Unit 1: Algebra, Relations and Functions — Complex Numbers, the Quadratic Equation, Sign of a Function and Quadratic Inequalities

Why number systems are extended

• We studied the natural, then integer, then rational, then real numbers; each system was enlarged so that equations unsolvable in the previous one could be solved.
• x² + 1 = 0 has no real solution, because the square of any real number is non-negative. That forced the introduction of the imaginary unit.

The imaginary unit

• i² = −1, i = √(−1), and √(−a) = i√a for a > 0.
• i¹ = i, i² = −1, i³ = −i, i⁴ = 1 — after i⁴ the values repeat, so i⁵ = i, i⁶ = −1, and so on.
• Rule for any positive integer n: divide n by 4 and use the remainder: iⁿ = i^remainder (i⁰ = 1 when the remainder is 0).

Worked example

• i²⁷ = i²⁴ · i³ = (i⁴)⁶ · (−i) = 1 · (−i) = −i.
• i⁴ + i⁴⁰ + i⁴⁰⁰ = 1 + 1 + 1 = 3.

Definition

• A complex number is z = a + bi where a, b ∈ ℝ and i² = −1.
• a is the real part, written Re(z); b is the imaginary part, written Im(z). Both parts are real numbers.
• If b = 0 the number is real; if a = 0 and b ≠ 0 it is purely imaginary.

Equality of two complex numbers

• a + bi = c + di ⟺ a = c and b = d: equate real parts to real parts and imaginary to imaginary, then solve the two equations.

Addition and subtraction

• (a + bi) + (c + di) = (a + c) + (b + d)i — add the real parts together and the imaginary parts together.

Multiplication

• Multiply as two algebraic expressions, then replace i² by −1.
• (a + bi)(c + di) = (ac − bd) + (ad + bc)i.
• Example: (2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i² = 2 − 5i + 12 = 14 − 5i.

The conjugate

• The conjugate of z = a + bi is z̄ = a − bi.
• z + z̄ = 2a — always real.
• z − z̄ = 2bi — purely imaginary.
• z · z̄ = a² + b² — real and non-negative.
• The conjugate of (z + w) is z̄ + w̄; the conjugate of (z · w) is z̄ · w̄.

Division

• Multiply both numerator and denominator by the conjugate of the denominator, so the denominator becomes a real number.
• Example: (3 + 2i)/(1 − i) = (3 + 2i)(1 + i) / (1 + 1) = (3 + 3i + 2i + 2i²)/2 = (1 + 5i)/2 = 1/2 + (5/2)i.

• For ax² + bx + c = 0 with a ≠ 0: first identify a, b and c with their signs, put the equation in standard form, then apply x = (−b ± √(b² − 4ac)) / (2a).
• Δ = b² − 4ac is called the discriminant.
• If the equation lacks a term its coefficient is zero (no x means b = 0, no constant means c = 0).

Worked example

• 2x² − 5x − 3 = 0: Δ = 25 + 24 = 49, √Δ = 7, so x = (5 ± 7)/4 → x = 3 or x = −1/2.

• Δ > 0: two different real roots.
• Δ = 0: two equal real roots, x = −b/(2a).
• Δ < 0: no real roots; the two roots are conjugate imaginary numbers.
• If a, b, c are rational: Δ > 0 and Δ a perfect square → two different rational roots; Δ > 0 but not a perfect square → two different irrational roots.
• Perfect squares to recognise at once: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 …

• Two equal roots → put Δ = 0 and solve for the unknown.
• Two different real roots → Δ > 0; no real roots → Δ < 0; two real roots (allowing equality) → Δ ≥ 0.
• To prove that the roots are real for every value of the parameter, compute Δ and rewrite it as a square plus a positive term, e.g. Δ = (k − 3)² + 4 ≥ 0 for all k — so the equation is never without real roots.
• If one root r is given, substitute it: ar² + br + c = 0 gives the unknown coefficient directly.

The rules

• α + β = −b/a and αβ = c/a.
• Proof sketch: from α = (−b + √Δ)/(2a) and β = (−b − √Δ)/(2a), adding cancels √Δ and multiplying gives (b² − Δ)/(4a²) = 4ac/(4a²).

Symmetric expressions

• A symmetric expression does not change when α and β are interchanged, so it is always evaluated through α + β and αβ only.
• α² + β² = (α + β)² − 2αβ.
• α³ + β³ = (α + β)³ − 3αβ(α + β).
• (α − β)² = (α + β)² − 4αβ, so |α − β| = √Δ / |a|.
• 1/α + 1/β = (α + β)/αβ ; α/β + β/α = (α² + β²)/αβ.

Special relations between the roots

• One root twice the other: roots α and 2α → α + β = 3α, αβ = 2α²; eliminate α.
• Roots of opposite signs → αβ < 0; roots that are reciprocals → αβ = 1; roots that are negatives of each other → α + β = 0.

• The equation whose roots are α, β is x² − (α + β)x + αβ = 0.
• Compute the sum, then the product, then write the rule; clear fractions by multiplying through by the L.C.M. if needed.

Roots linked to the roots of another equation

• If the new roots are 2α and 2β, then the new sum is 2(α + β) and the new product is 4αβ.
• If the new roots are α + 1 and β + 1: sum = (α + β) + 2, product = αβ + (α + β) + 1.
• Shortcut for a shift: in the original equation substitute x = y − 1 (for roots increased by 1) and expand.

The concept

• f is positive on a set of values of x when f(x) > 0, and negative when f(x) < 0; the zeros are the values that make f(x) = 0.

Sign of a linear function ax + b

• If a > 0 the function is negative to the left of its zero −b/a and positive to the right of it.
• If a < 0 the signs are reversed: positive on the left, negative on the right.

Sign of a quadratic ax² + bx + c with two different real zeros α < β

• If a > 0: positive outside the two zeros (x < α or x > β) and negative between them.
• If a < 0: negative outside and positive between.

When Δ ≤ 0

• Δ = 0: ax² + bx + c = a(x − r)² with r = −b/(2a), so the sign is the sign of a for all x except the zero at r.
• Δ < 0: the expression never changes sign; it keeps the sign of a for every real x.

Sign from the graph

• The x-intercepts of the curve are the zeros of the function; above the x-axis the function is positive, below it is negative.

The method

• Step 1: move all terms to one side and leave zero on the other.
• Step 2: solve the corresponding equation to get the zeros.
• Step 3: use the sign of the function to select the intervals, writing the solution set in interval or set notation.

Points of care

• Multiplying or dividing an inequality by a negative number reverses the sign.
• Δ < 0 with a > 0: the quadratic is always positive, so ">" and "≥" give ℝ while "<" and "≤" give ∅.
• Δ = 0 with a > 0: x²-type expression ≥ 0 on ℝ, but > 0 on ℝ − {r}.
• A non-strict inequality includes the zeros; a strict one excludes them.

• The domain of √f(x) is the solution set of f(x) ≥ 0.
• The denominator √f(x) requires f(x) > 0 (strict, because the denominator may not vanish).
• Verbal problems (area, consecutive numbers, projectile height): choose the unknown, express the quantities in terms of it, form the inequality or equation, solve, then reject any value that the situation forbids such as a negative length.