Loav · VCE Specialist Mathematics · Units 3 & 4
S34-C01 · Algebra, number and structure · Complex numbers
1. A larger number system with familiar algebra
The real equation x² = −1 has no solution because a real square is non-negative. Complex numbers extend the number system by introducing i with i² = −1. Every complex number can be written uniquely as z = a + bi, where a and b are real. The real part is Re(z) = a and the imaginary part is Im(z) = b. Notice that the imaginary part is the real coefficient b, not the term bi.
Equality is componentwise: a + bi = c + di exactly when a = c and b = d. This lets a complex equation supply two real equations. Real numbers are included when b = 0; purely imaginary numbers have a = 0. Zero is a legitimate complex number, but it has special geometric restrictions later because it has no direction.
Add and subtract corresponding parts. For multiplication, expand using ordinary distributive algebra, then replace i² by −1. Thus (a + bi)(c + di) = (ac − bd) + (ad + bc)i. There is no need to invent a new expansion rule. Powers of i repeat in a four-step cycle: i, −1, −i, 1. Reducing an integer exponent modulo 4 is usually faster and safer than repeated multiplication.
Be cautious with square roots. For positive real a, the equation z² = −a has the two solutions z = ±i√a. A complex equation generally needs both roots even if a calculator displays one square-root value. The familiar real identity √u√v = √(uv) cannot be used without conditions for arbitrary complex inputs. Substituting u = v = −1 would incorrectly turn i² into 1.
2. Conjugation makes division manageable
The conjugate of z = a + bi is z = a − bi. It preserves the real part and reverses the imaginary part. Adding a number to its conjugate gives 2a; subtracting gives 2bi. Their product is especially useful: zz = a² + b². The cross terms cancel, leaving a non-negative real number, which is zero only when z = 0.
To divide by c + di, multiply numerator and denominator by c − di. The denominator becomes c² + d², so the procedure is valid provided the original denominator is nonzero. Multiplying only the denominator would change the value. Keep the numerator bracketed, expand completely and collect real and imaginary parts before dividing each by the real denominator.
Conjugation respects addition and multiplication: the conjugate of z + w is z + w, and the conjugate of zw is zw. You can verify these facts by expansion. They will explain why non-real roots of real-coefficient polynomials come in conjugate pairs in C03. Conjugation is a reflection, not an operation that changes every sign: the real part stays fixed.
Worked example 1: divide, then check by multiplication
Write (3 + 2i)/(1 − 2i) in Cartesian form. The denominator is nonzero. Multiply both numerator and denominator by 1 + 2i. The numerator is 3 + 6i + 2i + 4i² = −1 + 8i, while the denominator is 1 + 4 = 5. Therefore the quotient is −1/5 + (8/5)i.
Check by multiplying the result by the original denominator. The real part is −1/5 + 16/5 = 3; the imaginary part is 2/5 + 8/5 = 2. This recovers 3 + 2i exactly. A decimal check is possible, but exact fractions reveal the cancellation more clearly and avoid rounding. The answer has a negative real part even though both parts of the original numerator were positive; division also changes direction in the complex plane.
3. The Argand plane links algebra and geometry
Plot z = a + bi as the point (a, b) on an Argand diagram. The horizontal axis is Re(z), and the vertical axis is Im(z). Addition behaves like coordinate addition, so it can be pictured by a parallelogram. The difference z − w represents the displacement from w to z. This explains why |z − w| is the ordinary distance between the two points.
The modulus is |z| = √(a² + b²), the distance from the origin. It is a real non-negative number, not a complex coordinate pair. Reflection across the real axis does not change distance, so |z| = |z|. Also zz = |z|². These identities connect the geometric distance with the denominator used in complex division.
For z ≠ 0, an argument θ measures the directed angle from the positive real axis to the ray through z. All arguments differ by integer multiples of 2π. We use the principal argument Arg(z) in the interval −π < θ ≤ π. In particular, a negative real number has principal argument π, not −π under this convention. Zero has modulus zero but no argument, because no direction is determined by the origin alone.
4. Convert between Cartesian and polar form
Define cis θ = cos θ + i sin θ. For a nonzero complex number, z = r cis θ with r = |z| > 0. Its Cartesian parts are a = r cos θ and b = r sin θ. The modulus is always taken non-negative; moving a negative sign into the angle is safer than using a “negative modulus”. All angles in the formulas here are in radians.
From Cartesian form, find r with Pythagoras and locate the quadrant before choosing θ. The equation tan θ = b/a alone cannot distinguish opposite quadrants, and it is unavailable when a = 0. For z = −3 − 4i, the reference angle is arctan(4/3), but the point is in the third quadrant. A principal argument is −π + arctan(4/3), approximately −2.2143 radians, and r = 5. Using arctan(4/3) alone would put the point in the first quadrant.
Exact circular values should remain exact when possible. For −1 + √3i, r = 2 and θ = 2π/3, so z = 2 cis(2π/3). A decimal angle can be useful for a sketch or numerical check, but it should not replace a requested exact answer. If technology provides an argument outside the chosen principal interval, add or subtract 2π until the convention is satisfied.
5. Multiplication combines lengths and angles
Expanding (cos α + i sin α)(cos β + i sin β) and using the angle-addition identities gives cis α cis β = cis(α + β). Consequently, if z = r cis α and w = s cis β, then zw = rs cis(α + β). Multiply the moduli and add the arguments. The argument of the product may need adjustment by a multiple of 2π to become principal.
