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Intermediate25 min19 / 68

Complex numbers

Complex numbers for the CSCA: the imaginary unit i, the algebraic form a + bi, real and imaginary parts, pure imaginary numbers and equality, the complex plane and quadrants, the conjugate and the modulus, the four operations and powers of i — fast and without a calculator.

Check yourself
In this lesson you will learn
  • 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

Definition
Imaginary unit (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.

Definition
Complex number (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.

z = a + bi (a, b ∈ R): real ⇔ b = 0; imaginary ⇔ b ≠ 0; pure imaginary ⇔ a = 0 and b ≠ 0 a + bi = c + di ⇔ a = c and b = d
where:
  • 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 >.

NumberRe zIm zType
550real
3 − 4i3−4imaginary (not pure imaginary)
−2i0−2pure imaginary
i²−10real
000real — not pure imaginary!
A pure imaginary number needs b ≠ 0: zero is not pure imaginary.
Example 1: real, pure imaginary, equal

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.

Show solution
1) Real: m + 2 = 0 ⇒ m = −2. Pure imaginary: m² − 4 = 0 and m + 2 ≠ 0. m² = 4 gives m = ±2, but m = −2 makes z = 0 ⇒ only m = 2 (z = 4i).
2) Equate the parts: x + y = 5 and x − 2y = −1. Subtract the second from the first: 3y = 6 ⇒ y = 2, x = 3.
Example 2 (CSCA format): a pure imaginary number

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

Show solution
a² − a − 2 = (a − 2)(a + 1) = 0 ⇒ a = 2 or a = −1. The condition a + 1 ≠ 0 rules out a = −1 (it gives z = 0) ⇒ answer B.
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.

z = a + bi ↔ Z(a, b) ↔ OZ⃗ z̄ = a − bi |z| = |OZ⃗| = √(a² + b²) |z₁ − z₂| = |Z₁Z₂|
where:
  • 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.

z = 2 + 3iz̄ = 2 − 3i|z|O23−3real axisimaginary axisIIIIIIIV
z and z̄ are symmetric about the real axis and have the same modulus: |z| = |z̄| = √13.
Example 3: points, conjugates and moduli

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.

Show solution
1) z₁ ↔ (−3, 2): quadrant II; z₂ ↔ (4, −1): quadrant IV; z₃ ↔ (0, −5): on the imaginary axis, in no quadrant.
2) z̄ = −6 − 8i, |z| = √(36 + 64) = 10, |z̄| = 10.
3) |z₁ − z₂| = |−3 + 4i| = √(9 + 16) = 5.
Example 4: a parameter and a circle

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?

Show solution
1) Quadrant II: real part < 0, imaginary part > 0: m − 3 < 0 and m + 1 > 0 ⇒ −1 < m < 3.
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.

(a + bi) ± (c + di) = (a ± c) + (b ± d)i (a + bi)(c + di) = (ac − bd) + (ad + bc)i (a + bi)(a − bi) = a² + b²
where:
  • ±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.

Example 5: adding and multiplying

Compute:
1) (3 − 2i) − (1 − 5i) + (−4 + i)
2) (2 + 3i)(1 − 4i)
3) (3 + 2i)² and (2 − i)(2 + i)

Show solution
1) Real parts: 3 − 1 − 4 = −2; imaginary parts: −2 + 5 + 1 = 4 ⇒ −2 + 4i.
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.
(a + bi)/(c + di) = (a + bi)(c − di) / ((c + di)(c − di)) = ((ac + bd) + (bc − ad)i) / (c² + d²)(a + bi)/(c + di) = (a + bi)(c − di) / ((c + di)(c − di)) = ((ac + bd) + (bc − ad)i) / (c² + d²)
where:
  • 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|.

  1. 1
    Write the conjugate

    Write the conjugate of the denominator: c + di → c − di.

  2. 2
    Multiply

    Multiply both the numerator and the denominator by it.

  3. 3
    Denominator

    The denominator becomes c² + d², a positive real number.

  4. 4
    Numerator

    Expand, replace i² by −1, and collect the real and the imaginary parts.

  5. 5
    Answer the question

    Divide both parts by c² + d² and read off what is asked: a part, the conjugate, the modulus or the quadrant.

Example 6: dividing

1) (5 + i)/(1 + i)
2) 10/(3 − i)
3) z(2 − i) = 5i. Find z and z̄.

Show solution
1) (5 + i)(1 − i)/((1 + i)(1 − i)) = (5 − 5i + i − i²)/2 = (6 − 4i)/2 = 3 − 2i.
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 ✓.
Example 7 (CSCA format): a modulus without dividing

z = (3 + i)/(1 − 2i). What is |z|?
A) √2 B) 2 C) √10 D) √50

Show solution
Fast way: |z| = |3 + i|/|1 − 2i| = √10/√5 = √2 ⇒ answer A. By dividing: z = (3 + i)(1 + 2i)/5 = (1 + 7i)/5 and |z| = √50/5 = √2 — the same.
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.
Example 8 (CSCA format): the quadrant of the conjugate

z = 2i/(1 + i). In which quadrant of the complex plane is the point of z̄?
A) I B) II C) III D) IV

Show solution
z = 2i(1 − i)/2 = i − i² = 1 + i, so z̄ = 1 − i ↔ (1, −1): quadrant IV, answer D.
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.

i¹ = i, i² = −1, i³ = −i, i⁴ = 1 i⁴ᵏ⁺ʳ = iʳ iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0
where:
  • 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.

Example 9: powers of i

Compute:
1) i³⁷, i⁵⁰ and i⁻³
2) i + i² + i³ + … + i¹⁰
3) (1 + i)⁸ and ((1 − i)/(1 + i))²⁰²⁶

Show solution
1) 37 = 4 · 9 + 1 ⇒ i³⁷ = i; 50 = 4 · 12 + 2 ⇒ i⁵⁰ = −1; i⁻³ = i⁻³⁺⁴ = i (check: i · i³ = i⁴ = 1 ✓).
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.
ax² + bx + c = 0 (a, b, c ∈ R), D = b² − 4ac < 0: x₁,₂ = (−b ± i√(−D)) / (2a)ax² + bx + c = 0 (a, b, c ∈ R), D = b² − 4ac < 0: x₁,₂ = (−b ± i√(−D)) / (2a)
where:
  • 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.

Example 10: equations with complex roots

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.

Show solution
1) D = 16 − 52 = −36, √(−D) = 6: x = (4 ± 6i)/2 = 2 ± 3i. Check with Vieta: sum 4, product 4 + 9 = 13 ✓.
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
The test is taken in English or Chinese — recognise the terms in both.

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

12 questions. Every correct answer earns XP.

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