- Read the real and imaginary parts and use the conditions “real”, “pure imaginary” and equality to find parameters
- Add, subtract, multiply and divide in algebraic form (division by the conjugate)
- Find the conjugate and the modulus and locate a number in the complex plane (quadrant, axis, distance)
- Simplify powers of i and sums of powers with the cycle of four
The equation x² + 1 = 0 has no real solution: no real number squared gives −1. Mathematicians simply added a number that does — i, with i² = −1 — and obtained a number system in which every quadratic equation has roots; engineers use it every day to describe alternating current and signals. In the CSCA syllabus complex numbers share the line “Vectors and complex numbers” with the lesson “Plane vectors”, and only the basics are tested: the algebraic form, the complex plane and the four operations. Know the handful of rules in this lesson, and a typical item is two lines of algebra.
The imaginary unit and the algebraic form
虚数单位)The number i with i² = −1. With it, square roots of negative numbers exist: √(−4) = 2i, and the equation x² = −9 has the two roots ±3i. The rules of algebra (expanding brackets, like terms, the formulas for squares) work as before — you only replace i² by −1.
复数)A number z = a + bi with real a and b (the algebraic form). a is the real part, Re z, and b is the imaginary part, Im z — a real number written without i: Im(3 − 4i) = −4. All complex numbers form the set C, and every real number is a complex number with b = 0: R ⊂ C.
- a = Re zthe real part
- b = Im zthe imaginary part — a real number, written without i
- a = c, b = done complex equation gives two real equations
Chinese textbooks: 实数 (b = 0), 虚数 (b ≠ 0), 纯虚数 (a = 0, b ≠ 0). Careful: many English books use “imaginary number” only for numbers of the form bi, while in the Chinese sense 3 + 2i is already 虚数. Complex numbers can be equal or unequal, but unless both are real they cannot be compared with < or >.
| Number | Re z | Im z | Type |
|---|---|---|---|
| 5 | 5 | 0 | real |
| 3 − 4i | 3 | −4 | imaginary (not pure imaginary) |
| −2i | 0 | −2 | pure imaginary |
| i² | −1 | 0 | real |
| 0 | 0 | 0 | real — not pure imaginary! |
1) z = (m² − 4) + (m + 2)i, m ∈ R. For which m is z real? Pure imaginary?
2) Find the real numbers x and y if (x + y) + (x − 2y)i = 5 − i.
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2) Equate the parts: x + y = 5 and x − 2y = −1. Subtract the second from the first: 3y = 6 ⇒ y = 2, x = 3.
The complex number z = (a² − a − 2) + (a + 1)i, a ∈ R, is pure imaginary. What is a?
A) −1 B) 2 C) −1 or 2 D) −2
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C is the classic trap — forgetting b ≠ 0. A is exactly the value that makes z = 0; D is a sign slip in the factorisation.
The complex plane, the conjugate and the modulus
Every complex number z = a + bi corresponds to the point Z(a, b) of the plane and to the vector OZ⃗ from the origin. The horizontal axis is the real axis, the vertical one the imaginary axis; every point of the imaginary axis except O is a pure imaginary number. Adding complex numbers means adding their vectors — the parallelogram rule from “Plane vectors” — so |z₁ − z₂| is the distance between the two points.
- z̄the conjugate (
共轭复数): the mirror image in the real axis - |z|the modulus (
模): the distance from O to Z; |z̄| = |z| - |z₁ − z₂|the distance between the points of z₁ and z₂
Quadrants: I — a > 0, b > 0; II — a < 0, b > 0; III — a < 0, b < 0; IV — a > 0, b < 0. A number on an axis lies in no quadrant.
1) In which quadrants are the points of z₁ = −3 + 2i, z₂ = 4 − i and z₃ = −5i?
2) z = −6 + 8i. Find z̄, |z| and |z̄|.
3) Find the distance between the points of z₁ = 1 + 2i and z₂ = 4 − 2i.
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2) z̄ = −6 − 8i, |z| = √(36 + 64) = 10, |z̄| = 10.
3) |z₁ − z₂| = |−3 + 4i| = √(9 + 16) = 5.
1) For which real m does the point of z = (m − 3) + (m + 1)i lie in quadrant II?
2) Which set of points satisfies |z − 2i| = 3?
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2) |z − 2i| is the distance from Z to the point (0, 2), so the points lie at distance 3 from it: the circle with centre (0, 2) and radius 3.
The four operations
Addition and subtraction work part by part, exactly as with vectors: z₁ + z₂ is the diagonal of the parallelogram built on OZ₁⃗ and OZ₂⃗. Multiplication is ordinary expansion of brackets plus one substitution, i² = −1. Division needs a trick, because i must not stay in a denominator: multiply the numerator and the denominator by the conjugate of the denominator, since (c + di)(c − di) = c² + d² is real.
- ±addition and subtraction: real parts separately, imaginary parts separately
- ac − bdmultiplication: expand like binomials, bi · di = bd · i² = −bd
- z · z̄ = a² + b²= |z|², always a non-negative real number
Squares worth remembering: (a + bi)² = (a² − b²) + 2abi, (1 + i)² = 2i, (1 − i)² = −2i.
Compute:
1) (3 − 2i) − (1 − 5i) + (−4 + i)
2) (2 + 3i)(1 − 4i)
3) (3 + 2i)² and (2 − i)(2 + i)
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2) 2 − 8i + 3i − 12i² = 2 + 12 − 5i = 14 − 5i.
3) (3 + 2i)² = 9 + 12i + 4i² = 5 + 12i; (2 − i)(2 + i) = 4 − i² = 5 — a real number, as z · z̄ always is.
