Complex numbers on the Argand diagram
Every complex number is a point and an arrow. Adding slides arrows end to end; multiplying multiplies the lengths and adds the angles.
读数
发生了什么
A complex number z = a + bi is the point (a, b) on the Argand diagram, or equally an arrow of length |z| = √(a² + b²) at angle arg z from the positive real axis. Adding complex numbers adds the arrows, so z + w is the far corner of a parallelogram. Multiplying is a rotation and a scaling: |zw| = |z||w| and arg zw = arg z + arg w, which is why the triangle 0, 1, z and the triangle 0, w, zw are similar, and why multiplying by i (length 1, angle 90°) is a quarter turn.
Dividing undoes it. Repeating the multiplication gives de Moivre's theorem: zⁿ has modulus |z|ⁿ and argument n arg z, so the powers spiral out (|z| > 1) or in (|z| < 1).
Going backwards, the n roots of z share the modulus |z|1/n and sit 2π/n apart, the corners of a regular polygon; for z = 1 they are the roots of unity, and they always add up to 0. The principal argument is the one in the interval (−π, π].
- Modulus-argument formz = a + bi = r(cos θ + i sin θ) = reiθ
- Multiplying|zw| = |z||w|, arg zw = arg z + arg w
- De Moivre's theoremzⁿ = rⁿ(cos nθ + i sin nθ)
- nth rootsz1/n = r1/n eiθ/n e2πik/n
考试大纲
- A-Level Further Maths AQA, Edexcel, OCR, OCR MEIArgand diagrams, modulus-argument form, loci, de Moivre's theorem and nth roots
- First-year university complex analysis
挑战
Predict first: z = 1 + i. What are the modulus and argument of z⁸? Use de Moivre, then switch to zⁿ, set n = 8 and check.
|z| = √2 and arg z = π/4, so z⁸ has modulus (√2)⁸ = 16 and argument 8 × π/4 = 2π, which is the same direction as 0. So z⁸ = 16, a real number.
系统地学
GCSE Maths 中讲这个主题的课：先看分步讲解，再做考试风格的题目，每次数字都不同。 课程为英文。
常见问题
How do I find the argument of a complex number?
Draw it on the Argand diagram and measure the angle from the positive real axis, anticlockwise positive. tan⁻¹(b/a) only gives the right answer in the first and fourth quadrants: in the second add π, in the third subtract π, so the principal argument lands in (−π, π].
Why does multiplying complex numbers add the arguments?
Write z = r(cos α + i sin α) and w = s(cos β + i sin β). Multiplying out and using the addition formulas gives zw = rs(cos(α + β) + i sin(α + β)). The lengths multiply and the angles add.
What are the roots of unity?
The n solutions of zⁿ = 1. They are e2πik/n for k = 0, 1, ..., n − 1, equally spaced round the unit circle as the corners of a regular n-sided polygon, and for n ≥ 2 they add up to 0.
What is the principal argument of −1, and of 0?
−1 lies on the negative real axis, so its principal argument is π (not −π, because the interval is (−π, π]). 0 has no argument at all, since it has no direction.
How do I find the cube roots of a complex number?
Write it in modulus-argument form and use de Moivre backwards. For 8i: |8i| = 8 and arg 8i = π/2, so every root has modulus 81/3 = 2 and the arguments are π/6, 5π/6 and −π/2 (adding 2π/3 each time). The roots are √3 + i, −√3 + i and −2i. Pick nth roots under Solve for to get every root with the working.