Exercise 12B.4: Complex Numbers and De Moivre's Theorem | 练习 12B.4: 复数与棣莫弗定理

📚 Exercise 12B.4: Complex Numbers and De Moivre's Theorem | 练习 12B.4: 复数与棣莫弗定理

In this IB Mathematics analysis article we work through a typical Exercise 12B.4 problem involving complex numbers. We are given a quotient of two complex numbers, asked to find its modulus and argument, and then to use De Moivre's theorem to compute a high power of the result. The solution below consolidates key techniques: simplifying complex fractions, converting between Cartesian and polar forms, and applying De Moivre's theorem efficiently. The problem is representative of the algebraic manipulation and geometric reasoning expected in the IB HL syllabus.

在这篇 IB 数学分析文章中，我们讲解一道典型的练习 12B.4 复数题。题目给出两个复数的商，要求计算该复数的模和辐角，然后利用棣莫弗定理求出它的高次幂。下面的解答整合了关键技巧：化简复数分式、在直角坐标形式与极坐标形式之间转换，以及高效应用棣莫弗定理。这道题体现了 IB HL 课程大纲所要求的代数运算和几何推理能力。

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1. Problem Statement | 问题陈述

The exercise states: Let z = (1 + i√3) / (1 - i). Determine the modulus and argument of z, giving the argument in radians. Hence, or otherwise, evaluate z⁶ and express your answer in the form a + bi.

题目陈述：设 z = (1 + i√3) / (1 - i)。求 z 的模和辐角（以弧度表示）。由此，或用其他方法，计算 z⁶，并将答案表示为 a + bi 的形式。

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2. First Principles: Complex Number Forms | 基本原理：复数的表示形式

x + iy can be written in Cartesian form with real part x and imaginary part y. The modulus |w| is √x² + y² x, y). For a non-zero complex number, the polar form is w = r (cos θ + i sin θ), where r = |w| and θ = arg(w).

复数 w = x + iy 可以写成直角坐标形式，其中实部为 x，虚部为 y。模 |w| = √x² + y²，辐角 θ = arg(w) 是从正实轴到连接原点与点 (x, y) 的有向角。对于非零复数，极坐标形式为 w = r (cos θ + i sin θ)，其中 r = |w|，θ = arg(w)。

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3. Simplifying the Complex Quotient | 化简复数的商

To simplify z = (1 + i√3 minator becomes (1 - i)(1 + i) = 1² - (i)² = 1 - (-1) = 2. The numerator expands as (1 + i√3)(1 + i) = 1 + i + i√3 + i²√3 = 1 + i + i√3 - √3, since i² = -1. Grouping real and imaginary parts: real = 1 - √3, imaginary = (1 + √3).

为化简 z = (1 + i√3) / (1 - i)，将分子分母同乘以分母的共轭复数 (1 + i)。分母变为 (1 - i)(1 + i) = 1² - (i)² = 1 - (-1) = 2。分子展开为 (1 + i√3)(1 + i) = 1 + i + i√3 + i²√3 = 1 + i + i√3 - √3（因为 i² = -1）。合并实部和虚部：实部 = 1 - √3，虚部 = (1 + √3)。

Thus, after simplification we obtain the Cartesian form: z = (1 - √3)/2 + i(1 + √3)/2.

因此，化简后我们得到直角坐标形式：z = (1 - √3)/2 + i(1 + √3)/2。

z = ½(1 - √3) + i·½(1 + √3)

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4. Computing the Modulus | 计算模

the square root of the sum of squares of the real and imaginary components. Write x = (1 - √3)/2, y = (1 + √3)/2. Then x² + y² = [(1 - √3)² + (1 + √3)²] / 4. Expand: (1 - 2√3 + 3) + (1 + 2√3 + 3) = (4 - 2√3) + (4 + 2√3) = 8. So |z| = √8/4 = √2.

模等于实部与虚部平方和的平方根。令 x = (1 - √3)/2，y = (1 + √3)/2。那么 x² + y² = [(1 - √3)² + (1 + √3)²] / 4。展开得：(1 - 2√3 + 3) + (1 + 2√3 + 3) = (4 - 2√3) + (4 + 2√3) = 8。因此 |z| = √8/4 = √2。

|z| = √2

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5. Determining the Argument | 确定辐角

ument θ satisfies tan θ = y/x, provided we account for the correct quadrant. Here x = (1 - √3)/2 is negative, because 1 < √3 ≈ 1.732, and y = (1 + √3 e. Hence z lies in the second quadrant. We compute the reference angle α such that tan α = |y/x| = (1+√3)/(√3-1). Rationalise: multiply numerator and denominator by (√3+1): ((1+√3)(√3+1)) / ((√3-1)(√3+1)) = (√3 + 1 + 3 + √3) / (3 - 1) = (2√3 + 4)/2 = √3 + 2. This is not a standard tangent, so we instead recognise that (1+√3)/(√3 However, working directly with polar conversion from the unsimplified z is often smarter.

