📚 Exercise I.3: Complex Numbers in Polar Form | 练习 I.3：极坐标形式的复数
In IB Mathematics, moving beyond the simple Cartesian form z = a + bi unlocks powerful tools for multiplication, division, and finding powers and roots of complex numbers. Exercise I.3 focuses on the polar representation r(cos θ + i sin θ), where r is the modulus and θ is the argument. Mastering this form is essential for tackling De Moivre's theorem and understanding complex functions. This article walks through the key concepts and practice questions you will encounter in a typical Exercise I.3, providing step-by-step methods to build your confidence.
在 IB 数学中，超越简单的直角坐标形式 z = a + bi 可以为复数的乘法、除法、求幂和求根提供强大的工具。练习 I.3 重点在于极坐标表示 r(cos θ + i sin θ)，其中 r 是模，θ 是辐角。掌握这一形式对于应用棣莫弗定理和理解复变函数至关重要。本文将梳理典型练习 I.3 中的关键概念和练习题，提供分步方法，帮助你建立信心。
1. Cartesian vs. Polar Form | 直角坐标与极坐标形式对比
A complex number can be expressed as z = a + bi, where a and b are real numbers, and i is the imaginary unit with i² = – on, it becomes cumbersome for multiplication, division, and exponentiation. The polar form z = r(cos θ + i sin θ) represents the same point using its distance from the origin (modulus r) and the angle it makes with the positive real axis (argument θ). This geometric perspective simplifies many operations.
复数可以表示为 z = a + bi，其中 a 和 b 为实数，i 为虚数单位且 i² = –1。虽然这种直角坐标形式对于加法和减法很直观，但在乘法、除法和乘方时会变得繁琐。极坐标形式 z = r(cos θ + i sin θ) 则利用点到原点的距离（模 r）和与正实轴的夹角（辐角 θ）来表示同一个点。这种几何视角能简化许多运算。
2. Modulus and Argument | 模与辐角
Given z = a + bi, the modulus r is calculated as r = √a² + b². It is always a non-negative real number. The argument θ satisfies tan θ = b/a, but you must use the quadrant of the point (a, b) to determine the correct angle. Arguments are usually given in radians or degrees, with the principal argument often restricted to (–π, π] or [0, 2π). Always draw a quick sketch to avoid sign errors.
给定 z = a + bi，模 r 的计算公式为 r = √a² + b²。它始终是一个非负实数。辐角 θ 满足 tan θ = b/a，但必须根据点 (a, b) 所在的象限来确定正确的角度。辐角通常以弧度或度给出，主辐角常限定在 (–π, π] 或 [0, 2π)。务必画一个简图以避免符号错误。
3. Converting to Polar Form | 转换为极坐标形式
To convert z = a + bi into polar form, first compute r = √a² + b². Then find the principal argument θ. For example, if z = 1 + i, then r = √1² + 1² = √2, and θ = arctan(1/1) = π/4. Thus z = √2 (cos π/4 + i sin π/4). For z = –√3 + i, r = √3 + 1 = 2, and because a < 0, b > 0 the angle lies in the second quadrant: θ = π – arctan(1/√3) = π – π/6 = 5π/6. The polar form becomes z = 2 (cos 5π/6 + i sin 5π/6).
要将 z = a + bi 转换为极坐标形式，先计算 r = √a² + b²，再求出主辐角 θ。例如，若 z = 1 + i，则 r = √1² + 1² = √2，θ = arctan(1/1) = π/4，因此 z = √2 (cos π/4 + i sin π/4)。对于 z = –√3 + i，r = √3 + 1 = 2，由于 a < 0 且 b > 0，角在第二象限：θ = π – arctan(1/√3) = π – π/6 = 5π/6。极坐标形式为 z = 2 (cos 5π/6 + i sin 5π/6)。
4. Multiplication in Polar Form | 极坐标形式乘法
When two complex numbers are in polar form, their product is surprisingly simple: multiply their moduli and add their arguments. If z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), then z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]. This rule follows from the compound angle identities for sine and cosine. Exercise I.3 typically asks you to verify this with both Cartesian and polar methods.
