📚 Exercise 6G.1: Mastering De Moivre's Theorem for IB Math | 练习 6G.1：掌握 IB 数学中的棣莫弗定理
Exercise 6G.1 in the IB Mathematics curriculum typically focuses on De Moivre's theorem, a powerful tool for working with complex numbers in polar form. Whether you are studying Analysis and Approaches HL or Applications and Interpretation HL, mastering this theorem enables you to raise complex numbers to integer powers, find multiple roots, and prove trigonometric identities with elegance. This article unpacks the essential theory, step-by-step methods, and exam-style problems that match the demands of Exercise 6G.1.
IB 数学课程中的练习 6G.1 通常围绕棣莫弗定理展开，这是在极坐标形式下处理复数的一大利器。无论你学的是分析与方法 (HL) 还是应用与解释 (HL)，掌握这一定理都能让你轻松计算复数的整数次幂、求多重复根以及优雅地证明三角恒等式。本文将逐一讲解核心理论、分步解题方法以及匹配练习 6G.1 要求的真题风格问题。
1. Complex Numbers in Polar Form | 复数的极坐标形式
A complex number z = a + b i can be represented in the complex plane by rectangular coordinates (a, b). Its polar form uses the modulus r and argument θ: z = r (cos θ + i sin θ). The modulus is the distance from the origin, r = √a² + b², and the argument is the directed angle from the positive real axis.
复数 z = a + b i 可以在复平面上用直角坐标 (a, b) 表示。其极坐标形式用模 r 和辐角 θ 表示：z = r (cos θ + i sin θ)。模是到原点的距离，r = √a² + b²，辐角是从正实轴出发的有向角度。
When working with IB Exercise 6G.1, you must be able to switch quickly between forms. For example, the number 1 + i√3 has a = 1, b = √3, giving r = √1² + 3 = 2 and θ = arctan(√3/1) = π/3. Thus, 1 + i√3 = 2(cos π/3 + i sin π/3).
在解答 IB 练习 6G.1 时，你必须能迅速在两种形式间转换。例如，数字 1 + i√3 有 a = 1，b = √3，得到 r = √1² + 3 = 2，θ = arctan(√3/1) = π/3。因此，1 + i√3 = 2(cos π/3 + i sin π/3)。
2. Multiplication in Polar Form | 极坐标形式下的乘法
add their arguments. If z₁ = r₁ (cos θ₁ + i sin θ₁) and z₂ = r₂ (cos θ₂ + i sin θ₂), then z₁ z₂ = (r₁ r₂) [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)].
两个复数在极坐标形式下的乘法十分直接：将模相乘，辐角相加。如果 z₁ = r₁ (cos θ₁ + i sin θ₁)，z₂ = r₂ (cos θ₂ + i sin θ₂)，那么 z₁ z₂ = (r₁ r₂) [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。
This multiplication rule naturally extends to repeated multiplication, forming the basis of De Moivre's theorem. Understanding this geometric interpretation—scaling by r and rotating by θ—is crucial for visualizing complex powers.
这个乘法规则可以自然地扩展到连乘，从而构成棣莫弗定理的基础。理解这一几何解释——按 r 缩放并按 θ 旋转——对于直观理解复数的幂至关重要。
3. De Moivre's Theorem Statement | 棣莫弗定理的陈述
De Moivre's theorem states that for any integer n, (cos θ + i sin θ)n = cos(nθ) + i sin(nθ). When combined with the modulus, we get the full form: [r (cos θ + i sin θ)]n = rn (cos nθ + i sin nθ).
棣莫弗定理指出，对任意整数 n，有 (cos θ + i sin θ)n = cos(nθ) + i sin(nθ)。当与模结合时，得到完整形式：[r (cos θ + i sin θ)]n = rn (cos nθ + i sin nθ)。
This theorem is valid for all real θ and any integer n. In IB exams, n is usually a positive integer, but the theorem also holds for negative integers, which is sometimes tested in Exercise 6G.1 when simplifying reciprocals or rational powers.
