Practice Maths

Complex Numbers in Cartesian Form

Key Terms

Complex number z
A number z = a + bi where a, b ∈ ℝ and i² = −1.
Real part Re(z)
The coefficient a in z = a + bi.
Imaginary part Im(z)
The coefficient b (NOT bi) in z = a + bi; a common error is to include the i.
Complex conjugate z̅
z̅ = a − bi; z · z̅ = a² + b² (always a non-negative real number).
Equal complex numbers
a + bi = c + di if and only if a = c AND b = d simultaneously (equate real and imaginary parts).
Powers of i
i¹ = i, i² = −1, i³ = −i, i&sup4; = 1; the cycle repeats with period 4.

Complex Numbers — Key Facts

Imaginary unit: i is defined by i² = −1, so i = √(−1)

Cartesian form: z = a + bi, where a = Re(z) is the real part, b = Im(z) is the imaginary part

Conjugate: If z = a + bi then z̅ = a − bi

Key product: z · z̅ = a² + b² (always a non-negative real number)

Equal complex numbers: a + bi = c + di if and only if a = c AND b = d

Powers of i: i¹ = i, i² = −1, i³ = −i, i&sup4; = 1, then the cycle repeats

Worked Example 1 — Identifying Parts and Conjugates

For each complex number, state the real part, imaginary part, and conjugate:

(a) z = 3 + 5i    (b) z = −2 − 7i    (c) z = 4 (a real number)    (d) z = −3i (a pure imaginary)

(a) Re(z) = 3, Im(z) = 5, z̅ = 3 − 5i

(b) Re(z) = −2, Im(z) = −7, z̅ = −2 + 7i

(c) Re(z) = 4, Im(z) = 0, z̅ = 4 (real numbers are their own conjugates)

(d) Re(z) = 0, Im(z) = −3, z̅ = 3i

Worked Example 2 — Simplifying Powers of i and Computing z·z̅

Simplify: (a) i³    (b) i10    (c) z · z̅ for z = 3 + 4i

(a) i³ = i² × i = −1 × i = −i

(b) i10 = (i&sup4;)² × i² = 1² × (−1) = −1

(c) z · z̅ = (3 + 4i)(3 − 4i) = 3² + 4² = 9 + 16 = 25

Hot Tip: Im(z) is the coefficient of i, not the term itself. For z = 3 − 5i, Im(z) = −5, NOT −5i. Many students lose marks by including the i in their answer.

Why Do We Need Complex Numbers?

The real number system has a fundamental gap: you cannot take the square root of a negative number. Consider the equation x² + 1 = 0. Rearranging gives x² = −1, but no real number satisfies this — squaring any real number always gives a non-negative result. For centuries, mathematicians dismissed such equations as “impossible.”

The solution was to define a new symbol i such that i² = −1. This is not a trick or a sleight of hand — it is a legitimate extension of the number system, just as fractions extended whole numbers and irrationals extended rationals. The resulting system of complex numbers is complete in a profound sense: every polynomial equation has a solution in ℂ (the Fundamental Theorem of Algebra).

The Cartesian Form a + bi

Every complex number can be written as z = a + bi where a and b are real numbers. We call this Cartesian form (or rectangular form). The number a is the real part Re(z), and b is the imaginary part Im(z) — note that Im(z) is the real number b, not the expression bi.

Special cases: when b = 0, z = a is a real number; when a = 0, z = bi is a pure imaginary number. This shows that the real numbers are a subset of the complex numbers.

The Imaginary Unit and Powers of i

Since i² = −1, we can simplify any power of i by reducing the exponent modulo 4:

PowerValueReasoning
idefinition
−1definition
−ii² × i = −1 × i
i&sup4;1i² × i² = (−1)(−1)
i&sup5;ii&sup4; × i = 1 × i

To simplify in: divide n by 4 and check the remainder. Remainder 0 → 1, remainder 1 → i, remainder 2 → −1, remainder 3 → −i.

The Conjugate and its Key Property

The conjugate of z = a + bi is z̅ = a − bi. Geometrically, taking the conjugate reflects the number across the real axis on the Argand diagram (covered in a later lesson).

The most important algebraic property is: z · z̅ = (a + bi)(a − bi) = a² + b². This is a real number, and it is always ≥ 0. This property is the engine behind dividing complex numbers — multiplying numerator and denominator by the conjugate of the denominator converts the denominator to a real number.

Equality of Complex Numbers

Two complex numbers are equal if and only if their real parts are equal AND their imaginary parts are equal. This gives us a powerful technique: if we know a + bi = c + di, we can equate real parts (a = c) and imaginary parts (b = d) to solve for unknowns. This technique appears constantly throughout Unit 2.

Common error: Students sometimes write i² = +1 instead of −1. Always remember: i² = −1 by definition.

Mastery Practice

  1. Fluency

    Q1 — Real and Imaginary Parts

    For each complex number, state Re(z), Im(z), and z̅:

    (a) z = 5 + 2i    (b) z = −3 + 4i    (c) z = 7i    (d) z = −6

  2. Fluency

    Q2 — Powers of i

    Simplify: (a) i6    (b) i11    (c) i20    (d) i35

  3. Fluency

    Q3 — Computing z · z̅

    Calculate z · z̅ for: (a) z = 2 + 3i    (b) z = 5 − i    (c) z = −1 + 2i

  4. Fluency

    Q4 — Simplifying Square Roots of Negatives

    Write in terms of i: (a) √(−9)    (b) √(−25)    (c) √(−7)

  5. Understanding

    Q5 — Equating Real and Imaginary Parts

    Find real numbers x and y if:   (a) x + yi = 5 − 3i    (b) (2x + 1) + (y − 3)i = 7 + 2i

  6. Understanding

    Q6 — Solve Using i

    Solve each equation over ℂ:   (a) x² + 9 = 0    (b) x² + 2x + 5 = 0

  7. Understanding

    Q7 — Conjugate Properties

    Given z = 3 − 4i, verify that: (a) Re(z) + Re(z̅) = 2 Re(z)    (b) z + z̅ is real    (c) z − z̅ is purely imaginary

  8. Understanding

    Q8 — Finding Unknown Complex Numbers

    Find z = a + bi given that z + 2z̅ = 9 − 6i.

  9. Problem Solving

    Q9 — Simplifying a Complex Expression

    Simplify (1 + i)² + (1 − i)² and hence explain the relationship between (1 + i) and (1 − i).

  10. Problem Solving

    Q10 — Proof Using Conjugates

    Prove that for any complex number z, z² + (z̅)² is always a real number.