📚 IB Mathematics: Cartesian Form of Complex Numbers | IB数学：复数的笛卡尔形式
Complex numbers are one of the most elegant and practical topics in the IB Mathematics curriculum. They extend our understanding of numbers beyond the real line and allow us to solve equations that were once considered impossible. In this article, we explore the Cartesian form z = a + bi, the foundation upon which all complex number operations are built.
复数是IB数学课程中最优雅且实用的主题之一。它将我们对数的理解从实数轴进一步扩展，使我们能够解出曾经被认为无解的方程。在本文中，我们将深入探讨笛卡尔形式 z = a + bi，这是构建所有复数运算的基石。
1. The Need for Complex Numbers | 复数的必要性
Consider the quadratic equation x² + 1 = 0. To solve it, we need a number whose square equals -1. No real number satisfies this condition, since the square of any real number is non-negative. Mathematicians resolved this by introducing the imaginary unit i, defined by i² = -1.
考虑二次方程 x² + 1 = 0。要解这个方程，我们需要一个平方等于 -1 的数。没有实数满足这个条件，因为任何实数的平方都是非负的。数学家通过引入虚数单位 i 解决了这个问题，定义 i² = -1。
The set of complex numbers, denoted ℂ, contains all numbers of the form a + bi, where a and b are real numbers. When b = 0, the complex number is purely real; when a = 0, it is purely imaginary. Thus the real numbers are a subset of the complex numbers.
复数集用 ℂ 表示，包含所有形如 a + bi 的数，其中 a 和 b 是实数。当 b = 0 时，复数为纯实数；当 a = 0 时，复数为纯虚数。因此，实数集是复数集的子集。
2. Definition of Cartesian Form | 笛卡尔形式的定义
The Cartesian form (also called rectangular form) of a complex number is written as z = a + bi. Here, a is the real part and b is the imaginary part. For example, in z = 3 – 4i, the real part is 3 and the imaginary part is -4.
复数的笛卡尔形式（也称直角坐标形式）写作 z = a + bi。其中 a 是实部，b 是虚部。例如，在 z = 3 – 4i 中，实部为 3，虚部为 -4。
It is crucial to note that the imaginary part is the real number b, not bi. We denote the real part as Re(z) = a and the imaginary part as Im(z) = b. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
必须注意，虚部是实数 b，而不是 bi。我们用 Re(z) = a 表示实部，用 Im(z) = b 表示虚部。两个复数相等，当且仅当它们的实部相等且虚部相等。
3. Representing Complex Numbers | 复数的表示
Just as real numbers can be placed on a number line, complex numbers can be placed on the complex plane (Argand diagram). The horizontal axis represents the real part, and the vertical axis represents the imaginary part. The number z = a + bi corresponds to the point (a, b).
正如实数可以放在数轴上一样，复数可以放在复平面（阿甘图）上。水平轴表示实部，垂直轴表示虚部。复数 z = a + bi 对应点 (a, b)。
This geometric view is powerful. The distance from the origin to the point (a, b) is called the modulus, written |z| = √(a² + b²). The angle measured counterclockwise from the positive real axis is called the argument, arg(z). These connections prepare the ground for polar form and Euler’s formula in later topics.
这种几何视角非常强大。从原点到点 (a, b) 的距离称为模，记作 |z| = √(a² + b²)。从正实轴逆时针方向测量的角度称为辐角，记作 arg(z)。这些联系为后续学习极坐标形式和欧拉公式奠定了基础。
4. Addition and Subtraction | 加法与减法
To add or subtract complex numbers, we combine the real parts and the imaginary parts separately. If z₁ = a + bi and z₂ = c + di, then:
要对复数进行加法或减法运算，我们分别合并实部和虚部。若 z₁ = a + bi，z₂ = c + di，则：
z₁ + z₂ = (a + c) + (b + d)i
z₁ – z₂ = (a – c) + (b – d)i
For example, (2 + 5i) + (3 – 7i) = 5 – 2i. Geometrically, addition corresponds to vector addition in the Argand diagram: the sum of two complex numbers is the diagonal of the parallelogram formed by the two vectors.
例如，(2 + 5i) + (3 – 7i) = 5 – 2i。几何上，加法对应阿甘图中的向量加法：两个复数的和是这两个向量构成的平行四边形的对角线。
5. Multiplication | 乘法
Multiplication of complex numbers follows the distributive law, with the key rule i² = -1. For z₁ = a + bi and z₂ = c + di:
复数的乘法遵循分配律，同时运用关键规则 i² = -1。对于 z₁ = a + bi 和 z₂ = c + di：
z₁z₂ = (a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i
For example, (2 + 3i)(1 – 4i) = 2 – 8i + 3i – 12i² = 2 – 5i + 12 = 14 – 5i. Notice that the term -12i² became +12 because i² = -1.
例如，(2 + 3i)(1 – 4i) = 2 – 8i + 3i – 12i² = 2 – 5i + 12 = 14 – 5i。注意 -12i² 变为 +12，因为 i² = -1。
A special case is multiplication by i, which rotates the point 90° counterclockwise about the origin. For instance, i(1 + 2i) = -2 + i, which is indeed a quarter-turn rotation.