该定理对所有实数 θ 和任意整数 n 均成立。IB 考试中，n 通常为正整数，但定理对负整数同样成立，在练习 6G.1 中有时会考查到，例如用于简化倒数或有理数幂。
4. Converting to Polar Form Before Applying | 应用前先转化为极坐标形式
A typical Exercise 6G.1 problem gives a complex number in rectangular form and asks for, say, z⁵. The first step is always to find r and θ. Write a + b i as r (cos θ + i sin θ), ensuring you place the angle in the correct quadrant using arctan(b/a) and adjusting for quadrant.
一道典型的练习 6G.1 题目会给出直角坐标形式的复数，然后要求计算例如 z⁵。第一步永远是求出 r 和 θ。将 a + b i 写成 r (cos θ + i sin θ)，务必根据 arctan(b/a) 并调整象限使辐角落在正确象限。
Always double-check the quadrant. For example, -1 - i has a = -1, b = -1, r = √2 θ = arctan(1) adjusted to the third quadrant, giving θ = -3π/4 or 5π/4. The polar form is √2 (cos 5π/4 + i sin 5π/4).
务必反复检查象限。例如，-1 - i 的 a = -1，b = -1，r = √2，θ = arctan(1) 调整到第三象限，得到 θ = -3π/4 或 5π/4。极坐标形式为 √2 (cos 5π/4 + i sin 5π/4)。
5. Applying De Moivre's Theorem for Integer Powers | 利用棣莫弗定理求整数次幂
1, you often need to express the final answer back in the form a + b i using exact values of trigonometric functions.
一旦得到 z = r (cos θ + i sin θ)，求整数 n 次幂便直接得出：zⁿ = rⁿ (cos nθ + i sin nθ)。在练习 6G.1 中，常常需要利用三角函数的精确值将最终答案表示回 a + b i 的形式。
Example: Find (1 + i√3)⁶.
r = 2, θ = π/3. Then (1 + i√3)⁶ = 2⁶ [cos(6 × π/3) + i sin(6 × π/3)] = 64 (cos 2π + i sin 2π) = 64 (1 + 0 i) = 64.
例：求 (1 + i√3)⁶。
r = 2, θ = π/3。则 (1 + i√3)⁶ = 2⁶ [cos(6 × π/3) + i sin(6 × π/3)] = 64 (cos 2π + i sin 2π) = 64 (1 + 0 i) = 64。
6. Negative and Fractional Powers (Introduction to Roots) | 负整数次幂与分数次幂（引入求根）
De Moivre' but appears in simplification of expressions like (cos θ + i sin θ)⁻² = cos 2θ - i sin 2θ.
棣莫弗定理可扩展到负整数：z⁻ⁿ = 1/zⁿ = r⁻ⁿ (cos(-nθ) + i sin(-nθ))。在练习 6G.1 中，这种情况较少见，但在简化表达式如 (cos θ + i sin θ)⁻² = cos 2θ - i sin 2θ 时会出现。
r non-integer exponents, but the idea naturally leads to finding complex roots using z = r1/n π)/n)], with k = 0, 1, ..., n-1. Exercise 6G.1 often includes cube roots and square roots.
对于非整数指数，定理不能直接应用，但其思路自然引向求复数的根：z = r1/n [cos((θ+2kπ)/n) + i sin((θ+2kπ)/n)]，其中 k = 0, 1, ..., n-1。练习 6G.1 常常包括立方根和平方根。
7. Using De Moivre to Derive Trigonometric Identities | 用棣莫弗定理推导三角恒等式
sin θ - 3 cos θ sin²θ - i sin³θ. Equating the real part gives cos 3θ = 4 cos³θ - 3 cos θ.
