Modulus-Argument Form
Every complex number z = x + iy can be written in modulus-argument (polar) form. This form makes multiplication, division, and powers dramatically easier.
Modulus |z|
|z| = r = √(x² + y²)
Distance from origin in the Argand plane. Always r ≥ 0.
Argument arg(z)
arg(z) = θ = arctan(y/x)
Angle from positive real axis. Principal argument: −π < θ ≤ π.
Modulus-argument form
z = r(cosθ + i sinθ) = r cisθ
Also written as z = re^(iθ) via Euler's formula: e^(iθ) = cosθ + i sinθ
Finding the Argument — Quadrant Rules
The formula arctan(y/x) only gives the correct angle in the first and fourth quadrants. Always check which quadrant z lies in:
Quadrant I x>0, y>0
θ = arctan(y/x)
0 < θ < π/2
Quadrant II x<0, y>0
θ = π + arctan(y/x)
π/2 < θ < π
Quadrant III x<0, y<0
θ = −π + arctan(y/x)
−π < θ < −π/2
Quadrant IV x>0, y<0
θ = arctan(y/x)
−π/2 < θ < 0
⛔ Most Common Error — Argument Calculation
For z = −1 + i: x = −1, y = 1 → arctan(1/−1) = arctan(−1) = −π/4. But z is in quadrant II, so arg(z) = π + (−π/4) = 3π/4. Never quote arctan directly without checking the quadrant.
Multiplication and Division in Polar Form
Multiplication
z₁z₂ = r₁r₂ · cis(θ₁+θ₂)
Moduli multiply, arguments add.
Division
z₁/z₂ = (r₁/r₂) · cis(θ₁−θ₂)
Moduli divide, arguments subtract.
Conjugate z* = x − iy
|z*| = |z| = r arg(z*) = −arg(z) = −θ
Reflection in the real axis. |z|² = z·z* = x² + y² = r².
Euler's Formula — The Most Beautiful Equation
Euler's Formula
e^(iθ) = cos θ + i sin θ
Setting θ = π: e^(iπ) + 1 = 0
Euler's identity — connects e, π, i, 1, and 0 in a single equation.
De Moivre's Theorem
De Moivre's Theorem
[r(cosθ + i sinθ)]ⁿ = rⁿ(cos nθ + i sin nθ)
Equivalently: (cisθ)ⁿ = cis(nθ) or (e^(iθ))ⁿ = e^(inθ)
Holds for all integers n (and in fact for rational n when finding roots). Proof by induction on positive integers, extend by conjugates.
Three Major Applications
Application 1
Powers of complex numbers
Convert to r·cisθ, apply de Moivre, convert back if needed.
Convert to r·cisθ, apply de Moivre, convert back if needed.
Application 2
Multiple-angle identities
Expand (cosθ+i sinθ)ⁿ by binomial theorem and equate real/imaginary parts to get cos nθ and sin nθ.
Expand (cosθ+i sinθ)ⁿ by binomial theorem and equate real/imaginary parts to get cos nθ and sin nθ.
Application 3
Roots of unity / equations
Solve zⁿ = a + bi by writing the RHS in polar form and applying the root version of de Moivre.
Solve zⁿ = a + bi by writing the RHS in polar form and applying the root version of de Moivre.
Application 1 — Powers
E
Finding (1+i)⁸
Express (1+i)⁸ in the form a + bi.
Polar form
|1+i| = √2, arg(1+i) = π/4
So 1+i = √2 · cis(π/4)
So 1+i = √2 · cis(π/4)
De Moivre
(1+i)⁸ = (√2)⁸ · cis(8·π/4) = 16 · cis(2π) = 16(cos 2π + i sin 2π)
Result
(1+i)⁸ = 16
Application 2 — Multiple Angle Formulae
E
Express cos 3θ in terms of cos θ
Use de Moivre's theorem to show cos 3θ = 4cos³θ − 3cosθ.