一个特殊情况是乘以 i，这会将点绕原点逆时针旋转 90°。例如，i(1 + 2i) = -2 + i，这确实是一个四分之一圈的旋转。
6. The Complex Conjugate | 共轭复数
For a complex number z = a + bi, its complex conjugate is defined as z̄ = a – bi. The conjugate has the same real part but the opposite imaginary part. Geometrically, z̄ is the reflection of z across the real axis.
对于复数 z = a + bi，它的共轭复数定义为 z̄ = a – bi。共轭复数具有相同的实部但虚部相反。几何上，z̄ 是 z 关于实轴的反射。
The product of a complex number and its conjugate is a real number:
复数与其共轭的乘积是一个实数：
z · z̄ = (a + bi)(a – bi) = a² + b² = |z|²
This property is extremely useful for simplifying quotients and for finding the modulus. It also explains why |z|² is always non-negative.
这个性质对于化简商以及求模非常有用。它也解释了为什么 |z|² 永远是非负的。
7. Division of Complex Numbers | 复数的除法
To divide two complex numbers, we multiply both numerator and denominator by the conjugate of the denominator. This eliminates the imaginary part from the denominator. If z₁ = a + bi and z₂ = c + di (with z₂ ≠ 0):
要计算两个复数的除法，我们将分子和分母同时乘以分母的共轭。这样可以消去分母中的虚部。若 z₁ = a + bi，z₂ = c + di（且 z₂ ≠ 0）：
z₁ / z₂ = (a + bi) / (c + di) = [(a + bi)(c – di)] / [(c + di)(c – di)]
Expanding gives:
展开后得到：
z₁ / z₂ = [(ac + bd) + (bc – ad)i] / (c² + d²)
For example, (3 + 2i) / (1 – i) = [(3 + 2i)(1 + i)] / [(1 – i)(1 + i)] = (1 + 5i) / 2 = 0.5 + 2.5i.
例如，(3 + 2i) / (1 – i) = [(3 + 2i)(1 + i)] / [(1 – i)(1 + i)] = (1 + 5i) / 2 = 0.5 + 2.5i。
8. Solving Quadratic Equations | 解二次方程
With complex numbers, every quadratic equation with real coefficients has exactly two solutions in ℂ (counting multiplicity). The quadratic formula works exactly as before:
有了复数，每个具有实系数的二次方程在 ℂ 中恰好有两个解（计入重数）。二次公式的用法与之前完全相同：
x = [-b ± √(b² – 4ac)] / 2a
When the discriminant Δ = b² – 4ac is negative, the solutions are complex conjugates. For example, x² + 4x + 13 = 0 has Δ = 16 – 52 = -36, so:
当判别式 Δ = b² – 4ac 为负数时，解为一对共轭复数。例如，x² + 4x + 13 = 0 的判别式 Δ = 16 – 52 = -36，因此：
x = [-4 ± √(-36)] / 2 = [-4 ± 6i] / 2 = -2 ± 3i
Thus the two roots are -2 + 3i and -2 – 3i. This conjugate root theorem applies whenever the coefficients of the polynomial are real.
因此两个根为 -2 + 3i 和 -2 – 3i。当多项式的系数为实数时，这个共轭根定理始终成立。
9. Square Roots of Complex Numbers | 复数的平方根
To find the square roots of a complex number z = a + bi, we set (x + yi)² = a + bi and equate real and imaginary parts:
要计算复数 z = a + bi 的平方根，我们设 (x + yi)² = a + bi，然后令实部和虚部分别相等：
x² – y² = a and 2xy = b
Together with x² + y² = √(a² + b²) = |z|, we can solve for x and y. As an example, let us find the square roots of 3 + 4i. We have x² – y² = 3 and 2xy = 4, so xy = 2. Also x² + y² = 5. Adding and subtracting these equations gives x² = 4 and y² = 1. Since xy = 2 is positive, x and y have the same sign, so the roots are 2 + i and -2 – i.
结合 x² + y² = √(a² + b²) = |z|，我们可以解出 x 和 y。例如，让我们求 3 + 4i 的平方根。我们有 x² – y² = 3 和 2xy = 4，因此 xy = 2。同时 x² + y² = 5。将这两个方程相加和相减，得到 x² = 4，y² = 1。由于 xy = 2 为正，x 和 y 同号，所以根为 2 + i 和 -2 – i。
10. Geometric Interpretation of Operations | 运算的几何意义
In the Argand diagram, addition and subtraction are vector operations, while multiplication by a complex number combines a rotation and a scaling. Specifically, multiplying by a complex number of modulus r and argument θ rotates the original vector by θ and scales it by r.
