Modulus, argument, polar form and de Moivre's theorem for integral powers

1. Modulus and principal argument

For z=a+bi, the modulus |z|=√(a²+b²) is the distance from the origin to the point (a,b) on the Argand plane. The principal argument Arg z is the angle measured from the positive real axis to the ray through z, with −π<Arg z≤π. The WACE formula sheet and marking keys use this interval, so an answer of 7π/4 where −π/4 is required is not in principal form even though it names the same ray. Arg 0 is undefined.

Always locate the quadrant before using a reference angle. The ratio b/a alone cannot separate opposite quadrants: −3−3i and 3+3i both give b/a=1. Work with the reference angle α=tan⁻¹|b/a| and then place it:

Worked example: for z=−√3+i, |z|=√(3+1)=2 and the reference angle is tan⁻¹(1/√3)=π/6. The point is in the second quadrant, so Arg z=π−π/6=5π/6. Write the answer in the form the question asks for; z=2cis(5π/6) is the polar form.

Points on the axes need no reference angle: a positive real number has argument 0, a negative real number has argument π, and purely imaginary numbers have ±π/2. A quick sketch of the point on the Argand plane, even a rough one in the margin, prevents the most common error in this topic, which is a correct reference angle placed in the wrong quadrant.

2. Converting between Cartesian and polar form

The polar form is z=r(cos θ+i sin θ)=r cis θ, where r=|z| and θ is an argument. Converting from polar to Cartesian form needs only exact values: 4cis(−2π/3)=4(−½−(√3/2)i)=−2−2√3i. Converting from Cartesian to polar form needs the modulus and a correctly placed argument, as in the previous section. A calculator-free paper expects the exact values of sine and cosine for multiples of π/6 and π/4 to be automatic; SCSA reports from 2021 and 2023 both name weak recall of exact trigonometric values as a cause of lost marks.

Three identities that follow straight from the definitions are worth proving once so that they can be quoted with confidence:

• z·z̄=a²+b²=|z|², so 1/z=z̄/|z|² for z≠0.
• |z̄|=|z| and, for z not on the negative real axis, Arg z̄=−Arg z.
• Arg(−z)=Arg z−π when Arg z>0, and Arg z+π when Arg z≤0. The 2021 report singled out the argument of the opposite of a complex number as a weak routine skill. For z=2cis(2π/3), −z=2cis(−π/3).

An argument proof is usually a short chain: write z=r cis θ, apply the definition or a polar-form rule, and finish with the required statement. For example, to show |z²|=|z|², write z²=r²cis 2θ, so |z²|=r²=|z|². State every step; a line that just repeats the result does not count as a proof.

The same approach proves |z₁z₂|=|z₁||z₂| in Cartesian form if a question insists: expand (ac−bd)²+(ad+bc)², observe that the cross terms cancel, and factorise the remainder as (a²+b²)(c²+d²). Polar form makes the result immediate, which is a good reason to switch forms.

3. Multiplication and division in polar form

If z₁=r₁cis θ₁ and z₂=r₂cis θ₂, expanding the product and using the compound-angle formulas gives z₁z₂=r₁r₂cis(θ₁+θ₂). Moduli multiply and arguments add. Division follows in the same way: z₁/z₂=(r₁/r₂)cis(θ₁−θ₂) for z₂≠0. These rules give the geometric meaning used in the next note: multiplying by r cis θ dilates by factor r and rotates anticlockwise by θ about the origin.

Worked example: let z₁=2cis(π/3) and z₂=3cis(3π/4).

• z₁z₂=6cis(π/3+3π/4)=6cis(13π/12). Since 13π/12>π, subtract 2π: the principal form is 6cis(−11π/12).
• z₁/z₂=(2/3)cis(π/3−3π/4)=(2/3)cis(−5π/12), already in the principal interval.

The argument rule holds for arguments in general, but Arg(z₁z₂)=Arg z₁+Arg z₂ can fail for principal arguments because the sum may leave (−π, π]. Adjusting by a multiple of 2π at the end is part of the answer, not an optional tidy-up.

When numbers arrive in Cartesian form, decide which form makes the task shorter. Products of several factors, powers and quotients are quicker in polar form; sums and differences need Cartesian form. Converting each factor, combining, then converting back is a standard three-step pattern in calculator-free questions.

4. Proving de Moivre's theorem

De Moivre's theorem states that (cis θ)ⁿ=cis(nθ) for every integer n, so (r cis θ)ⁿ=rⁿcis(nθ). The 2025 syllabus wording says it is proved and used for integral powers, so be ready to give the induction proof.

Positive integers, by induction. Let P(n) be the statement (cis θ)ⁿ=cis(nθ).

• Initial case: (cis θ)¹=cis θ, so P(1) is true.
• Inductive step: assume P(k) is true for some positive integer k. Then (cis θ)^k+1=(cis θ)^kcis θ=cis(kθ)cis θ. Expanding, this is cos kθ cos θ−sin kθ sin θ+i(sin kθ cos θ+cos kθ sin θ)=cos(k+1)θ+i sin(k+1)θ by the compound-angle formulas. So P(k+1) is true.
• Conclusion: P(1) is true and P(k) implies P(k+1), so P(n) is true for all positive integers n.

Zero and negative integers. (cis θ)⁰=1=cis 0. For n=−m with m a positive integer, (cis θ)^−m=1/cis(mθ). Multiplying numerator and denominator by the conjugate cis(−mθ) and using cis(mθ)cis(−mθ)=cis 0=1 gives cis(−mθ)=cis(nθ).