De Moivre
(cosθ + i sinθ)³ = cos 3θ + i sin 3θ
Binomial expand
LHS = cos³θ + 3cos²θ(i sinθ) + 3cosθ(i sinθ)² + (i sinθ)³
= cos³θ + 3i cos²θ sinθ − 3cosθ sin²θ − i sin³θ
= cos³θ + 3i cos²θ sinθ − 3cosθ sin²θ − i sin³θ
Equate real parts
cos 3θ = cos³θ − 3cosθ sin²θ
= cos³θ − 3cosθ(1−cos²θ)
= cos³θ − 3cosθ(1−cos²θ)
cos 3θ = 4cos³θ − 3cosθ
Bonus — sin 3θ
Imaginary parts: sin 3θ = 3cos²θ sinθ − sin³θ = 3sinθ − 4sin³θ
Powers of cos θ and sin θ — The z + z⁻¹ Trick
📌 The Key Identities
Let z = cisθ = e^(iθ)
z + z⁻¹ = 2cosθ
z − z⁻¹ = 2i sinθ
Powers (from de Moivre)
zⁿ + z⁻ⁿ = 2cos nθ
zⁿ − z⁻ⁿ = 2i sin nθ
Use these to express cosⁿθ or sinⁿθ as a sum of multiple angles — essential for integration questions in 9231.
E
Express cos⁴θ in terms of multiple angles
Show that cos⁴θ = ⅛(cos 4θ + 4cos 2θ + 3).
Setup
2cosθ = z + z⁻¹, so 2⁴cos⁴θ = (z + z⁻¹)⁴
Expand (z+z⁻¹)⁴
= z⁴ + 4z² + 6 + 4z⁻² + z⁻⁴
= (z⁴+z⁻⁴) + 4(z²+z⁻²) + 6
= 2cos 4θ + 4·2cos 2θ + 6
= 2cos 4θ + 8cos 2θ + 6
= (z⁴+z⁻⁴) + 4(z²+z⁻²) + 6
= 2cos 4θ + 4·2cos 2θ + 6
= 2cos 4θ + 8cos 2θ + 6
Divide by 16
cos⁴θ = ⅛(cos 4θ + 4cos 2θ + 3)
Integration use
∫cos⁴θ dθ = ⅛[sin4θ/4 + 4sin2θ/2 + 3θ] + C = ⅛[sin4θ/4 + 2sin2θ + 3θ] + C
Roots of Unity & Complex Equations
nth Roots of Unity
The equation zⁿ = 1 has exactly n solutions in ℂ, equally spaced around the unit circle at angles 2πk/n for k = 0, 1, …, n−1.
nth roots of unity
zₖ = e^(2πik/n) = cos(2πk/n) + i sin(2πk/n), k = 0, 1, …, n−1
Denote the primitive root ω = e^(2πi/n). Then the roots are 1, ω, ω², …, ωⁿ⁻¹.
💡 Key Symmetry Properties
- Sum of all nth roots of unity = 0 (for n ≥ 2). This is used to simplify expressions.
- Product of all nth roots of unity = (−1)ⁿ⁺¹
- The roots come in conjugate pairs: if ω is a root, so is ω̄ = ωⁿ⁻¹.
- The non-trivial roots satisfy: 1 + ω + ω² + ··· + ωⁿ⁻¹ = 0.
Roots of Unity — Interactive
n =
Solving zⁿ = a + bi — General Method
E
Solve z³ = −8
Find all three cube roots of −8, giving answers in modulus-argument form.
Polar form
−8 = 8·cis(π) [|−8|=8, arg(−8)=π]
General form
−8 = 8·cis(π + 2πk) for k ∈ ℤ
Apply de Moivre
z = 8^(1/3) · cis((π+2πk)/3) = 2·cis((π+2πk)/3)
k = 0, 1, 2
k=0: z₀ = 2·cis(π/3) = 2(cos π/3 + i sin π/3) = 1 + i√3
k=1: z₁ = 2·cis(π) = −2
k=2: z₂ = 2·cis(5π/3) = 2·cis(−π/3) = 1 − i√3
k=1: z₁ = 2·cis(π) = −2
k=2: z₂ = 2·cis(5π/3) = 2·cis(−π/3) = 1 − i√3
z = 1+i√3, −2, 1−i√3
Note: z₀ and z₂ are conjugates ✓ z₀+z₁+z₂ = 0 ✓
Loci in the Argand Diagram
A locus is the set of all points z satisfying a given condition. Cambridge 9231 tests four standard types — you must recognise each immediately from its algebraic form.
|z − a| = r
Circle
Centre at the point representing a, radius r. The set of all z at distance r from the fixed point a.
|z − a| = |z − b|
Perpendicular Bisector
The perpendicular bisector of the line segment joining a to b. Equal distance from two fixed points.
arg(z − a) = θ
Half-line (ray)
A ray from the point a (not including a itself) at angle θ from the positive real direction. The endpoint is excluded.
|z − a| ≤ r
Disc (region)
The closed disc including the boundary circle. Use < for open disc (boundary excluded). Shade the appropriate region.
|z − a| / |z − b| = k (k ≠ 1)
Circle of Apollonius
A circle (not a perpendicular bisector) — the Apollonius circle. Convert to Cartesian form by writing z = x+iy and squaring both sides.
