🌀 Complex Numbers
From AC circuit phasors to control-system stability, complex numbers let engineers handle two-dimensional quantities — magnitude and phase — with simple algebra. Work through each tab: rules first, then the interactive tools, then worked examples and practice.
3Topics
12Worked Examples
4Eng. Cases
12Practice Qs
➕Rectangular Forma + jb, Argand diagram
📐Polar & Exponentialr∠θ, re^(jθ)
🔁De Moivre's TheoremPowers & roots
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Represent complex numbers in rectangular (a + jb), polar (r∠θ) and exponential (rejθ) forms; perform the four operations; plot and interpret Argand diagrams; apply De Moivre's theorem to powers and roots; and use complex numbers to analyse AC circuits.
💡 Engineers write j = √−1 (not i) because i is reserved for electric current. Everything else works exactly like the maths you already know.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
➕
Rectangular Form & Operations
A complex number as a point on a 2D plane: a real part and an imaginary part.
🔧Complex Calculator — the four operationsEnter two numbers, choose an operation
Enter z₁ and z₂, choose an operation, then press Calculate to see the full working…
📖 Key Vocabulary
Imaginary unit (j) — defined by j² = −1; engineers use j to avoid clashing with current i.
Real part, Re(z) — the a in a + jb; imaginary part, Im(z) — the b (itself a real number).
Complex conjugate (z̄) — a − jb; multiplying z × z̄ always gives a real number.
Argand diagram — a plane with real numbers on the horizontal axis and imaginary numbers on the vertical axis.
👁️ Visual aid: a complex number is a vector on the Argand plane — its horizontal shadow is Re(z), its vertical shadow is Im(z).
✏️Worked Example 1 — Addition and subtractionType: Add/Subtract
Question If z₁ = 5 + j3 and z₂ = 2 − j4, find z₁ + z₂ and z₁ − z₂.
Add
(5+2) + j(3+(−4)) = 7 − j1Subtract
(5−2) + j(3−(−4)) = 3 + j7Answer Sum = 7 − j, Difference = 3 + j7 — combine real and imaginary parts separately, exactly like vectors.
✖️Worked Example 2 — Multiplication (FOIL with j² = −1)Type: Multiply
Question Find (3 + j2)(1 − j4).
Expand
3(1) + 3(−j4) + j2(1) + j2(−j4)Simplify
= 3 − j12 + j2 − j²8 = 3 − j10 − (−1)(8)Answer
= 3 + 8 − j10 = 11 − j10➗Worked Example 3 — Division using the conjugateType: Divide
Question Find (4 + j3) ÷ (2 − j1).
Multiply top & bottom by conjugate
(4+j3)(2+j1) ÷ (2−j1)(2+j1)Denominator
2² + 1² = 5 (real, as guaranteed)Numerator
8 + j4 + j6 + j²3 = 8 − 3 + j10 = 5 + j10Answer
(5 + j10) ÷ 5 = 1 + j2⚙️Worked Example 4 — Engineering context (impedance)Type: Engineering
Question A resistor (5 Ω) and inductor (j3 Ω reactance) are in series with a capacitor (−j7 Ω reactance). Find the total impedance Z.
Model Series impedances simply add:
Z = 5 + j3 + (−j7)Combine
Z = 5 + j(3 − 7) = 5 − j4Answer Z = (5 − j4) Ω — the negative imaginary part shows the circuit is net capacitive at this frequency.
🧠 Common slip: forgetting j² = −1 mid-calculation. Always finish multiplying first, then replace every j² with −1 as a separate clean-up step.
📐
Polar & Exponential Form
The same number, described by distance from the origin and angle turned — perfect for multiplying and dividing.
🔧Form Converter — rectangular ⇄ polarEnter a and b, then press Convert
Enter a and b, then press Convert to see the modulus, argument, and all three forms of the same number…
📖 Key Vocabulary
Modulus |z| — the length r of the vector; always positive.
Argument arg(z) — the angle θ from the positive real axis, usually given as −180° < θ ≤ 180°.
Euler's formula — ejθ = cos θ + j sin θ, the bridge between exponential and rectangular form.
Phasor — an engineering complex number representing amplitude and phase of a sinusoidal signal.
👁️ Visual aid: r is how far you travel from the origin; θ is the angle you turn through to get there.
✏️Worked Example 1 — Rectangular → PolarType: Conversion
Question Convert z = 3 + j4 to polar form.
Modulus
r = √(3² + 4²) = √25 = 5Argument
θ = tan⁻¹(4/3) = 53.13° (first quadrant, no adjustment needed)Answer
z = 5∠53.13° = 5ej53.13°🧭Worked Example 2 — Handling a negative quadrantType: Quadrant care
Question Convert z = −5 + j2 to polar form.