在阿甘图中，加法和减法是向量运算，而乘以一个复数则结合了旋转和缩放。具体来说，乘以一个模为 r、辐角为 θ 的复数，会将原向量旋转 θ 并缩放 r 倍。
The modulus and argument also satisfy important properties:
模和辐角还满足重要的性质：
- |z₁z₂| = |z₁||z₂| and |z₁/z₂| = |z₁|/|z₂| (for z₂ ≠ 0)
- arg(z₁z₂) = arg(z₁) + arg(z₂) mod 2π
- arg(z₁/z₂) = arg(z₁) – arg(z₂) mod 2π
- |z₁z₂| = |z₁||z₂| 以及 |z₁/z₂| = |z₁|/|z₂|（z₂ ≠ 0）
- arg(z₁z₂) = arg(z₁) + arg(z₂)（模 2π）
- arg(z₁/z₂) = arg(z₁) – arg(z₂)（模 2π）
These properties are often tested in IB exams, especially when combined with trigonometric identities to prove De Moivre’s theorem later in the course.
这些性质在IB考试中经常出现，尤其是在课程后期结合三角恒等式证明棣莫弗定理时。
11. Common Exam Pitfalls | 常见考试陷阱
Students frequently make errors when working with the Cartesian form. Below is a table of common mistakes and their corrections.
学生在处理笛卡尔形式时经常犯错。下表列出了一些常见错误及其修正。
| Common Mistake | 常见错误 | Correct Approach | 正确做法 |
| Writing i = √(-1) and applying √(ab) = √a√b to negative numbers, e.g., √(-4)√(-9) = 6 | Use i² = -1 only; do not separate square roots of negative numbers. Write √(-4) = 2i, √(-9) = 3i, so (2i)(3i) = 6i² = -6. |
| Confusing the imaginary part: saying Im(3 – 4i) = -4i | The imaginary part is the coefficient of i, so Im(3 – 4i) = -4. |
| Forgetting to multiply both numerator and denominator by the conjugate in division. | Always multiply numerator and denominator by z̄₂ to keep the fraction equivalent. |
| Writing |z| = √(a² + b²) as a + b or a² + b². | Remember the square root covers the entire sum: |z| = √(a² + b²). |
In addition, when solving equations involving complex numbers, always equate real and imaginary parts separately. This is a fundamental method that appears throughout the topic.
此外，在解涉及复数的方程时，一定要将实部和虚部分别相等。这是贯穿整个主题的基本方法。
12. Practice Problems | 练习题
Below are four problems to test your understanding of the Cartesian form. Try them before reading the brief answers.
下面是四道练习题，用于检验你对笛卡尔形式的理解。先试着自己做，再参考答案。
Problem 1: Find the real numbers x and y such that (x + 2yi)(2 – i) = 8 – i.
习题1：求实数 x 和 y，使得 (x + 2yi)(2 – i) = 8 – i。
Solution: Expanding gives (2x + 2y) + (4y – x)i = 8 – i. Hence 2x + 2y = 8 and 4y – x = -1. Solving simultaneously yields x = 3, y = 1.
解答：展开得 (2x + 2y) + (4y – x)i = 8 – i。因此 2x + 2y = 8，4y – x = -1。联立解得 x = 3，y = 1。
Problem 2: Express (1 + 2i) / (2 – i) in the form a + bi.
习题2：将 (1 + 2i) / (2 – i) 化为 a + bi 的形式。
Solution: Multiply numerator and denominator by 2 + i: (1 + 2i)(2 + i) = 2 + i + 4i + 2i² = 0 + 5i, and (2 – i)(2 + i) = 5. Thus the quotient is i.
解答：分子分母同乘以 2 + i：(1 + 2i)(2 + i) = 2 + i + 4i + 2i² = 0 + 5i，而 (2 – i)(2 + i) = 5。因此商为 i。
Problem 3: Solve z² – 6z + 25 = 0.
习题3：解方程 z² – 6z + 25 = 0。
Solution: Using the quadratic formula, Δ = 36 – 100 = -64, so z = [6 ± √(-64)] / 2 = [6 ± 8i] / 2 = 3 ± 4i.
解答：使用二次公式，Δ = 36 – 100 = -64，所以 z = [6 ± √(-64)] / 2 = [6 ± 8i] / 2 = 3 ± 4i。
Problem 4: Given that z = 5 – 12i, find |z| and z · z̄.
习题4：已知 z = 5 – 12i，求 |z| 和 z · z̄。
Solution: |z| = √(25 + 144) = √169 = 13, and z · z̄ = 25 + 144 = 169 = |z|².
解答：|z| = √(25 + 144) = √169 = 13，且 z · z̄ = 25 + 144 = 169 = |z|²。
Mastering the Cartesian form gives you a solid foundation for all later complex number topics, including polar form, De Moivre’s theorem, and complex roots of unity. Practice drawing Argand diagrams, verify every algebra step, and always separate real and imaginary parts carefully.
掌握笛卡尔形式为你学习后续所有复数主题奠定了坚实的基础，包括极坐标形式、棣莫弗定理以及单位复根。练习绘制阿甘图，验证每一步代数运算，并始终仔细区分实部和虚部。
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