BTEC Level 3 · Further Engineering Maths Complex Numbers

🏠 Complex Numbers

Complex numbers let you handle quantities that ordinary real numbers can't — most famously, the square root of a negative number. In engineering they're not just an abstract curiosity: they're the standard language for describing AC circuits, oscillations, and anything with both a size and a direction (or phase) baked into one value.

j-notation & Argand Diagrams Rectangular Form Polar Form Multiplying & Dividing AC Circuit Applications

📋 The Big Picture

1
A complex number has a real part and an imaginary part: a + jb
2
j is defined so that j² = −1
3
Every complex number can be written two ways — rectangular (a + jb) or polar (r∠θ)
4
Rectangular is best for adding/subtracting; polar is best for multiplying/dividing
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Key Words You'll Meet Throughout

Real PartThe ordinary number part, a
Imaginary PartThe j-multiplied part, b
ModulusThe size/length of a complex number
ArgumentThe angle of a complex number
ConjugateSame real part, flipped-sign imaginary part

🗺️ How these topics connect

  • j-numbers & Argand diagrams — the notation and the picture: every complex number is a point (or an arrow) on a 2D plane.
  • Rectangular form — a + jb. Easiest form for addition, subtraction, and understanding what's "real" vs "imaginary" about a value.
  • Polar form — r∠θ. Same number, described instead by its distance from the origin and its angle.
  • Multiplying & dividing in polar form — this is where polar form earns its place: multiply moduli and add angles, instead of a messy four-term expansion.
  • Engineering applications — AC circuit impedance is where all of this becomes genuinely useful, not just theoretical.
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j-Numbers & Argand Diagrams

What j actually means, how its powers cycle, and how to picture a complex number as a point on a plane.

📋 j-Numbers — Quick Reference

1
j = √(−1), so j² = −1
2
Powers of j cycle every 4: j, −1, −j, 1, then repeat
3
On an Argand diagram: real part → x-axis, imaginary part → y-axis
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Key Words — What Do They Mean?

j-operatorRepresents √(−1); "i" in pure maths, "j" in engineering
Argand DiagramA graph plotting complex numbers

Why j exists

Ordinary (real) numbers can't produce a negative number when squared — so √(−9) has no answer among the real numbers. Engineers and mathematicians solve this by defining a new number, j, with the property:

j = √(−1)    so    j² = −1

Once j exists, √(−9) = √(9 × −1) = √9 × √(−1) = 3j. Any complex number is then written a + jb, where a is the real part and b is the imaginary part (both a and b are ordinary real numbers).

📌 In pure mathematics this number is usually called i. Engineers use j instead, because i is already used for electrical current!

Powers of j

Because j² = −1, the powers of j cycle through just four values, then repeat forever:

j¹ = j    j² = −1    j³ = −j    j⁴ = 1    j⁵ = j    …
✏️ Worked example
Problem: Find j¹¹.
Divide the power by 4: 11 ÷ 4 = 2 remainder 3
Use the remainder: j¹¹ = j³ = −j
Answer: j¹¹ = −j

The Argand diagram

A complex number a + jb can be plotted as a single point (a, b) — the real part along the horizontal axis, the imaginary part along the vertical axis. Often it's drawn as an arrow from the origin to that point, which becomes very useful once you get to polar form.

Re Im 4 + 3j

The point 4 + 3j, plotted 4 units along the real axis and 3 units up the imaginary axis

🔁
j Power Finder
n
Enter a power n (≥ 0), then click Find jⁿ.
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Rectangular Form Operations

Adding, subtracting, multiplying and dividing complex numbers written as a + jb — and the conjugate trick that makes division possible.

📋 Rectangular Form — Quick Reference

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Add/subtract: combine real with real, imaginary with imaginary
2
Multiply: expand normally, then replace j² with −1
3
Divide: multiply top and bottom by the conjugate of the denominator
4
Conjugate of a + jb is a − jb (flip the sign of the imaginary part)
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Key Words — What Do They Mean?

ConjugateSame real part, opposite-sign imaginary part
RationaliseRemove j from the denominator

Addition and subtraction

Treat the real parts and imaginary parts as two completely separate totals:

(a + jb) ± (c + jd) = (a ± c) + j(b ± d)
✏️ Worked example
Problem: Simplify (5 + 3j) + (2 − 7j).
Real parts: 5 + 2 = 7
Imaginary parts: 3 + (−7) = −4
Answer: = 7 − 4j

Multiplying by a real constant, and multiplying two complex numbers

Multiplying by a real number just scales both parts. Multiplying two complex numbers means expanding like any two brackets — but remember j² = −1:

(a + jb)(c + jd) = (ac − bd) + j(ad + bc)
✏️ Worked example
Problem: Simplify (3 + 2j)(1 − 4j).
Expand: 3(1) + 3(−4j) + 2j(1) + 2j(−4j)
Simplify each term: 3 − 12j + 2j − 8j²
Replace j² = −1: 3 − 12j + 2j − 8(−1) = 3 − 12j + 2j + 8
Answer: = 11 − 10j

Division — using the conjugate

You can't leave j in a denominator. The fix: multiply top and bottom by the conjugate of the denominator, which always turns the denominator into a real number, because (c + jd)(c − jd) = c² + d².