Modulus
r = √(25+4) = √29 = 5.385Raw angle
tan⁻¹(2/−5) gives −21.8°, but a is negative, b is positive → 2nd quadrantAdjust
θ = 180° − 21.8° = 158.2°Answer
z = 5.385∠158.2° — always sketch the quadrant before trusting a calculator's raw tan⁻¹.🔄Worked Example 3 — Polar → RectangularType: Conversion back
Question Convert z = 10∠30° to rectangular form.
Real part
a = r cos θ = 10 cos 30° = 8.660Imaginary part
b = r sin θ = 10 sin 30° = 5.000Answer
z = 8.660 + j5.000⚙️Worked Example 4 — Engineering context (multiplying phasors)Type: Engineering
Question A voltage phasor V = 240∠0° V is applied to an impedance Z = 15∠36.87° Ω. Find the current I = V/Z.
Rule In polar form, divide moduli and subtract arguments:
I = (240/15) ∠(0° − 36.87°)Compute
I = 16∠−36.87° AAnswer Current lags the voltage by 36.87° — polar form turns circuit phase shifts into simple subtraction.
🧠 Common slip: using the calculator's tan⁻¹ blindly. It only returns −90° to 90°; you must inspect the signs of a and b to place θ in the correct quadrant.
🔁
De Moivre's Theorem & Roots
Raising a complex number to a power — or splitting it into roots — becomes simple multiplication of angles.
🔧Power & Root FinderEnter r, θ and n, choose Power or Roots
Enter r, θ and n, choose Power or Roots, then press Calculate…
📖 Key Vocabulary
De Moivre's theorem — (cos θ + j sin θ)ⁿ = cos(nθ) + j sin(nθ).
nth root — a number which, raised to the power n, returns the original z.
Principal root — the root corresponding to k = 0, usually with the smallest positive argument.
Roots of unity — the special case z = 1: n roots evenly spaced around the unit circle.
👁️ Visual aid: every one of the n roots sits on the same circle, equally spaced by 360°/n.
✏️Worked Example 1 — Raising to a powerType: Power
Question Find (2∠30°)⁵.
Apply theorem
rⁿ = 2⁵ = 32, nθ = 5 × 30° = 150°Answer
32∠150° = 32(cos150° + j sin150°) = −27.71 + j16🧮Worked Example 2 — Power directly from rectangular formType: Convert-then-power
Question Find (1 + j)⁸ using De Moivre.
Convert
r = √2, θ = 45°Apply theorem
rⁿ = (√2)⁸ = 2⁴ = 16, nθ = 8 × 45° = 360°Answer
16∠360° = 16∠0° = 16 — a purely real result, since 360° wraps back to the start.🌐Worked Example 3 — Finding all cube rootsType: Roots
Question Find all cube roots of z = 8∠60°.
Root modulus
r1/3 = 81/3 = 2Root angles
θk = (60° + 360°k)/3 for k = 0, 1, 2Compute k=0: 20°; k=1: 140°; k=2: 260°
Answer Three roots:
2∠20°, 2∠140°, 2∠260° — same radius, 120° apart.⚙️Worked Example 4 — Engineering context (three-phase supply)Type: Engineering
Question A balanced three-phase supply has line voltages 400∠0°, 400∠−120° and 400∠−240° V. Confirm these are the cube roots of a single quantity and find it.
Check spacing Each phasor is 120° apart — matching the n = 3 root pattern exactly.
Find z Cube each:
(400∠0°)³ = 400³∠0° = 6.4×10⁷∠0°; the other two phasors cube to the same value (angles become multiples of 360°).Answer All three phasors are cube roots of
z = 6.4×10⁷∠0° — this is exactly why three-phase systems are modelled as roots of unity scaled by voltage.🧠 Common slip: forgetting to add 360°k before dividing by n when finding roots — without it you only ever find the one principal root, missing the other n − 1.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🔌 AC Impedance — Series R-L-C Circuit
Resistance, inductive reactance and capacitive reactance combine as a complex number: Z = R + j(X_L − X_C).
Z = R + j(X_L − X_C) Ω
📟 Try it — Circuit Impedance
Enter R, X_L and X_C, then press Calculate…
⚡ Power Triangle — Complex Power
Apparent power S = P + jQ combines real (working) power and reactive power into one complex quantity, whose modulus is what generators must actually supply.
|S| = √(P² + Q²) VA
📟 Try it — Apparent Power
Enter P and Q, then press Calculate…
🌀 Phasor Addition — Combining Two Signals
Two sinusoidal sources of the same frequency add as phasors: convert each to rectangular form, add, then convert the result back to polar.
