Lesson 1: Complex Numbers

The equation x^2 + 1 = 0 has no real solution, because every real square is non-negative. Rather than declare the equation unsolvable, mathematicians of the 16th century invented a new number whose square is −1 and discovered that the resulting system — the complex numbers — is consistent, complete, and indispensable: every polynomial of degree n turns out to have exactly n complex roots, and engineers use complex numbers daily to describe alternating current, waves and control systems. This lesson builds the arithmetic of complex numbers and shows how they complete the story of quadratic equations.

The imaginary unit and the form a + bi

Define i by i^2 = −1, so i = √(−1). A complex number is an expression a + bi where a and b are real; a is the real part and b is the imaginary part (note: the imaginary part is the real number b, not the term bi). The complex conjugate of a + bi is a − bi, written z̄. The modulus |a + bi| = √(a^2 + b^2) is the distance of the point (a, b) from the origin in the complex plane, where the horizontal axis is real and the vertical axis is imaginary. Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal.

Square roots of negative numbers follow one rule: for b > 0, √(−b) = i√b. So √(−36) = 6i and √(−8) = 2i√2. Always rewrite √ of a negative in i-form before multiplying — otherwise you can "prove" falsehoods, as the Common Mistakes section shows.

Arithmetic: addition, subtraction, multiplication

Add and subtract by combining like parts. Multiplication works exactly like expanding brackets, with one extra step: replace every i^2 by −1.

Worked Example 1 — multiplying. Expand (3 + 2i)(1 − 4i).

(3 + 2i)(1 − 4i) = 3·1 + 3·(−4i) + 2i·1 + 2i·(−4i)
                 = 3 − 12i + 2i − 8i^2
                 = 3 − 10i − 8(−1)
                 = 11 − 10i

Graphical and numerical view. As points, 3 + 2i is (3, 2) and 1 − 4i is (1, −4), with moduli √13 ≈ 3.61 and √17 ≈ 4.12. The product 11 − 10i has modulus √(121 + 100) = √221 ≈ 14.87 — and indeed √13 · √17 = √221: the modulus of a product is the product of the moduli. Multiplication by a complex number rotates and scales the plane, which is why complex numbers are the natural language of anything that oscillates.

Conjugates and division

The product of conjugates is always a non-negative real number: (a + bi)(a − bi) = a^2 + b^2 = |z|^2. That single fact makes division possible: multiply numerator and denominator by the conjugate of the denominator to make the denominator real.

Worked Example 2 — dividing. Write (5 + 3i)/(2 − i) in the form a + bi.

(5 + 3i)/(2 − i) · (2 + i)/(2 + i) = (10 + 5i + 6i + 3i^2)/(4 + 1)
                                 = (7 + 11i)/5
                                 = 7/5 + (11/5)i

Check by multiplying back: (7/5 + (11/5)i)(2 − i) = 14/5 − (7/5)i + (22/5)i − (11/5)i^2 = 14/5 + 11/5 + (15/5)i = 25/5 + 3i = 5 + 3i. ✓

Powers of i and the modulus

The powers of i cycle with period 4: i^1 = i, i^2 = −1, i^3 = −i, i^4 = 1, then repeat. To evaluate any power, divide the exponent by 4 and use the remainder.

Worked Example 3 — powers and modulus. (a) i^23: 23 = 4·5 + 3, so i^23 = i^3 = −i. (b) |3 − 4i| = √(9 + 16) = √25 = 5 — the 3-4-5 triangle in disguise. (c) The distance between 1 + 2i and 4 + 6i in the plane is |(4 + 6i) − (1 + 2i)| = |3 + 4i| = 5.

Solving quadratics: the Fundamental Theorem in miniature

The quadratic formula x = (−b ± √(b^2 − 4ac))/(2a) does not fail when the discriminant is negative; it simply returns a conjugate pair of complex roots.

Worked Example 4 — a quadratic with no real roots. Solve x^2 − 4x + 13 = 0.

Discriminant: (−4)^2 − 4(1)(13) = 16 − 52 = −36
x = (4 ± √(−36))/2 = (4 ± 6i)/2 = 2 ± 3i

The roots 2 + 3i and 2 − 3i are conjugates — for quadratics with real coefficients, non-real roots always arrive in conjugate pairs. Check with the sum-and-product relations: the sum of the roots is 4 = −b/a ✓ and the product is (2 + 3i)(2 − 3i) = 4 + 9 = 13 = c/a ✓. Graphically, the parabola y = x^2 − 4x + 13 has vertex (2, 9) and never touches the x-axis — no real roots — yet the complex roots still encode the vertex: the real part 2 is the axis of symmetry, and |root|^2 = 13 is the y-intercept.

Modelling: impedance in an AC circuit

In alternating-current circuits, resistance and reactance (from coils and capacitors) combine as a complex number called impedance, Z = R + Xi, measured in ohms. A circuit has a 6 Ω resistor and an 8 Ω inductive reactance, so Z = 6 + 8i. The modulus |Z| = √(36 + 64) = 10 Ω is the total opposition to current. With a supply voltage of 100 V, Ohm's law I = V/Z gives

I = 100/(6 + 8i) · (6 − 8i)/(6 − 8i) = 100(6 − 8i)/100 = 6 − 8i

The current has magnitude |I| = √(36 + 64) = 10 A. The imaginary part is not decoration: its sign tells engineers whether the current lags or leads the voltage, which determines how motors, transformers and power grids behave. This is algebra with a pay cheque — every electrical engineering degree begins here.