(a + jb) / (c + jd)   =   (a + jb)(c − jd) / (c² + d²)
✏️ Worked example
Problem: Simplify (4 + 3j) / (1 − 2j).
Multiply by the conjugate (1 + 2j) top and bottom: (4+3j)(1+2j) / (1−2j)(1+2j)
Expand the denominator: 1² + 2² = 5
Expand the numerator: 4 + 8j + 3j + 6j² = 4 + 11j − 6 = −2 + 11j
Divide through by 5: (−2 + 11j) / 5
Answer: = −0.4 + 2.2j
➕
Rectangular Form Calculator
a b (j) c d (j)
Enter (a+jb) and (c+jd), pick an operation, then click Calculate.
📐

Polar Form & Conversion

Describing the same complex number by its distance from the origin (modulus) and its angle (argument), and converting between the two forms.

📋 Polar Form — Quick Reference

1
Modulus: r = √(a² + b²)
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Argument: θ = tan⁻¹(b/a) — check the quadrant!
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Rectangular → Polar: a = r cos θ, b = r sin θ
4
Written as r∠θ or r(cos θ + j sin θ)
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Key Words — What Do They Mean?

ModulusThe distance from the origin (size)
ArgumentThe angle from the positive real axis
r∠θShorthand for polar form

Converting rectangular → polar

The modulus is just Pythagoras' theorem applied to the real and imaginary parts. The argument is the angle the arrow makes with the positive real axis:

r = √(a² + b²)     θ = tan⁻¹(b/a)
✏️ Worked example
Problem: Convert 4 + 3j to polar form.
Modulus: r = √(4² + 3²) = √(16+9) = √25
Argument: θ = tan⁻¹(3/4) = 36.87°
Answer: 4 + 3j = 5∠36.87°
📌 Watch the quadrant! A calculator's tan⁻¹ only gives answers between −90° and 90°. If a is negative, add 180° to the calculator's answer to get the correct angle in the 2nd or 3rd quadrant.

Converting polar → rectangular

a = r cos θ     b = r sin θ
✏️ Worked example
Problem: Convert 10∠53.13° to rectangular form.
Real part: a = 10 cos(53.13°) = 10 × 0.6 = 6
Imaginary part: b = 10 sin(53.13°) = 10 × 0.8 = 8
Answer: = 6 + 8j
➡️
Rectangular → Polar
a b (j)
Enter a and b, then click Convert.
⬅️
Polar → Rectangular
r θ (°)
Enter r and θ (degrees), then click Convert.
✖️

Multiplying & Dividing in Polar Form

This is where polar form really pays off — no expansion, no conjugates, just two simple rules.

📋 Polar Operations — Quick Reference

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Multiply: multiply the moduli, add the angles
2
Divide: divide the moduli, subtract the angles
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This is far faster than expanding in rectangular form
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Key Words — What Do They Mean?

De Moivre's Theorem(r∠θ)ⁿ = rⁿ∠(nθ)

Multiplying in polar form

r₁∠θ₁ × r₂∠θ₂ = (r₁r₂) ∠(θ₁ + θ₂)
✏️ Worked example
Problem: Multiply 5∠30° by 2∠45°.
Multiply moduli: 5 × 2 = 10
Add angles: 30° + 45° = 75°
Answer: = 10∠75°

Dividing in polar form

r₁∠θ₁ ÷ r₂∠θ₂ = (r₁/r₂) ∠(θ₁ − θ₂)
✏️ Worked example
Problem: Divide 12∠80° by 3∠20°.
Divide moduli: 12 ÷ 3 = 4
Subtract angles: 80° − 20° = 60°
Answer: = 4∠60°

De Moivre's theorem — raising to a power

Multiplying a complex number by itself n times, in polar form, is a direct extension of the multiplication rule:

(r∠θ)ⁿ = rⁿ ∠(nθ)
✏️ Worked example
Problem: Find (2∠15°)⁴.
Raise modulus to the power: 2⁴ = 16
Multiply angle by the power: 15° × 4 = 60°
Answer: = 16∠60°
✖️
Polar Multiply / Divide Calculator
r₁ θ₁ (°) r₂ θ₂ (°)
Enter r₁∠θ₁ and r₂∠θ₂, pick an operation, then click Calculate.
⚙️

Engineering Applications — AC Circuits

Complex numbers describe AC circuit behaviour precisely, because they naturally carry both a size and a phase (timing offset) in a single value.

📋 AC Circuits — Quick Reference

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Impedance Z = R + jX (resistance + reactance)
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Inductive reactance is +j; capacitive reactance is −j
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Ohm's law still applies: V = IZ (all as complex numbers)
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Key Words — What Do They Mean?