📌 Cambridge Loci — What You Must Show
- Circle |z−a|=r: state centre and radius. Mark the centre with a dot and label the radius.
- Perpendicular bisector: find the midpoint of ab, verify the gradient is perpendicular to ab, draw the line.
- Half-line arg(z−a)=θ: mark the fixed point a (open circle — excluded), draw the ray at angle θ. State the direction and that the endpoint is excluded.
- Region questions: shade the correct side, use solid line for ≤ and dashed for <.
E
Locus — Circle and Perpendicular Bisector
Describe and sketch the locus of z satisfying both:
(i) |z − 2| = |z − 2i| (ii) |z − 2| ≤ 3
(i) |z − 2| = |z − 2i| (ii) |z − 2| ≤ 3
Part (i)
|z−2| = |z−2i| means z is equidistant from 2 (point (2,0)) and 2i (point (0,2)).
Locus: perpendicular bisector of the segment from (2,0) to (0,2).
Midpoint: (1,1). Slope of segment: (2−0)/(0−2) = −1. Perp slope = 1.
Locus: perpendicular bisector of the segment from (2,0) to (0,2).
Midpoint: (1,1). Slope of segment: (2−0)/(0−2) = −1. Perp slope = 1.
Line: y − 1 = 1(x − 1) → y = x
Part (ii)
|z−2| ≤ 3: closed disc, centre (2, 0), radius 3 (boundary included).
The region satisfying both conditions is the arc of y = x inside or on the disc.
The region satisfying both conditions is the arc of y = x inside or on the disc.
E
Locus — Half-line to Cartesian
Describe the locus of z satisfying arg(z − 1 − i) = 2π/3.
Identify fixed point
The fixed point is a = 1 + i, i.e. the point (1, 1) in the Argand plane.
Describe
Half-line (ray) starting at (1, 1), not including (1, 1), at angle 2π/3 to the positive real direction.
Direction: cos(2π/3) = −½, sin(2π/3) = √3/2. The ray goes into quadrant II from (1,1).
Direction: cos(2π/3) = −½, sin(2π/3) = √3/2. The ray goes into quadrant II from (1,1).
Ray from (1,1) at angle 120° to positive real axis, endpoint excluded.
Worked Examples
1
Division in Polar Form
z₁ = 4(cos π/3 + i sin π/3), z₂ = 2(cos π/6 + i sin π/6). Find z₁/z₂ and z₁·z₂.
Division
|z₁/z₂| = 4/2 = 2 arg = π/3 − π/6 = π/6
z₁/z₂ = 2(cos π/6 + i sin π/6) = 2·(√3/2 + i/2) = √3 + i
Multiplication
|z₁z₂| = 4×2 = 8 arg = π/3 + π/6 = π/2
z₁z₂ = 8(cos π/2 + i sin π/2) = 8i
2
Express sin⁵θ in terms of sin nθ
Using the substitution z = cisθ, show that sin⁵θ = (1/16)(sin5θ − 5sin3θ + 10sinθ).
Setup
2i sinθ = z − z⁻¹, so (2i)⁵ sin⁵θ = (z − z⁻¹)⁵
32i sin⁵θ = (z − z⁻¹)⁵
32i sin⁵θ = (z − z⁻¹)⁵
Binomial expansion
(z−z⁻¹)⁵ = z⁵ − 5z³ + 10z − 10z⁻¹ + 5z⁻³ − z⁻⁵
= (z⁵−z⁻⁵) − 5(z³−z⁻³) + 10(z−z⁻¹)
= 2i sin5θ − 5·2i sin3θ + 10·2i sinθ
= 2i(sin5θ − 5sin3θ + 10sinθ)
= (z⁵−z⁻⁵) − 5(z³−z⁻³) + 10(z−z⁻¹)
= 2i sin5θ − 5·2i sin3θ + 10·2i sinθ
= 2i(sin5θ − 5sin3θ + 10sinθ)
Divide by 32i
sin⁵θ = (1/16)(sin5θ − 5sin3θ + 10sinθ) ✓
3
Solve z⁴ = −4 + 4√3 i
Find all four roots of z⁴ = −4 + 4√3 i, giving answers in the form re^(iθ).