Common mistakes

• Writing i^2 = 1. The defining property is i^2 = −1. Forgetting the sign poisons every expansion.
• Expanding (a + bi)^2 as a^2 + b^2. (2 + 3i)^2 = 4 + 12i + 9i^2 = −5 + 12i, not 13. The middle term exists, and the last term flips sign.
• Misusing square-root rules on negatives. √(−4)·√(−9) = (2i)(3i) = −6, but √((−4)(−9)) = √36 = +6. The rule √(ab) = √a√b is only guaranteed for non-negative numbers — convert to i-form first.
• Answering "no solution" to quadratics with negative discriminant. In the reals, yes; in the complex numbers, there are two conjugate solutions, and Algebra II expects them.
• Dividing without the conjugate. (5 + 3i)/(2 − i) is not finished until the denominator is real.

Exam Experience

1. (4 − 3i)(2 + i) is equal to

• (A) 5 − 2i
• (B) 11 − 2i
• (C) 11 − 10i
• (D) 5 − 10i

Answer: (B). Expand: 8 + 4i − 6i − 3i^2 = 8 − 2i + 3 = 11 − 2i. Choice (C) comes from writing −3i^2 = −3; choice (A) from then also dropping the real correction.

2. The solutions of x^2 + 4x + 20 = 0 are

• (A) −2 ± 4i
• (B) 2 ± 4i
• (C) −4 ± 2i
• (D) no solutions

Answer: (A). x = (−4 ± √(16 − 80))/2 = (−4 ± √(−64))/2 = (−4 ± 8i)/2 = −2 ± 4i. Choice (D) ignores the complex number system; (C) swaps the parts.

3. If z = 6 + 8i, then z · z̄ equals

• (A) 100
• (B) 36 + 64i
• (C) 10
• (D) 100i

Answer: (A). z · z̄ = |z|^2 = 6^2 + 8^2 = 100. Choice (C) is |z| itself, not its square.

Constructed response. The quadratic x^2 + kx + 25 = 0, where k > 0, has complex roots. (a) Find the value of k for which the roots are −3 + 4i and −3 − 4i. (b) Using your value of k, verify the roots satisfy the equation. (c) State the modulus of each root and explain its connection to the constant term 25.

Scoring guide. (a) 2 points: 1 point for using the sum of roots = −k (or equating the real part of the roots to −k/2, so −k/2 = −3); 1 point for k = 6. (b) 2 points: 1 point for substituting −3 + 4i (squaring correctly: (−3 + 4i)^2 = 9 − 24i + 16i^2 = −7 − 24i); 1 point for completing the check: (−7 − 24i) + 6(−3 + 4i) + 25 = −7 − 24i − 18 + 24i + 25 = 0 ✓. (c) 2 points: 1 point for |−3 + 4i| = √(9 + 16) = 5; 1 point for noting the product of the roots is |root|^2 = 25, which equals c/a — the constant term.

Going deeper: multiplication by i is a rotation

Plot z = 3 + 2i at the point (3, 2). Multiply by i: i(3 + 2i) = 3i + 2i^2 = −2 + 3i, the point (−2, 3). The new point is the old one rotated 90° anticlockwise about the origin — the lengths agree (|z| = √13 both times) and the coordinates swap with one sign change. Multiply by i again: −3 − 2i, the point opposite the original (180°). Four multiplications by i return to the start, mirroring i^4 = 1 exactly. This is the geometric soul of complex arithmetic: multiplication by a unit-modulus complex number rotates the plane, which is why signal engineers describe phase shifts in this language, and why the powers of i cycle instead of growing.

The Fundamental Theorem of Algebra — every degree-n polynomial has exactly n complex roots, counting multiplicity — becomes believable once quadratics surrender. Solve x^4 − 16 = 0 by factoring twice: (x^2 − 4)(x^2 + 4) = 0, giving x = ±2 from the first factor and x = ±2i from the second (since x^2 = −4). Four roots for a degree-4 equation: two real, two purely imaginary, all visible as points on a circle of radius 2 in the complex plane.

Worked Example 5 — recovering a complex number from its square. Find a + bi (a, b real) whose square is −7 + 24i. Expand: (a + bi)^2 = (a^2 − b^2) + 2abi. Match parts: a^2 − b^2 = −7 and 2ab = 24, so ab = 12. Try factor pairs of 12: a = 3, b = 4 gives 9 − 16 = −7 ✓ and ab = 12 ✓. So the square roots of −7 + 24i are ±(3 + 4i) — check the modulus route too: |−7 + 24i| = √(49 + 576) = 25, and |3 + 4i| = 5 = √25 ✓, since squaring doubles... precisely, |z^2| = |z|^2. Equating real and imaginary parts is a system of real equations (Lesson 9's territory) hiding inside one complex equation — a pattern that recurs throughout advanced mathematics: one complex statement is always two real statements.