The imaginary unit i & complex arithmetic
The imaginary unit satisfies i² = −1, which is the single rule everything else follows from. Complex numbers a + bi add componentwise and multiply by expanding brackets and then replacing i² with −1. Powers of i cycle with period 4 — i, −1, −i, 1 — so any power reduces by taking the exponent mod 4.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Add and subtract componentwise(a + bi) + (c + di) = (a+c) + (b+d)i. Real parts with real, imaginary with imaginary.
- Multiply by expandingTreat i as a symbol, expand fully, then substitute i² = −1 and collect.
- Reduce powers of i mod 4i⁴ = 1, so i^n depends only on n mod 4: 1, i, −1, −i for remainders 0, 1, 2, 3.
- Present in the form a + biReal part first, and combine all the i terms into one coefficient.
Worked example
Compute (3 + 2i)(1 − 4i) and simplify i²⁷.
- Expand: 3(1) + 3(−4i) + 2i(1) + 2i(−4i) = 3 − 12i + 2i − 8i².
- Replace i² = −1: −8i² = +8.
- Collect: (3 + 8) + (−12 + 2)i = 11 − 10i.
- For i²⁷: 27 = 4(6) + 3, so i²⁷ = i³ = −i.
Answer. (3 + 2i)(1 − 4i) = 11 − 10i, and i²⁷ = −i.
Where marks get dropped
These are the specific errors that cost credit on the imaginary unit i & complex arithmetic questions — QED's rubric penalises each of them separately.
- Forgetting that i² = −1 turns a subtraction into an addition. The 2i × (−4i) term contributed +8, not −8.
- Writing √(−4) as −2. It is 2i, and the square root of a negative is always imaginary.
- Applying √a·√b = √(ab) to negatives. √(−1)·√(−1) = i² = −1, not √1 = 1 — the identity requires non-negative arguments.
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The imaginary unit i & complex arithmetic — frequently asked questions
Why do we need complex numbers?
So every polynomial has a root. x² + 1 = 0 has no real solution, and adjoining i produces an algebraically closed field where every polynomial factors completely.
Is i "imaginary" in a meaningful sense?
The name is historical and misleading. Complex numbers model rotations and oscillations concretely, and are indispensable in electrical engineering and quantum mechanics.
How do I simplify i to a large power?
Reduce the exponent mod 4. i^100 = i^0 = 1 since 100 is divisible by 4.
The rest of Complex Numbers
Arithmetic, the Argand plane, polar form and De Moivre. Each subtopic below has its own method, worked example and mark-losing traps.
- 1The imaginary unit i & complex arithmetic
- 2Conjugates & division
- 3Modulus, argument & the Argand diagram
- 4Polar & exponential form re^{iθ}
- 5Multiplying & dividing in polar form
- 6De Moivre’s theorem & powers
- 7nth roots & roots of unity
- 8Solving polynomial equations over ℂ
- 9The fundamental theorem of algebra & conjugate roots
- 10Euler’s formula & trigonometric identities
- 11Loci & regions in the complex plane
- 12Complex multiplication as rotation & scaling
- 13Applications: phasors & AC circuits
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