当两个复数表示为极坐标形式时，它们的乘积出奇地简单：模相乘，辐角相加。若 z₁ = r₁(cos θ₁ + i sin θ₁) 且 z₂ = r₂(cos θ₂ + i sin θ₂)，则 z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。这一规律源自正弦和余弦的和角公式。练习 I.3 通常会要求你用直角坐标法和极坐标法进行验证。
Example: Multiply z₁ = 2(cos π/3 + i sin π/3) and z₂ = 3(cos π/6 + i sin π/6). Product modulus: 2 × 3 = 6. Sum of arguments: π/3 + π/6 = π/2. Result: 6(cos π/2 + i sin π/2) = 6i.
例子：将 z₁ = 2(cos π/3 + i sin π/3) 与 z₂ = 3(cos π/6 + i sin π/6) 相乘。模的乘积：2 × 3 = 6。辐角之和：π/3 + π/6 = π/2。结果为 6(cos π/2 + i sin π/2) = 6i。
5. Division in Polar Form | 极坐标形式除法
Division works in a complementary way: divide the moduli and subtract the arguments. For z₁ / z₂, the result is (r₁ / r₂)[cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)], provided r₂ ≠ 0. This makes division far more elegant than rationalizing the denominator in Cartesian form.
除法以互补的方式运作：模相除，辐角相减。对于 z₁ / z₂，结果为 (r₁ / r₂)[cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)]，前提是 r₂ ≠ 0。这使得除法远比在直角坐标形式中分母有理化要优雅得多。
When solving Exercise I.3 problems, be careful with the argument difference: you may need to add or subtract 2π to keep the result within the principal range.
在解答练习 I.3 的题目时，要注意辐角之差：可能需要加上或减去 2π 以使结果保持在主值范围内。
6. De Moivre's Theorem | 棣莫弗定理
De Moivre's theorem extends polar multiplication to integer powers. It states that for any integer n, [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ). This theorem is the cornerstone of computing high powers of complex numbers and is heavily tested in IB exams. A common Exercise I.3 task is proving the theorem by induction for positive integers n.
棣莫弗定理将极坐标乘法推广到整数次幂。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)。该定理是计算复数高次幂的基石，在 IB 考试中考查频繁。练习 I.3 中常见的一个任务是利用数学归纳法对正整数 n 证明该定理。
Proof sketch (induction): Base case n = 1 is trivial. Assume true for n = k, then for k+1 multiply the induction hypothesis by r(cos θ + i sin θ) using the multiplication rule to obtain rk+1[cos((k+1)θ) + i sin((k+1)θ)].
证明梗概（归纳法）：基础情形 n = 1 显然成立。假设对 n = k 成立，则对 k+1，将归纳假设乘以 r(cos θ + i sin θ) 并利用乘法法则，即得 rk+1[cos((k+1)θ) + i sin((k+1)θ)]。
7. Powers of Complex Numbers | 复数的乘方
Armed with De Moivre's theorem, you can quickly evaluate expressions like (1 + i)⁸. First convert 1 + i to polar form: √2(cos π/4 + i sin π/4). Then raise to the 8th power: (√2)⁸ (cos 2π + i sin 2π) = 16(1 + 0i) = 16. Without De Moivre, expanding binomially would be tedious.
借助棣莫弗定理，你可以快速计算如 (1 + i)⁸ 的表达式。首先将 1 + i 转换为极坐标形式：√2(cos π/4 + i sin π/4)。然后求 8 次幂：(√2)⁸ (cos 2π + i sin 2π) = 16(1 + 0i) = 16。若不用棣莫弗定理，二项式展开将非常繁琐。
Many questions in Exercise I.3 also combine powers with algebraic simplifications, such as showing that (cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ and then expanding the left-hand side using binomial theorem to derive multiple-angle identities for cos 4θ and sin 4θ.