Polar form of RHS
r = √(16 + 48) = √64 = 8
arg: x=−4<0, y=4√3>0 → Quadrant II
arctan(4√3/−4) = arctan(−√3) = −π/3 → arg = π − π/3 = 2π/3
So: −4+4√3i = 8·cis(2π/3)
arg: x=−4<0, y=4√3>0 → Quadrant II
arctan(4√3/−4) = arctan(−√3) = −π/3 → arg = π − π/3 = 2π/3
So: −4+4√3i = 8·cis(2π/3)
General form
8·cis(2π/3 + 2πk), k = 0,1,2,3
Apply de Moivre (n=4)
z = 8^(1/4) · cis((2π/3 + 2πk)/4) = √2 · cis(π/6 + πk/2)
Four roots
k=0: √2 · e^(iπ/6)
k=1: √2 · e^(i·2π/3)
k=2: √2 · e^(i·7π/6) = √2 · e^(−i·5π/6)
k=3: √2 · e^(i·5π/3) = √2 · e^(−i·π/3)
k=1: √2 · e^(i·2π/3)
k=2: √2 · e^(i·7π/6) = √2 · e^(−i·5π/6)
k=3: √2 · e^(i·5π/3) = √2 · e^(−i·π/3)
Roots equally spaced at π/2 intervals on circle |z|=√2
Practice Questions
Question 1 — Modulus and Argument
[4 marks]
Find the modulus and principal argument of z = (−1 − i√3).
Hence write z⁶ in the form a + bi.
Hence write z⁶ in the form a + bi.
z is in Quadrant III. |z| = √(1+3) = 2. arctan(−√3/−1) = arctan(√3) = π/3, but Quadrant III → arg = −π + π/3 = −2π/3.
✓ Solution
|z| = √(1+3) = 2. x<0, y<0 → Quadrant III.arg = −π + arctan(√3/1) = −π + π/3 = −2π/3
z = 2·cis(−2π/3)
z⁶ = 2⁶·cis(6·(−2π/3)) = 64·cis(−4π) = 64·cis(0)
z⁶ = 64
Question 2 — De Moivre: cos 4θ
[5 marks]
Use de Moivre's theorem to show that cos 4θ = 8cos⁴θ − 8cos²θ + 1.
Expand (cosθ + i sinθ)⁴ by binomial theorem. Collect real parts and use sin²θ = 1−cos²θ.
✓ Solution
(cosθ+i sinθ)⁴ = cos4θ+i sin4θ (de Moivre)LHS = cos⁴θ + 4cos³θ(i sinθ) + 6cos²θ(i sinθ)² + 4cosθ(i sinθ)³ + (i sinθ)⁴
= cos⁴θ + 4i cos³θ sinθ − 6cos²θ sin²θ − 4i cosθ sin³θ + sin⁴θ
Real part: cos4θ = cos⁴θ − 6cos²θ sin²θ + sin⁴θ
= cos⁴θ − 6cos²θ(1−cos²θ) + (1−cos²θ)²
= cos⁴θ − 6cos²θ + 6cos⁴θ + 1 − 2cos²θ + cos⁴θ
= 8cos⁴θ − 8cos²θ + 1 ✓
Question 3 — Roots of Unity
[5 marks]
Find all solutions to z⁵ = 1, expressing each in the form e^(iθ) where −π < θ ≤ π.
Show that their sum is zero.
Show that their sum is zero.
The 5th roots of unity are e^(2πik/5) for k=0,1,2,3,4. Adjust k=3 and k=4 to give principal arguments.
✓ Solution
zₖ = e^(2πik/5) for k = 0,1,2,3,4k=0: e^(0) = 1
k=1: e^(2πi/5)
k=2: e^(4πi/5)
k=3: e^(6πi/5) = e^(−4πi/5) [subtract 2π to get principal arg]
k=4: e^(8πi/5) = e^(−2πi/5)
Sum = 1 + e^(2πi/5) + e^(4πi/5) + e^(−4πi/5) + e^(−2πi/5)
= 1 + 2cos(2π/5) + 2cos(4π/5)
= 1 + 2(0.309) + 2(−0.809) = 1 + 0.618 − 1.618 = 0 ✓
(Or: sum of roots of z⁵−1=0 equals −(coeff of z⁴)/(coeff of z⁵) = 0 by Vieta's)
Question 4 — Locus
[5 marks]
The complex number z satisfies |z − 3 − 4i| = 5.
(i) Describe the locus geometrically.
(ii) Find the Cartesian equation of the locus.