辐角 θ 满足 tan θ = y/x，但需确定正确象限。此处 x = (1 - √3)/2 为负（因为 1 < √3 ≈ 1.732），y = (1 + √3)/2 为正，因此 z 位于第二象限。我们计算参考角 α 使得 tan α = |y/x| = (1+√3)/(√3-1)。有理化：分子分母同乘 (√3+1)：((1+√3)(√3+1)) / ((√3-1)(√3+1)) = (√3 + 1 + 3 + √3) / (3 - 1) = (2√3 + 4)/2 = √3 + 2。这不是标准角正切值，所以更明智的做法是从未化简的 z 直接进行极坐标转换。

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6. Polar Conversion Directly from the Quotient | 直接从商式转换为极坐标形式

the modulus and argument of numerator and denominator separately. For numerator w₁ = 1 + i√3: modulus r₁ = √(1² + (√3)²) = √1+3 = 2. Its argument φ₁ = arctan(√3 nd imaginary parts are positive). For denominator w₂ = 1 - i: modulus r₂ = √(1² + (-1)²) = √2. Its argument φ₂ = arctan(-1/1) = -π/4. We use the division property: |z| = r₁/r₂ = 2/√2 = √2, which matches our earlier result. For the argument: arg(z) = φ₁ – φ₂ = π/3 – (-π/4) = π/3 + π/4 = 7π/12.

与从直角坐标形式出发不同，我们可以分别求出分子与分母的模和辐角。分子 w₁ = 1 + i√3：模 r₁ = √(1² + (√3)²) = √1+3 = 2；辐角 φ₁ = arctan(√3/1) = π/3（因为实部和虚部均为正）。分母 w₂ = 1 - i：模 r₂ = √(1² + (-1)²) = √2；辐角 φ₂ = arctan(-1/1) = -π/4。利用复数除法的性质：|z| = r₁/r₂ = 2/√2 = √2，与之前结果一致。辐角：arg(z) = φ₁ – φ₂ = π/3 – (-π/4) = π/3 + π/4 = 7π/12。

arg(z) = 7π/12

This is a much cleaner method and avoids messy algebra with the Cartesian form. Note that 7π/12 radians equals 105°, and it falls in the second quadrant as expected.

这种方法简洁得多，也避免了直角坐标形式下繁琐的代数运算。注意 7π/12 弧度等于 105°，正如预期位于第二象限。

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7. Expressing z in Polar Form | 将 z 表示为极坐标形式

With modulus and argument determined, the polar representation is z = √2 etric values if required. For completeness, cos(7π/12) = cos(π/3 + π/4) = cos(π/3)cos(π/4) – sin(π/3)sin(π/4) = (1/2)(√2/2) – (√3/2)(√2/2) = √2/4 – √6/4 = (√2 – √6)/4. Similarly, sin(7π/12) = sin(π/3 + π/4) = sin(π/3)cos(π/4) + cos(π/3)sin(π/4) = (√3/2)(√2/2) + (1/2)(√2/2) = √6/4 + √2/4 = (√6 + √2)/4. These match the Cartesian form after multiplying by √2, confirming consistency.

确定了模和辐角后，极坐标表示为 z = √2 [cos(7π/12) + i sin(7π/12)]。我们可以保留角度形式，但计算精确的三角函数值有时也很有用。完整起见，cos(7π/12) = cos(π/3 + π/4) = cos(π/3)cos(π/4) – sin(π/3)sin(π/4) = (1/2)(√2/2) – (√3/2)(√2/2) = √2/4 – √6/4 = (√2 – √6)/4。类似地，sin(7π/12) = sin(π/3 + π/4) = sin(π/3)cos(π/4) + cos(π/3)sin(π/4) = (√3/2)(√2/2) + (1/2)(√2/2) = √6/4 + √2/4 = (√6 + √2)/4。将它们乘以 √2 即与直角坐标形式吻合，验证了结果的一致性。

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8. Applying De Moivre's Theorem | 应用棣莫弗定理

De Moivre's theorem states that for any integer n, (r (cos θ + i sin θ))ⁿ = rⁿ (cos(nθ) + i sin(nθ)). We apply this with r = √2, θ = 7π/12, and n = 6. Then z⁶ = (√2)⁶ [cos(6 × 7π/12) + i sin(6 × 7π/12)]. Compute r⁶: (√2)⁶ = (2^1/2)⁶ = 2³ = 8. Compute the new argument: 6 × 7π/12 = 42π/12 = 7π/2.