V_total = V₁∠θ₁ + V₂∠θ₂ (add in rectangular, report in polar)
📟 Try it — Phasor Sum
Enter both phasors, then press Calculate…
🎛️ Control Systems — Pole Location & Stability
A system pole at s = σ + jω is stable only if σ < 0 (the real part is negative) — the pole sits in the left half of the complex s-plane.
Stable if Re(s) < 0; oscillation frequency = |Im(s)| rad/s
📟 Try it — Pole Stability Check
Enter σ and ω, then press Calculate…
🎮
Quick-Fire Complex Numbers Game
12 questions — forms, operations and De Moivre mixed. How many can you get?
⚡ Complex Numbers Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
If z₁ = 6 + j2 and z₂ = 1 − j5, find z₁ + z₂.
(6+1) + j(2−5) = 7 − j3
Question 2 🥉2 marks
Simplify j³ and j⁴.
j³ = j²·j = −1·j = −j
j⁴ = (j²)² = (−1)² = 1
j⁴ = (j²)² = (−1)² = 1
Question 3 🥉2 marks
Find the modulus and argument of z = 6 + j8.
r = √(36+64) = √100 = 10
θ = tan⁻¹(8/6) = 53.13°
θ = tan⁻¹(8/6) = 53.13°
Question 4 🥉2 marks
Convert z = 5∠90° to rectangular form.
a = 5cos90° = 0, b = 5sin90° = 5
z = 0 + j5
z = 0 + j5
Question 5 🥈3 marks
Multiply (2 + j5)(3 − j2).
6 − j4 + j15 − j²10 = 6 + 10 + j11
= 16 + j11
= 16 + j11
Question 6 🥈4 marks
Divide (6 + j8) ÷ (1 + j).
Multiply by conjugate (1−j)/(1−j):
Denom: 1+1 = 2
Numer: 6 − j6 + j8 − j²8 = 6+8 + j2 = 14 + j2
= (14+j2)/2 = 7 + j1
Denom: 1+1 = 2
Numer: 6 − j6 + j8 − j²8 = 6+8 + j2 = 14 + j2
= (14+j2)/2 = 7 + j1
Question 7 🥈3 marks
Convert z = −4 − j3 to polar form (state θ correctly for its quadrant).
r = √(16+9) = 5
Both a,b negative → 3rd quadrant
Reference angle tan⁻¹(3/4) = 36.87°
θ = 180° + 36.87° = 216.87° (or equivalently −143.13°)
Both a,b negative → 3rd quadrant
Reference angle tan⁻¹(3/4) = 36.87°
θ = 180° + 36.87° = 216.87° (or equivalently −143.13°)
Question 8 🥈4 marks
Find (1∠45°)⁴ using De Moivre's theorem.
rⁿ = 1⁴ = 1
nθ = 4×45° = 180°
1∠180° = −1
nθ = 4×45° = 180°
1∠180° = −1
Question 9 🥇4 marks
Find (3∠40°)³ and express the answer in rectangular form to 3 s.f.
rⁿ = 27, nθ = 120°
27∠120° = 27(cos120° + jsin120°)
= 27(−0.5) + j27(0.866)
= −13.5 + j23.4
27∠120° = 27(cos120° + jsin120°)
= 27(−0.5) + j27(0.866)
= −13.5 + j23.4
Question 10 🥇5 marks
Find all square roots of z = 16∠80°.
r1/2 = 4
θ_k = (80+360k)/2, k=0,1
k=0: 40°; k=1: 220°
4∠40° and 4∠220° (180° apart, as expected for square roots)
θ_k = (80+360k)/2, k=0,1
k=0: 40°; k=1: 220°
4∠40° and 4∠220° (180° apart, as expected for square roots)
Question 11 🥇5 marks
An impedance is Z = 12 + j16 Ω. A 240∠0° V supply is connected across it. Find the current I = V/Z in polar form.
Convert Z: r=√(144+256)=√400=20, θ=tan⁻¹(16/12)=53.13°
Z = 20∠53.13°
I = (240/20)∠(0−53.13°)
= 12∠−53.13° A
Z = 20∠53.13°
I = (240/20)∠(0−53.13°)
= 12∠−53.13° A
Question 12 🥇5 marks
Find all cube roots of unity (z = 1∠0°) and verify their sum is zero.
r1/3=1, θ_k=(0+360k)/3: 0°, 120°, 240°
Roots: 1, (−0.5+j0.866), (−0.5−j0.866)
Sum = 1 −0.5−0.5 + j(0.866−0.866) = 0 ✓
Roots: 1, (−0.5+j0.866), (−0.5−j0.866)
Sum = 1 −0.5−0.5 + j(0.866−0.866) = 0 ✓