Markers look for the induction hypothesis to be used explicitly in the step, and for the conclusion to be stated. Writing 'and so on' in place of the inductive step earns little. If the question asks you to use the theorem rather than prove it, a one-line citation is enough.

Two consequences are worth stating. First, |zⁿ|=|z|ⁿ, so the moduli of powers grow or shrink geometrically. Second, the argument of zⁿ is n times an argument of z, but nArg z may not be the principal argument of zⁿ; it must be reduced by a multiple of 2π, which is the same adjustment seen with products.

5. Powers: worked examples

Example 1. Evaluate (1+i√3)⁸ in Cartesian form. Here 1+i√3=2cis(π/3), so the power is 2⁸cis(8π/3)=256cis(2π/3) after removing 2π. Then 256(−½+(√3/2)i)=−128+128√3i.

Example 2. Evaluate (1−i)¹⁰. Since 1−i=√2cis(−π/4), the power is (√2)¹⁰cis(−10π/4)=32cis(−5π/2). Adding 2π gives 32cis(−π/2)=−32i.

Example 3 (negative power). Evaluate (√3−i)⁻⁴. With √3−i=2cis(−π/6), the power is 2⁻⁴cis(4π/6)=(1/16)cis(2π/3)=−1/32+(√3/32)i.

Example 4 (condition on n). Find the smallest positive integer n for which (1+i)ⁿ is a positive real number. Since (1+i)ⁿ=2^n/2cis(nπ/4), we need nπ/4 to be a multiple of 2π, so n is a multiple of 8 and the answer is n=8. If the question asked for a real number of either sign, nπ/4 would need to be a multiple of π, giving n=4.

Set out each power in four visible steps: polar form of the base, apply de Moivre, reduce the argument into (−π, π], convert back. Under exam pressure the reduction step is the one most often skipped, and a wrong exact value then follows. Check the size of the answer too: |(1+i√3)⁸| must be 256, and √(128²+(128√3)²)=256 confirms Example 1.

When the base has an awkward argument, such as 3+4i, de Moivre still applies but the answer stays in polar form, for instance 5⁶cis(6θ) with θ=tan⁻¹(4/3). Exam bases are chosen to give exact values, so an ugly argument is a hint to re-check the conversion.

6. Using de Moivre to derive trigonometric identities

Equating two expansions of (cis θ)ⁿ gives multiple-angle identities. For n=3, de Moivre gives cos 3θ+i sin 3θ. The binomial expansion gives (c+is)³=c³+3c²(is)+3c(is)²+(is)³=c³−3cs²+i(3c²s−s³), with c=cos θ and s=sin θ.

• Real parts: cos 3θ=c³−3cs²=c³−3c(1−c²)=4cos³θ−3cos θ.
• Imaginary parts: sin 3θ=3c²s−s³=3(1−s²)s−s³=3sin θ−4sin³θ.

Note the powers of i in the expansion: i²=−1 and i³=−i. The 2021 report recorded errors in exactly this kind of bracket work.

A second family of results comes from z=cis θ, for which z⁻¹=cis(−θ)=z̄. Adding and subtracting:

• zⁿ+z⁻ⁿ=2cos nθ
• zⁿ−z⁻ⁿ=2i sin nθ

These express powers of cos θ in terms of multiple angles, which is the form needed for integration. For example, (2cos θ)³=(z+z⁻¹)³=z³+3z+3z⁻¹+z⁻³=2cos 3θ+6cos θ. Dividing by 8 gives cos³θ=¼(cos 3θ+3cos θ). This links to note 12, where integrating cos³θ by this route is an alternative to the substitution u=sin θ.

In a 'show that' question the result is given, so marks are for the visible link between the binomial expansion, the comparison of real or imaginary parts, and the use of sin²θ+cos²θ=1.

A quick numerical check is always available: at θ=0.7 radians a calculator gives the same value for cos 2.1 and for 4cos³0.7−3cos 0.7. A test like this does not prove the identity, but it catches a sign error before you rely on the result in a later part.

7. Exam technique and common traps

Complex numbers appear in both sections of every recent paper. Typical calculator-free items ask for a power in Cartesian form, a polar-form product or quotient, or a short proof. Calculator-assumed items mix polar form with geometry and loci. The following habits protect marks.

• Show the reference angle and the quadrant. A bare answer of 5π/6 with no working can lose a method mark in a question worth more than two marks, since SCSA requires valid working or justification above that threshold.
• Keep exact values exact. If a CAS gives 1.0472, recognise it as π/3 and write π/3.
• Reduce the argument at the end. Write cis(13π/12)=cis(−11π/12) as a separate line.
• Distinguish r cis θ from r cos θ+i sin θ. The modulus multiplies both parts: 4cis(π/6)=2√3+2i, not 2√3+i/2.
• Copy signs carefully. Every report from 2021 to 2025 tells candidates to check working line by line because negatives disappear between lines.

A model line for a polar-form product reads: 'z₁z₂=2×3 cis(π/3+3π/4)=6cis(13π/12)=6cis(−11π/12), since the principal argument lies in (−π, π].' It names the rule, shows the arithmetic and gives the reason for the adjustment. A model induction conclusion reads: 'Since P(1) is true, and P(k) true implies P(k+1) true, P(n) is true for all positive integers n by mathematical induction.'

For checking on the calculator-assumed section, a CAS converts between forms instantly. Use it to confirm your exact answer, but write the exact working when the question says 'show' or 'using de Moivre's theorem'.