棣莫弗定理指出，对于任意整数 n，有 (r (cos θ + i sin θ))ⁿ = rⁿ (cos(nθ) + i sin(nθ))。我们将 r = √2，θ = 7π/12，n = 6 代入。则 z⁶ = (√2)⁶ [cos(6 × 7π/12) + i sin(6 × 7π/12)]。计算 r⁶：(√2)⁶ = (2^1/2)⁶ = 2³ = 8。计算新的辐角：6 × 7π/12 = 42π/12 = 7π/2。

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9. Simplifying the Resulting Argument | 化简得到的辐角

π, so we subtract full revolutions of 2π to find a coterminal angle between 0 and 2π. 7π/2 – 2π = 7π/2 – ernatively, 3π/2 is equivalent to -π/2. Thus cos(3π/2) = 0 and sin(3π/2) = -1. Therefore z⁶ = 8 (0 + i(-1)) = -8i.

辐角 7π/2 大于 2π，因此我们减去整数倍的 2π（即 4π/2）得到 0 到 2π 之间的同界角：7π/2 – 4π/2 = 3π/2。3π/2 等同于 -π/2。于是 cos(3π/2) = 0，sin(3π/2) = -1。因此 z⁶ = 8 (0 + i(-1)) = -8i。

z⁶ = –8i

In Cartesian form this corresponds to a = 0, b = –8. The result is a pure imaginary number, which is an elegant simplification.

在直角坐标形式下，相当于 a = 0，b = –8。结果是一个纯虚数，这是一个优雅的简化。

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10. Verification Using an Alternative Method | 使用替代方法验证

e this is not the most efficient, it reassures. From Cartesian form z = x + iy with x = (1–√3)/2, y = (1+√3)/2, compute z² = (x² – y²) + i(2xy). With x² – y² = (x–y)(x+y). x+y = (1–√3 + 1+√3)/2 = 1. x–y = (1–√3 – 1 – √3)/2 = –√3. So z² = (1)(–√3) + i(2xy). Now 2xy = 2 × (1–√3)/2 × (1+√3)/2 = (1–3)/2 = –1. Therefore z² = –√3 – i. Cubing this gives z⁶ = (z²)³. It is manageable but the De Moivre route is clearly more systematic. The direct algebra confirms the final answer –8i.

我们可以通过对 z 进行三次平方来验证结果。尽管非最高效，但可增加确信度。从直角坐标形式 z = x + iy（x = (1–√3)/2，y = (1+√3)/2）出发，计算 z² = (x² – y²) + i(2xy)。注意到 x² – y² = (x–y)(x+y)。x+y = (1–√3 + 1+√3)/2 = 1，x–y = (1–√3 – 1 – √3)/2 = –√3。因此 z² = –√3 + i(2xy)。计算 2xy = 2 × (1–√3)/2 × (1+√3)/2 = (1–3)/2 = –1。于是 z² = –√3 – i。对其立方得 z⁶ = (z²)³，过程虽可行，但棣莫弗方法显然更系统。直接代数运算也证实了最终答案 –8i。

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11. Key Takeaways for IB Exam Questions | IB 考试要点

When faced with a fraction of complex numbers, immediately consider the conjugate method or the modulus-argument quotient rule. The latter often gives the argument more cleanly. Always check the quadrant of the resulting complex number and adjust the argument accordingly. De Moivre's theorem is a powerful tool for evaluating powers and should be accompanied by simplification of the angle to a principal value. Remember to express final answers in the requested form, typically a + bi.

遇到复数分式时，立即考虑使用共轭法或模-辐角除法规则。后者通常能更清晰地给出辐角。务必检查结果复数所在的象限，并相应调整辐角。棣莫弗定理是计算高次幂的强有力工具，同时应该将角度化简为主值。记住按题目要求表达最终答案，通常为 a + bi 形式。

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12. Common Pitfalls and Further Practice | 常见错误与拓展练习

ified. To deepen understanding, try similar exercises with different powers, such as z = (√3 + i)/(1 - i) and find z⁴. Linking polar forms to exponential form e^iθ can also enrich your toolkit.

常见错误之一是当分量含有根号时误判辐角象限，导致符号错误；另一常见错误是忘记取主辐角（通常为 (-π, π] 或 [0, 2π)），除非另有说明。为加深理解，可尝试不同幂次的类似练习，例如 z = (√3 + i)/(1 - i) 并求 z⁴。将极坐标形式与指数形式 e^iθ 关联起来也能丰富你的解题工具箱。

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