ImpedanceAC "resistance" — combines resistance and reactance
ReactanceOpposition to current from inductors/capacitors
Phase AngleTiming offset between voltage and current
Power Factorcos of the phase angle

Why AC circuits need complex numbers

In a DC circuit, resistance is just one real number. In an AC circuit, resistors, inductors and capacitors all oppose current differently — and inductors/capacitors also shift the timing between voltage and current, not just the size. A single complex number, impedance Z, captures both effects at once:

Z = R + jX

where R is resistance (real part) and X is reactance (imaginary part). Positive j means inductive (current lags voltage); negative j means capacitive (current leads voltage).

Combining impedances in series

✏️ Worked example
Problem: A resistor (R = 30 Ω) and an inductor (reactance X_L = 40 Ω) are in series. Find the total impedance in rectangular and polar form.
Rectangular form: Z = 30 + 40j Ω
Modulus: |Z| = √(30² + 40²) = √(900+1600) = √2500
Argument: θ = tan⁻¹(40/30) = 53.13°
Answer: Z = 50∠53.13° Ω

Using V = IZ

✏️ Worked example
Problem: A supply of 100∠0° V is connected across the impedance above (50∠53.13° Ω). Find the current.
Rearrange Ohm's law: I = V / Z
Divide in polar form: I = (100/50) ∠(0° − 53.13°)
Answer: I = 2∠−53.13° A

Notice the current's angle is negative relative to the voltage — this is exactly the "current lags voltage" behaviour expected from an inductive circuit, and the maths captures it automatically.

Power factor

The power factor tells you how much of the supplied power is actually doing useful work, versus being tied up in the reactive (timing) part of the circuit:

Power factor = cos θ   (θ = argument of Z)

For the circuit above, power factor = cos(53.13°) ≈ 0.6 — meaning a fair amount of the supply is reactive rather than useful, which is exactly why power factor correction matters in industrial electrical systems.

✏️

Practice

A mixed set of problems across all the topics — working from straightforward substitution through to multi-step engineering scenarios.

💡 Show your working for every problem. Click "Show answer" to check once you've had a genuine attempt — the value is in the attempt, not the answer.
Q1
Simplify j¹⁴.
14 ÷ 4 = 3 remainder 2, so j¹⁴ = j² = −1
Q2
Simplify (6 − 2j) − (3 + 5j).
Real: 6−3 = 3. Imaginary: −2−5 = −7. Answer: 3 − 7j
Q3
Simplify (2 + j)(3 − j).
Expand: 6 − 2j + 3j − j² = 6 + j − (−1) = 6 + j + 1 = 7 + j
Q4
Simplify (5 + 2j) ÷ (1 + j), giving your answer in the form a + jb.
Multiply by conjugate (1−j): numerator (5+2j)(1−j) = 5−5j+2j−2j² = 5−3j+2 = 7−3j. Denominator (1+j)(1−j) = 1+1 = 2. Answer: 3.5 − 1.5j
Q5
Find the modulus and argument of 6 + 8j.
r = √(6²+8²) = √100 = 10. θ = tan⁻¹(8/6) = 53.13°. Answer: 10∠53.13°
Q6
Convert 8∠120° to rectangular form.
a = 8cos(120°) = 8×(−0.5) = −4. b = 8sin(120°) = 8×0.866 = 6.93. Answer: −4 + 6.93j
Q7
Find −3 + 4j in polar form, taking care with the quadrant.
r = √(9+16) = 5. Calculator gives tan⁻¹(4/−3) = −53.13°, but a is negative so the point is in the 2nd quadrant: add 180° → 126.87°. Answer: 5∠126.87°
Q8
Multiply 7∠40° by 3∠−10°.
Moduli: 7×3 = 21. Angles: 40°+(−10°) = 30°. Answer: 21∠30°
Q9
Divide 20∠90° by 4∠30°.
Moduli: 20÷4 = 5. Angles: 90°−30° = 60°. Answer: 5∠60°
Q10
Use De Moivre's theorem to find (3∠20°)³.
Modulus: 3³ = 27. Angle: 20°×3 = 60°. Answer: 27∠60°
Q11
An impedance has resistance 8 Ω and capacitive reactance 6 Ω. Write Z in rectangular form, then convert to polar form.
Capacitive reactance is −j, so Z = 8 − 6j Ω. Modulus: √(64+36) = √100 = 10. Argument: tan⁻¹(−6/8) = −36.87°. Answer: Z = 10∠−36.87° Ω
Q12
A supply of 50∠0° V is applied across an impedance of 10∠−36.87° Ω (from Q11). Find the current, and state whether the power factor circuit is inductive or capacitive.
I = V/Z = (50/10)∠(0°−(−36.87°)) = 5∠36.87° A. The current's angle is positive relative to voltage, meaning current leads voltage — this confirms a capacitive circuit, consistent with the negative j reactance given.