一个经典应用是通过二项式定理展开 (cos θ + i sin θ)ⁿ，并令实部与虚部分别相等，从而推导出 cos(nθ) 和 sin(nθ) 的公式。例如 n = 3 时：(cos θ + i sin θ)³ = cos³θ + 3 i cos²θ sin θ - 3 cos θ sin²θ - i sin³θ。令实部相等得 cos 3θ = 4 cos³θ - 3 cos θ。
Exercise 6G.1 may ask you to prove an identity or find sin 3θ in terms of sin θ and cos θ. The symmetry of the binomial expansion and the alternating signs of i make this a favourite exam topic.
练习 6G.1 可能会要求证明某个恒等式，或求用 sin θ 和 cos θ 表示 sin 3θ。二项式展开的对称性与 i 的符号交替变化使这成为考卷上的热门题目。
8. Finding nth Roots of a Complex Number | 求复数的 n 次方根
= w, write w in polar form w = R (cos φ + i sin φ). Then the n distinct roots are given by zₖ = R1/n [cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)] for k = 0, 1, ..., n-1. These roots are equally spaced on a circle of radius R1/n in the complex plane.
为求解 zⁿ = w，先将 w 写成极坐标形式 w = R (cos φ + i sin φ)。则 n 个不同的根由 zₖ = R1/n [cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)] 给出，其中 k = 0, 1, ..., n-1。这些根在复平面上均匀分布在半径为 R1/n 的圆上。
A typical Exercise 6G.1 problem: Find the cube roots of 8 (cos 2π/3 + i sin 2π/3). The modulus of each root is 81/3 2π/3 + 2kπ)/3 for k = 0,1,2. Thus the roots are 2 cis (2π/9), 2 cis (8π/9), 2 cis (14π/9).
一道典型的练习 6G.1 题目：求 8 (cos 2π/3 + i sin 2π/3) 的立方根。每个根的模为 81/3 = 2，辐角为 (2π/3 + 2kπ)/3，k = 0,1,2。因此根为 2 cis (2π/9), 2 cis (8π/9), 2 cis (14π/9)。
9. Solving Equations with Complex Coefficients | 解含有复系数的方程
= -8 means you must express -8 as 8 (cos π + i sin π) or 8 cis π before taking cube roots.
练习 6G.1 有时涉及 z³ + 8 = 0 或 z⁴ - 16 = 0 等方程。求解时先将幂独立出来，并将两边视为复数。例如 z³ = -8，须先将 -8 表示为 8 (cos π + i sin π) 或 8 cis π，再求立方根。
π, so the roots are 2 cis (π/3), 2 cis (π), and 2 cis (5π/3). Always remember to list all n solutions, unless the question restricts the range of arguments.
对于 z³ = -8，模 r = 8，辐角为 π，因此根为 2 cis (π/3)、2 cis (π) 以及 2 cis (5π/3)。务必记住列出所有 n 个解，除非题目对辐角范围有额外限制。
10. Common Pitfalls and How to Avoid Them | 常见错误与防范
Forgetting to consider the quadrant: Using arctan(b/a) without quadrant adjustment gives the wrong argument, leading to completely wrong powers or roots. Always sketch the complex number on the Argand diagram.
忘记考虑象限： 仅使用 arctan(b/a) 而不进行象限调整会得出错误的辐角，导致幂或根完全错误。每次都要在阿尔冈图上画出该复数。
Modulus errors: Raising r to the power n is straightforward, but when taking roots, students often forget to take the nth root of r (R1/n, not R/n). Double-check your arithmetic.
模运算错误： 对 r 乘 n 次方很简单，但求根时学生常常忘记是对 r 开 n 次方（即 R1/n，而非 R/n）。务必仔细核对运算。
Missing solutions: When finding nth roots, remember there are exactly n distinct roots. In an IB exam, leaving out one root can cost you several marks.
