identities (a + b)², (a − b)², (a + b)³ etc. all hold for complex numbers.
Express (5 − 3i)³ in the form a + ib.
Solution: 125 − 3(25)(3i) + 3(5)(9i²) − 27i³ = 125 − 225i − 135 + 27i = −10 − 198i.
Express (3 + 2i)/(1 − i) as a + ib.
Solution: Multiply by (1 + i)/(1 + i): (3 + 3i + 2i + 2i²)/2 = (1 + 5i)/2 = ½ + (5/2)i.
4. Conjugate and Modulus
- Conjugate: z̄ = a − ib.
- Modulus: |z| = √(a² + b²).
- z z̄ = |z|² = a² + b².
- z⁻¹ = z̄ / |z|².
| Property | Statement |
|---|---|
| Product | |z₁z₂| = |z₁||z₂| |
| Quotient | |z₁/z₂| = |z₁|/|z₂| |
| Conjugate of sum/product | (z₁ + z₂)‾ = z̄₁ + z̄₂; (z₁z₂)‾ = z̄₁z̄₂ |
| Real/imaginary parts | Re z = (z + z̄)/2; Im z = (z − z̄)/(2i) |
| Triangle inequality | |z₁ + z₂| ≤ |z₁| + |z₂| |
5. Argand Plane and Polar Form
Represent z = x + iy by the point P(x, y) in the plane: the x-axis is the real axis and the y-axis the imaginary axis. This is the Argand plane (after Jean-Robert Argand). |z| is the distance OP; z̄ is the reflection of P in the real axis.
x = r cos θ, y = r sin θ, so z = r(cos θ + i sin θ), with r = |z| and θ = arg z.
The principal argument θ lies in (−π, π].
Write z = −1 + i√3 in polar form.
Solution: r = √(1 + 3) = 2. cos θ = −½, sin θ = √3/2 → θ = 2π/3 (quadrant II). z = 2(cos 2π/3 + i sin 2π/3).
| Quadrant of z | Principal argument (α = tan⁻¹|y/x|) |
|---|---|
| I (x > 0, y > 0) | α |
| II (x < 0, y > 0) | π − α |
| III (x < 0, y < 0) | −(π − α) |
| IV (x > 0, y < 0) | −α |
6. Quadratic Equations
For ax² + bx + c = 0 with real coefficients and D = b² − 4ac < 0, the roots are
Such roots occur in conjugate pairs. (A polynomial equation of degree n has n roots, counted with multiplicity, in the complex numbers.)
Solve x² + x + 1 = 0.
Solution: D = 1 − 4 = −3. x = (−1 ± i√3)/2.
Solve √5x² + x + √5 = 0.
Solution: D = 1 − 20 = −19. x = (−1 ± i√19)/(2√5).
7. Solved Problems (Board Pattern)
Find the modulus and multiplicative inverse of 4 − 3i.
Solution: |z| = 5. z⁻¹ = (4 + 3i)/25 = 4/25 + (3/25)i.
If (x + iy)(2 − 3i) = 4 + i, find x and y.
Solution: x + iy = (4 + i)/(2 − 3i) = (4 + i)(2 + 3i)/13 = (8 + 12i + 2i − 3)/13 = (5 + 14i)/13. x = 5/13, y = 14/13.
Find the real θ such that (3 + 2i sin θ)/(1 − 2i sin θ) is purely real.
Solution: Multiply by (1 + 2i sin θ): numerator = 3 + 6i sin θ + 2i sin θ − 4sin²θ; imaginary part 8 sin θ = 0 → sin θ = 0 → θ = nπ.
8. Master Formula Table
| Result | Formula |
|---|---|
| Powers of i | i⁴ᵏ = 1, i⁴ᵏ⁺¹ = i, i⁴ᵏ⁺² = −1, i⁴ᵏ⁺³ = −i |
| Product | (a + ib)(c + id) = (ac − bd) + i(ad + bc) |
| Inverse | (a − ib)/(a² + b²) |
| Modulus | |z| = √(a² + b²) |
| Conjugate product | z z̄ = |z|² |
| Polar form | z = r(cos θ + i sin θ) |
| Roots (D < 0) | [−b ± i√(4ac − b²)]/2a |
| Sum of four consecutive powers | iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0 |
9. Master Summary Table
| Concept | Key idea | Exam focus |
|---|---|---|
| i | i² = −1, cycle of 4 | Simplify iⁿ |
| Algebra | Treat i like a variable, then i² = −1 | Write in a + ib form |
| Division | Multiply by conjugate | Inverse, quotient |
| Modulus, conjugate | Distance; reflection | Properties, proofs |
| Polar form | r and θ from x, y | Quadrant-correct argument |
| Quadratics | D < 0 → conjugate complex roots | Solve equations |
10. Mind Map
- i² = −1
- Period 4
- +, −, ×, ÷
- Inverse
- Argand plane
- |z|, arg z, polar
- D < 0
- Conjugate roots
11. Tips and Tricks
- To simplify a fraction, multiply top and bottom by the conjugate of the denominator.