练习 I.3 中的许多题目还将乘方与代数化简相结合，例如证明 (cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ，然后通过二项式定理展开左边，导出 cos 4θ 和 sin 4θ 的多倍角恒等式。
8. Roots of Complex Numbers | 复数的根
Extracting roots is the inverse of raising to a power. For z = r(cos θ + i sin θ), the nth roots are given by:
z1/n = r1/n ( cos( (θ + 2kπ)/n ) + i sin( (θ + 2kπ)/n ) ), k = 0, 1, 2, ..., n–1
There are exactly n distinct roots, equally spaced around a circle of radius r1/n in the complex plane. This symmetry is a beautiful geometric result frequently explored in Exercise I.3.
求根是乘方的逆运算。对于 z = r(cos θ + i sin θ)，n 次方根由下式给出：
z1/n = r1/n ( cos( (θ + 2kπ)/n ) + i sin( (θ + 2kπ)/n ) ), k = 0, 1, 2, ..., n–1
恰好有 n 个不同的根，它们在复平面上均匀分布在半径为 r1/n 的圆上。这种对称性是一种优美的几何结果，在练习 I.3 中常有探究。
9. Examples: Finding Cube Roots | 例题：求立方根
Consider finding the cube roots of –8. First, express –8 in polar form: –8 = 8(cos π + i sin π). Here r = 8, θ = π. For n = 3, the modulus of each root is ∛8 = 2. The angles are (π + 2kπ)/3 for k = 0, 1, 2:
- k = 0: 2(cos π/3 + i sin π/3) = 2(1/2 + i√3/2) = 1 + i√3
- k = 1: 2(cos π + i sin π) = 2(–1 + 0i) = –2
- k = 2: 2(cos 5π/3 + i sin 5π/3) = 2(1/2 – i√3/2) = 1 – i√3
These three roots form an equilateral triangle on the complex plane. In an IB exam, you might also be asked to verify that the sum of all nth roots is zero, an elegant property linked to symmetry.
考虑求 –8 的立方根。首先将 –8 表示为极坐标形式：–8 = 8(cos π + i sin π)。这里 r = 8，θ = π。对于 n = 3，每个根的模为 ∛8 = 2。角度为 (π + 2kπ)/3，k = 0, 1, 2：
- k = 0: 2(cos π/3 + i sin π/3) = 2(1/2 + i√3/2) = 1 + i√3
- k = 1: 2(cos π + i sin π) = 2(–1 + 0i) = –2
- k = 2: 2(cos 5π/3 + i sin 5π/3) = 2(1/2 – i√3/2) = 1 – i√3
这三个根在复平面上构成一个等边三角形。在 IB 考试中，你可能还会被要求验证所有 n 次方根之和为零，这是与对称性相关的一个优雅性质。
10. Exponential Form and Euler's Formula | 指数形式与欧拉公式
IB syllabi often introduce Euler's formula eiθ = cos θ + i sin θ. This leads to the compact exponential form z = reiθ. Multiplication, division, and De Moivre's theorem become even more intuitive: (reiθ)ⁿ = rⁿ ei nθ. Many Exercise I.3 problem sets conclude with exercises that ask you to switch between polar, Cartesian, and exponential forms with ease.
IB 教学大纲通常会引入欧拉公式 eiθ = cos θ + i sin θ。由此得到简洁的指数形式 z = reiθ。乘法、除法和棣莫弗定理变得更加直观：(reiθ)ⁿ = rⁿ ei nθ。许多练习 I.3 的题组以要求你自如地在极坐标、直角坐标和指数形式之间切换的题目收尾。
A typical question: Express 3eiπ/4 in Cartesian form. Solution: 3(cos π/4 + i sin π/4) = 3(√2/2 + i√2/2) = (3√2/2) + i(3√2/2). The exponential form also elegantly explains why multiplying by i corresponds to a rotation by 90°: i = eiπ/2.
一个典型问题：将 3eiπ/4 表示为直角坐标形式。解答：3(cos π/4 + i sin π/4) = 3(√2/2 + i√2/2) = (3√2/2) + i(3√2/2)。指数形式也优雅地解释了为何乘以 i 相当于旋转 90°：i = eiπ/2。
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