遗漏解： 求 n 次方根时，切记正好有 n 个不同的根。IB 考卷上漏掉一个根可能导致丢掉好几档分数。
11. IB Exam-Style Problems and Solutions Walkthrough | IB 真题风格演练
Problem 1: Given z = -1 + √3 i, find z⁵ and express your answer in the form a + b i.
第 1 题： 已知 z = -1 + √3 i，求 z⁵ 并以 a + b i 的形式表达答案。
Step 1: Modulus r = √((-1)² + (√3)²) = √1+3 = 2. Argument θ: arctan(-√3 ut point is in second quadrant, so θ = 2π/3 (or 120°). Thus z = 2 (cos 2π/3 + i sin 2π/3).
步骤 1：模 r = √((-1)² + (√3)²) = √1+3 = 2。辐角 θ：点位于第二象限，因此 θ = 2π/3。故 z = 2 (cos 2π/3 + i sin 2π/3)。
/3). Reduce argument: 10π/3 - 2π = 4π/3. So z⁵ = 32 (cos 4π/3 + i sin 4π/3) = 32 (-1/2 - i √3/2) = -16 - 16√3 i.
步骤 2：z⁵ = 2⁵ [cos(10π/3) + i sin(10π/3)] = 32 (cos 10π/3 + i sin 10π/3)。辐角化简：10π/3 - 2π = 4π/3。因此 z⁵ = 32 (cos 4π/3 + i sin 4π/3) = 32 (-1/2 - i √3/2) = -16 - 16√3 i。
Problem 2: Find all complex numbers z such that z⁴ = 1 - i. Give answers in exact cartesian form.
第 2 题： 找出所有满足 z⁴ = 1 - i 的复数 z。以精确直角坐标形式给出答案。
Write 1 - i in polar form. Modulus R = √(1²+(-1)²) = √2. Argument φ = arctan(-1/1) = -π/4 (or 7π/4). So w = √2 (cos(7π/4) + i sin(7π/4)).
将 1 - i 写成极坐标形式。模 R = √2，辐角 φ = -π/4（或 7π/4）。所以 w = √2 (cos(7π/4) + i sin(7π/4))。
Then z = (√2)1/4 [cos((7π/4 + 2kπ)/4) + i sin((7π/4 + 2kπ)/4)], k = 0,1,2,3. (√2)1/4 = 21/8 3: 31π/16. Express each root using exact trig values; these are left as expressions like 21/8 (cos 7π/16 + i sin 7π/16), etc.
则 z = (√2)1/4 [cos((7π/4 + 2kπ)/4) + i sin((7π/4 + 2kπ)/4)]，k = 0,1,2,3。(√2)1/4 = 21/8。辐角分别为：k=0：7π/16；k=1：15π/16；k=2：23π/16；k=3：31π/16。每个根用精确三角函数值表达，如 21/8 (cos 7π/16 + i sin 7π/16) 等。
12. Conclusion and Final Tips for Exercise 6G.1 | 总结与练习 6G.1 终极技巧
De Moivre's theorem is a fundamental tool in the IB Math HL toolkit. Approaching Exercise 6G.1 with a clear method—convert to polar form, apply the theorem, simplify the angle, and convert back to rectangular form—will make even complex-looking problems manageable. Practice identifying when the question requires the theorem, not just direct algebraic expansion.
棣莫弗定理是 IB 数学 HL 工具箱中的基本工具。用清晰的方法应对练习 6G.1——转化为极坐标形式、应用定理、化简角度、再转回直角坐标形式——即使看似复杂的题目也能迎刃而解。多加练习，学会识别何时需要定理，而非直接进行代数展开。
Always check that your final argument lies within a principal range if specified, and never forget the cyclic nature of roots. Mastering these skills from Exercise 6G.1 will solidify your understanding and prepare you for higher-level exam questions.
如果题目指定了主值范围，务必检查最终辐角是否落在该区间内，并永远不要忘记根的周期性。掌握练习 6G.1 的这些技巧将巩固你的理解，为应对更高级别的考题做好充分准备。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图