For w ≠ 0, z/w = (r/s) cis(α − β), provided z is also nonzero when an argument is used. Divide the moduli and subtract the angles. If the numerator is zero, the quotient is simply zero and has no argument. Multiplication by a fixed nonzero w therefore rotates every nonzero point by β and scales every distance from the origin by s. Division reverses those actions.
The modulus rules |zw| = |z||w| and |z/w| = |z|/|w| provide independent checks. Addition has no comparable rule that simply adds moduli or arguments. For example, 1 + (−1) = 0, although both summands have modulus 1. Choose the representation that matches the operation: Cartesian form is usually best for sums, and polar form is often best for products, quotients and powers.
Worked example 2: choose polar form for a product and quotient
Let z = −1 + √3i and w = √3 + i. Their polar forms are 2 cis(2π/3) and 2 cis(π/6). Thus zw = 4 cis(5π/6) = −2√3 + 2i. Both the modulus 4 and the second-quadrant position follow before expansion.
The quotient is z/w = cis(2π/3 − π/6) = cis(π/2) = i. A quick Cartesian check is iw = i(√3 + i) = −1 + √3i = z. This confirms the subtraction order. Reversing the angles would give −i, which is the reciprocal quotient w/z. The principal angles here already lie in the required interval.
Worked example 3: interpret a repeated transformation
A point begins at z₀ = √3 + i = 2 cis(π/6). Each step multiplies its complex coordinate by m = 1 + √3i = 2 cis(π/3). Find its positions after one and two steps. The first is z₁ = 4 cis(π/2) = 4i. The second is z₂ = 8 cis(5π/6) = −4√3 + 4i.
Each step doubles the distance from the origin and rotates anticlockwise by π/3. It does not add π/3 to the Cartesian coordinates. To check the second step, multiply 4i(1 + √3i) = −4√3 + 4i. The scale factors accumulate multiplicatively, giving radii 2, 4 and 8. This small transformation model uses exactly the same arithmetic as a bare product calculation, while geometry helps detect a wrong sign or quadrant.
6. Build a reliable checking routine
Before computing, decide whether an exact form is available and whether division or an argument excludes zero. After computing, compare the expected modulus and quadrant with the result. On a calculator, check radian mode and use an argument function that handles quadrants, rather than relying only on inverse tangent. Plotting both representations should produce the same point. A rounded numerical match is supporting evidence; exact expansion establishes an identity.
Label an Argand diagram with complex coordinates or clear real and imaginary coordinates. Write Arg(z) for the selected principal value and explain when you are instead giving all arguments. Avoid treating an argument as a unique unrestricted real number. These distinctions become essential when finding all roots in C02 and when interpreting rays and sectors in C04.
Practise, then check
Keep answers exact unless a decimal is explicitly requested.
- Simplify i37 + i22.
- For z = −5 + 12i, find its conjugate, modulus, real part and imaginary part.
- Express (4 − i)/(2 + i) in Cartesian form.
- Find the polar form of −√2 − √2i using its principal argument.
- Let z = 3 cis(5π/6) and w = 2 cis(π/3). Find zw and z/w in polar form with principal arguments.
- Real numbers a and b satisfy (a + bi)(1 + i) = 5 + i. Find a and b.
- A student says Arg(zw) = Arg(z) + Arg(w) always, as an equality of principal values. Give a counterexample and the correct rule.
Reasoned answers
- The powers repeat every four. Since 37 leaves remainder 1 and 22 leaves remainder 2, i37 = i and i22 = −1. The sum is −1 + i. Reducing the exponents works because i⁴ = 1.
- The conjugate is −5 − 12i. The modulus is √(25 + 144) = 13. Re(z) = −5 and Im(z) = 12. In particular, Im(z) is not 12i. The conjugate has the same modulus because reflection preserves distance from the origin.
- Multiplication by 2 − i gives numerator (4 − i)(2 − i) = 7 − 6i and denominator 5. The quotient is 7/5 − (6/5)i. Multiplying this by 2 + i recovers 4 − i, checking both the real and imaginary components.
- The modulus is √(2 + 2) = 2. Both parts are negative, so the principal argument is −3π/4. Hence z = 2 cis(−3π/4). The angle π/4 has the correct tangent but the wrong quadrant.
- The product initially has angle 7π/6, so its principal form is 6 cis(−5π/6). The quotient has modulus 3/2 and angle π/2, giving (3/2)cis(π/2). Subtracting 2π from the product angle changes its representation, not its value.
- Expansion gives (a − b) + (a + b)i. Equating parts yields a − b = 5 and a + b = 1. Adding gives 2a = 6, so a = 3 and b = −2. Substitution gives (3 − 2i)(1 + i) = 5 + i as required.
- Take z = w = −1. Both principal arguments are π, but zw = 1 has principal argument 0, not 2π. Arguments add modulo 2π. For principal values, first add the two angles and then adjust by an integer multiple of 2π into −π < θ ≤ π. Both factors must be nonzero.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4. Scope follows the VCAA Mathematics Study Design 2023–2027, Specialist Mathematics Areas of Study 1 and 3 as applicable, and the current Specialist Mathematics FAQ. Unit 2 complex-number foundations are retained where required. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 6A–6D, pp. 238–261. All explanations, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.