📚 Solving IB HL Exercise 14B.2: Powers of Complex Numbers with De Moivre | 用棣莫弗定理求解IB HL练习14B.2中的复数的幂
In IB Mathematics HL, Chapter 14 typically covers complex numbers, and Exercise 14B focuses on applying De Moivre's theorem to compute powers and roots. A typical problem, Exercise 14B.2, asks you to simplify expressions like (1 + i√3)⁶. This article will guide you through the core theory and provide a step-by-step solution, helping you tackle similar questions with confidence.
在 IB 数学 HL 中，第 14 章通常围绕复数展开，练习 14B 的重点是利用棣莫弗定理计算复数的幂与根。其中一道典型题目——练习 14B.2——要求你化简形如 (1 + i√3)⁶ 的表达式。本文将带你梳理核心理论并逐步求解，助你自信地应对同类问题。
1. Understanding the Question | 理解题目
Exercise 14B.2 in many IB textbooks presents you with a complex number like z = 1 + i√3 and asks you to compute z⁶, writing the result in Cartesian form a + bi. Direct binomial expansion would be lengthy and prone to error; De Moivre's theorem provides an elegant shortcut that is expected in the HL syllabus.
许多 IB 教材中的练习 14B.2 会给出一个如 z = 1 + i√3 的复数，要求你计算 z⁶ 并将结果写成笛卡尔形式 a + bi。直接使用二项式展开将极为冗长且容易出错；棣莫弗定理则提供了一条简洁优美的捷径，这也是 HL 大纲所要求的。
2. Modulus and Argument | 模与辐角
st step is to convert the complex number into modulus-argument (polar) form. For z = 1 + i√3:
第一步是将复数转换为模-辐角（极坐标）形式。对于 z = 1 + i√3：
- Modulus r = √(1² + (√3)²) = √1+3 = √4 = 2.
模 r = √(1² + (√3)²) = √1+3 = √4 = 2。 - Argument θ satisfies tan θ = (√3)/1 = √3 e both real and imaginary parts are positive, θ lies in the first quadrant: θ = π/3 (60°).
辐角 θ 满足 tan θ = (√3)/1 = √3。由于实部和虚部均为正，θ 位于第一象限：θ = π/3 (60°)。
Thus, the polar form is z = 2(cos(π/3) + i sin(π/3)).
因此，极坐标形式为 z = 2(cos(π/3) + i sin(π/3))。
3. De Moivre's Theorem | 棣莫弗定理
De Moivre's theorem is a powerful tool for raising complex numbers to integer powers. It states that for any integer n:
棣莫弗定理是将复数提升到整数次幂的强大工具。它指出，对于任意整数 n：
[r(cos θ + i sin θ)]ⁿ = rⁿ (cos(nθ) + i sin(nθ))
This formula works because multiplication of complex numbers in polar form multiplies moduli and adds arguments. Repeated multiplication naturally leads to exponentiation of the modulus and multiplication of the argument.
该公式之所以成立，是因为极坐标形式下复数的乘法会将模相乘、辐角相加。重复乘法自然导致模的指数运算与辐角的倍增。
4. Applying De Moivre to (1 + i√3)⁶ | 将棣莫弗定理应用于 (1 + i√3)⁶
Now substitute r = 2, θ = π/3, and n = 6 into the theorem:
现在将 r = 2，θ = π/3 和 n = 6 代入定理：
z⁶ = [2(cos(π/3) + i sin(π/3))]⁶
z⁶ = 2⁶ [cos(6 × π/3) + i sin(6 × π/3)]
z⁶ = 64 [cos(2π) + i sin(2π)]
We know cos(2π) = 1 and sin(2π) = 0, so:
我们知道 cos(2π) = 1 且 sin(2π) = 0，因此：
z⁶ = 64 (1 + i·0) = 64
The result is the real number 64, showing how a seemingly complex expression collapses nicely.
结果是实数 64，这表明一个看似复杂的表达式可以简洁地塌缩为实数。
5. Verification via Binomial Expansion (Optional) | 用二项式展开验证（可选）
For those who like to cross-check, expanding (1 + i√3)⁶ using the binomial theorem yields 64 after grouping real and imaginary parts and simplifying powers of i. This confirms the De Moivre result, though the algebra is significantly more laborious.
对于喜欢交叉验证的同学，使用二项式定理展开 (1 + i√3)⁶ 后，通过归类实部与虚部并化简 i 的幂次同样可得 64。这验证了棣莫弗的结果，尽管代数运算要繁琐得多。
6. Exponential Form Alternative | 指数形式的替代方法
Complex numbers can also be written in Euler's exponential form: z = r eiθ. Then:
复数也可写成欧拉指数形式：z = r eiθ。于是：
z⁶ = (2 eiπ/3)⁶ = 2⁶ ei·6·π/3 = 64 ei2π = 64(cos 2π + i sin 2π) = 64
This approach is even more compact and is fully accepted in IB exams. It highlights why many HL students prefer the exponential notation for powers and roots.
这种方法更加紧凑，且在 IB 考试中完全被接受。它解释了为什么许多 HL 学生更倾向于使用指数记法处理幂与根。
7. Extending to Complex Roots | 扩展到复数的根
De Moivre' arts of Exercise 14B often require roots. The general formula for the k-th root (k = 0, 1, …, n-1) is:
棣莫弗定理还用于求复数的所有 n 次根。虽然练习 14B.2 关注的是幂，但练习 14B 后续部分常要求求根。第 k 个根（k = 0, 1, …, n-1）的一般公式为：
zₖ = r1/n [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]
For instance, the sixth roots of 64 are equally spaced on a circle of radius 2, with arguments 0, π/3, 2π/3, π, 4π/3, and 5π/3. Understanding powers naturally leads to understanding roots.
例如，64 的六个六次方根均匀分布在半径为 2 的圆上，辐角分别为 0, π/3, 2π/3, π, 4π/3 和 5π/3。理解了幂，自然也会理解根。
8. Visualising on the Argand Diagram | 在阿根图上可视化
Plotting 1 + i√3 gives the point (1, √3 origin by a factor of 2⁶ = 64. The final point lands at (64, 0) on the positive real axis.
绘制 1 + i√3 得到点 (1, √3)，模长为 2，夹角 60°。将其升至 6 次幂
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