(iii) Find the maximum value of |z|.
(i) Describe the locus geometrically.
(ii) Find the Cartesian equation of the locus.
(iii) Find the maximum value of |z|.
For (iii): |z| is the distance from z to the origin. The maximum distance from the origin to a point on the circle equals the distance from origin to centre plus the radius.
✓ Solution
(i) Circle with centre (3, 4) and radius 5.(ii) Write z = x+iy:
|x+iy−3−4i| = 5 → |(x−3)+i(y−4)| = 5
(x−3)² + (y−4)² = 25
(iii) Distance from origin to centre = √(9+16) = √25 = 5.
Max |z| = distance to centre + radius = 5 + 5 =
10
(achieved at z = 6+8i, the point on the circle furthest from origin)
Question 5 — cos⁶θ reduction (z+z⁻¹ method)
[6 marks]
Use the substitution z = e^(iθ) to show that cos⁶θ = (1/32)(cos6θ + 6cos4θ + 15cos2θ + 10).
2cosθ = z+z⁻¹, so 2⁶cos⁶θ = (z+z⁻¹)⁶. Expand using binomial coefficients ₆C₀ through ₆C₆ and pair zᵏ+z⁻ᵏ = 2cos kθ.
✓ Solution
64cos⁶θ = (z+z⁻¹)⁶= z⁶ + 6z⁴ + 15z² + 20 + 15z⁻² + 6z⁻⁴ + z⁻⁶
= (z⁶+z⁻⁶) + 6(z⁴+z⁻⁴) + 15(z²+z⁻²) + 20
= 2cos6θ + 6·2cos4θ + 15·2cos2θ + 20
= 2cos6θ + 12cos4θ + 30cos2θ + 20
Divide by 64:
cos⁶θ = (1/32)(cos6θ + 6cos4θ + 15cos2θ + 10) ✓
Interactive Complex Number Tool
Enter a complex number and explore its polar form, powers, and roots — with a live Argand diagram.
Complex Number Explorer
Formula Sheet — Complex Numbers
Modulus-Argument Form
z = x+iyr(cosθ+i sinθ)
|z| = r√(x²+y²)
arg(z) = θUse quadrant rules
x = r cosθy = r sinθ
Operations in Polar
z₁z₂r₁r₂ · cis(θ₁+θ₂)
z₁/z₂(r₁/r₂) · cis(θ₁−θ₂)
z*r · cis(−θ)
|z|² = z·z*x²+y²
De Moivre's Theorem
[r·cisθ]ⁿrⁿ·cis(nθ)
z+z⁻¹2cosθ
z−z⁻¹2i sinθ
zⁿ+z⁻ⁿ2cos nθ
zⁿ−z⁻ⁿ2i sin nθ
Roots of zⁿ = w
Write w = r·cis α—
zₖ = r^(1/n) · cis((α+2πk)/n)—
k = 0,1,…,n−1n roots total
Roots of unity sum= 0 (n≥2)
Euler's Formula
e^(iθ)cosθ + i sinθ
e^(iπ) + 1= 0
cosθ(e^(iθ)+e^(−iθ))/2
sinθ(e^(iθ)−e^(−iθ))/2i
Loci
|z−a| = rCircle, centre a, radius r
|z−a| = |z−b|Perp bisector of ab
arg(z−a) = θHalf-line from a
|z−a|/|z−b| = k≠1Apollonius circle
📋 Cambridge Exam Strategy — Complex Numbers
- Argument calculation: Always state which quadrant first, then compute arctan, then adjust. Never quote arctan(y/x) without this check.
- Powers: Convert to polar, apply de Moivre, convert back. Never attempt to expand (a+bi)ⁿ directly for n ≥ 3.
- Multiple angle formulae: Expand (cosθ+i sinθ)ⁿ by binomial, equate real part for cos nθ, imaginary for sin nθ. Then use sin²θ = 1−cos²θ to reduce.
- z+z⁻¹ trick: For cosⁿθ use (z+z⁻¹)ⁿ = 2ⁿcosⁿθ. For sinⁿθ use (z−z⁻¹)ⁿ = (2i)ⁿsinⁿθ. Always divide correctly by 2ⁿ or (2i)ⁿ at the end.
- Roots: Write the RHS in general form r·cis(α+2πk) before dividing by n. State explicitly how many roots and give all of them.
- Loci: Identify the type from the algebraic form, state centre/radius or endpoints, draw and shade correctly. Solid boundary for ≤, dashed for <.