- c − dithe conjugate of the denominator
- c² + d²a real denominator: |c + di|²
Multiplying by the conjugate makes the denominator real — like removing a root from a denominator. Moduli behave simply: |z₁z₂| = |z₁| · |z₂| and |z₁/z₂| = |z₁|/|z₂|.
- 1Write the conjugate
Write the conjugate of the denominator: c + di → c − di.
- 2Multiply
Multiply both the numerator and the denominator by it.
- 3Denominator
The denominator becomes c² + d², a positive real number.
- 4Numerator
Expand, replace i² by −1, and collect the real and the imaginary parts.
- 5Answer the question
Divide both parts by c² + d² and read off what is asked: a part, the conjugate, the modulus or the quadrant.
1) (5 + i)/(1 + i)
2) 10/(3 − i)
3) z(2 − i) = 5i. Find z and z̄.
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2) 10(3 + i)/(9 + 1) = 3 + i.
3) z = 5i/(2 − i) = 5i(2 + i)/5 = 2i + i² = −1 + 2i, z̄ = −1 − 2i. Check: (−1 + 2i)(2 − i) = −2 + i + 4i − 2i² = 5i ✓.
z = (3 + i)/(1 − 2i). What is |z|?
A) √2 B) 2 C) √10 D) √50
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B is |z|², C forgets the modulus of the denominator, and D is the modulus of the numerator 1 + 7i before dividing by 5.
z = 2i/(1 + i). In which quadrant of the complex plane is the point of z̄?
A) I B) II C) III D) IV
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A is the quadrant of z itself — but the question asks about z̄. B and C come from sign slips with i² = −1.
Powers of i and quadratic equations
Because i⁴ = 1, the powers of i come back to where they started every four steps, like the hand of a clock. So any power can be reduced to i⁰, i¹, i² or i³ by dividing the exponent by 4 — a favourite one-line CSCA item. Powers of 1 + i and 1 − i work the same way: square first, since (1 ± i)² = ±2i.
- rthe remainder of n divided by 4 (0, 1, 2, 3); i⁰ = 1
- kan integer
The powers go round the cycle i → −1 → −i → 1 with period 4, so four consecutive powers always add up to 0. Negative exponents too: i⁻¹ = −i, i⁻² = −1.
Compute:
1) i³⁷, i⁵⁰ and i⁻³
2) i + i² + i³ + … + i¹⁰
3) (1 + i)⁸ and ((1 − i)/(1 + i))²⁰²⁶
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2) The first 8 terms form two blocks with sum 0; i⁹ + i¹⁰ = i − 1 ⇒ −1 + i.
3) (1 + i)⁸ = ((1 + i)²)⁴ = (2i)⁴ = 16i⁴ = 16. (1 − i)/(1 + i) = −i, and (−i)²⁰²⁶ = i²⁰²⁶ (an even power), 2026 = 4 · 506 + 2 ⇒ −1.
- Dthe discriminant; for D < 0 there are no real roots but two conjugate complex roots
- √(−D)an ordinary square root, since −D > 0
Vieta's formulas still hold: x₁ + x₂ = −b/a, x₁x₂ = c/a. If a + bi is a root of an equation with real coefficients, so is a − bi.
1) Solve x² − 4x + 13 = 0.
2) 1 + i is a root of x² + px + q = 0 with real p and q. Find p and q.
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2) The other root is the conjugate 1 − i. Sum 2 = −p ⇒ p = −2; product (1 + i)(1 − i) = 2 = q ⇒ q = 2.
How CSCA asks about this
The CSCA mathematics paper has 48 questions in 60 minutes — about 75 seconds each, and a complex-number item should take less. The usual patterns:
- simplify a product or a quotient, then give the real or imaginary part, the conjugate, the modulus or the quadrant;
- a parameter: z is real, pure imaginary, or its point lies in a given quadrant;
- equations for z (z times an expression, z together with z̄) and the equality of two complex numbers;
- powers of i and sums of consecutive powers;
- the complex plane: the distance |z₁ − z₂| and simple sets such as |z − z₀| = r;
- the roots of a real quadratic with D < 0 come in a conjugate pair.
| Term | 中文 | Pinyin |
|---|---|---|
| complex number | 复数 | fùshù |
| imaginary unit | 虚数单位 | xūshù dānwèi |
| real part / imaginary part | 实部 / 虚部 | shíbù / xūbù |
| real number / imaginary number | 实数 / 虚数 | shíshù / xūshù |
| pure imaginary number | 纯虚数 | chún xūshù |
| algebraic form | 代数形式 | dàishù xíngshì |
| equality of complex numbers | 复数相等 | fùshù xiāngděng |
| conjugate complex number | 共轭复数 | gòng'è fùshù |
| modulus | 模 | mó |
| complex plane | 复平面 | fù píngmiàn |
| real axis / imaginary axis | 实轴 / 虚轴 | shízhóu / xūzhóu |
| quadrant | 象限 | xiàngxiàn |
| four arithmetic operations | 四则运算 | sìzé yùnsuàn |
Key points
- i² = −1; z = a + bi: Re z = a, Im z = b (a real number, without i).
- Real ⇔ b = 0; pure imaginary ⇔ a = 0 and b ≠ 0; a + bi = c + di ⇔ a = c and b = d.
- z ↔ Z(a, b) ↔ OZ⃗; z̄ = a − bi is the mirror image in the real axis; |z| = √(a² + b²), |z₁ − z₂| is a distance.
- Multiply like binomials and replace i² by −1; to divide, multiply the numerator and the denominator by the conjugate of the denominator; z · z̄ = |z|².
- iⁿ repeats with period 4 (i, −1, −i, 1); four consecutive powers add up to 0; (1 ± i)² = ±2i.
Check yourself
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Topic test: 20 questions · 25 min
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