- For the argument, find the reference angle from |y/x| first, then adjust for the quadrant.
- (1 + i)² = 2i and (1 − i)² = −2i — handy for higher powers.
- √−4 × √−9 = √36 = 6 (wrong); correct value is 2i × 3i = −6.
- Taking Im(a + ib) as ib instead of b.
- Giving arg(−1 − i) as π/4 instead of −3π/4.
12. Important One-liners
- i² = −1 and powers of i repeat every 4.
- Every real number is a complex number with imaginary part 0.
- z z̄ is always a non-negative real number.
- |z₁z₂| = |z₁||z₂|.
- Non-real roots of a real quadratic come in conjugate pairs.
13. Exam-pattern Questions with Detailed Answers
The value of i⁹ + i¹⁹ is:
- 0
- 2i
- −2i
- 1
Show answer
i⁹ = i, i¹⁹ = i³ = −i. Sum 0.
Answer: (A)
Express (1 − i) − (−1 + 6i) in the form a + ib.
Show answer
2 − 7i.
Answer: 2 − 7i
Find the conjugate and modulus of −3 + 4i.
Show answer
Conjugate −3 − 4i; modulus 5.
Answer: −3 − 4i; 5
Find the multiplicative inverse of √5 + 3i.
Show answer
(√5 − 3i)/(5 + 9) = √5/14 − (3/14)i.
Answer: √5/14 − 3i/14
Convert 1 + i to polar form.
Show answer
r = √2, θ = π/4: √2(cos π/4 + i sin π/4).
Answer: √2(cos π/4 + i sin π/4)
Solve 3x² − 4x + 20/3 = 0.
Show answer
Multiply by 3: 9x² − 12x + 20 = 0. D = 144 − 720 = −576. x = (12 ± 24i)/18 = 2/3 ± (4/3)i.
Answer: 2/3 ± (4/3)i
Evaluate (1 + i)⁴ and (1 − i)⁴.
Show answer
(1 + i)² = 2i → (2i)² = −4. (1 − i)² = −2i → (−2i)² = −4.
Answer: −4 and −4
Find the modulus of (1 + i)/(1 − i) − (1 − i)/(1 + i).
Show answer
(1 + i)/(1 − i) = (1 + i)²/2 = i; the second is −i. Difference = 2i; modulus 2.
Answer: 2
If (a + ib)/(c + id) = x + iy, prove that (a² + b²)/(c² + d²) = x² + y².
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Take modulus: |a + ib|/|c + id| = |x + iy| → √(a² + b²)/√(c² + d²) = √(x² + y²). Square both sides. ∎
Answer: Proved
If z₁ = 2 − i and z₂ = 1 + i, find |(z₁ + z₂ + 1)/(z₁ − z₂ + 1)|.
Show answer
Numerator 4; denominator 2 − 2i. Modulus = 4/|2 − 2i| = 4/(2√2) = √2.
Answer: √2
Find the real numbers x and y if (x − iy)(3 + 5i) is the conjugate of −6 − 24i.
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(3x + 5y) + i(5x − 3y) = −6 + 24i. Solve 3x + 5y = −6, 5x − 3y = 24: x = 3, y = −3.
Answer: x = 3, y = −3
Convert (−16)/(1 + i√3) to polar form.
Show answer
Multiply by (1 − i√3)/4: −4(1 − i√3) = −4 + 4√3 i. r = 8; quadrant II, reference angle π/3 → θ = 2π/3. 8(cos 2π/3 + i sin 2π/3).
Answer: 8(cos 2π/3 + i sin 2π/3)
14. Quick Revision Summary
- i² = −1; iⁿ repeats with period 4.
- z = a + ib; operate like algebra, divide using the conjugate.
- |z| = √(a² + b²), z z̄ = |z|², z⁻¹ = z̄/|z|².
- Polar form r(cos θ + i sin θ); principal argument in (−π, π].
- D < 0: roots [−b ± i√(4ac − b²)]/2a, a conjugate pair.