📚 Algebraic Form of Complex Numbers | 复数的代数形式表示
Complex numbers extend the real number system to include solutions of equations such as x² = -1. In their algebraic form, a complex number is written as z = a + bi, where a and b are real numbers and i is the imaginary unit. This form provides a powerful framework for solving problems in algebra, geometry, and calculus, and it forms a core part of the IB Mathematics Higher Level syllabus.
复数将实数系扩展到包含形如 x² = -1 的方程的解。在代数形式中，复数写作 z = a + bi，其中 a 和 b 是实数，i 是虚数单位。这一形式为代数、几何和微积分中的问题提供了强大的求解框架，也是 IB 数学高级水平课程的核心内容。
1. The Imaginary Unit i | 虚数单位 i
The imaginary unit is defined by the property i² = -1. It allows us to take square roots of negative numbers. For example, √(-4) = 2i, because (2i)² = 4i² = -4. In IB mathematics, you must be comfortable with powers of i: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, and then the pattern repeats every four powers.
虚数单位由性质 i² = -1 定义。它使我们能够对负数开平方。例如，√(-4) = 2i，因为 (2i)² = 4i² = -4。在 IB 数学中，你必须熟练掌握 i 的幂：i¹ = i，i² = -1，i³ = -i，i⁴ = 1，此后每四个幂循环一次。
i² = -1, i³ = -i, i⁴ = 1
2. Definition of Algebraic Form | 代数形式的定义
A complex number in algebraic form is expressed as z = a + bi, where a and b are real numbers. Here, a is called the real part, denoted Re(z), and b is called the imaginary part, denoted Im(z). Note that the imaginary part is the real coefficient of i, not the term bi itself.
代数形式的复数表示为 z = a + bi，其中 a 和 b 是实数。这里，a 称为实部，记作 Re(z)；b 称为虚部，记作 Im(z)。注意虚部是 i 的实数系数，而不是整个 bi 项。
For example, in z = 3 – 2i, we have Re(z) = 3 and Im(z) = -2. A complex number with b = 0 is purely real, while one with a = 0 is purely imaginary.
例如，在 z = 3 – 2i 中，Re(z) = 3，Im(z) = -2。当 b = 0 时复数为纯实数；当 a = 0 时复数为纯虚数。
3. Equality of Complex Numbers | 复数相等条件
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. That is, for z₁ = a + bi and z₂ = c + di, z₁ = z₂ ⇔ a = c and b = d.
两个复数相等当且仅当它们的实部相等且虚部相等。即对于 z₁ = a + bi 和 z₂ = c + di，z₁ = z₂ ⇔ a = c 且 b = d。
This condition is often used to solve equations involving complex numbers. For instance, if x + yi = 5 – 3i, then x = 5 and y = -3.
这个条件常用于求解含复数的方程。例如，若 x + yi = 5 – 3i，则 x = 5，y = -3。
4. Addition and Subtraction | 复数的加减法
To add two complex numbers, add their real parts and their imaginary parts separately:
两个复数相加时，分别将实部与实部、虚部与虚部相加：
(a + bi) + (c + di) = (a + c) + (b + d)i
Similarly, subtraction is performed component-wise:
类似地，减法也按分量进行：
(a + bi) – (c + di) = (a – c) + (b – d)i
For example, (3 + 2i) + (1 – 4i) = 4 – 2i, and (3 + 2i) – (1 – 4i) = 2 + 6i. These operations follow the usual commutative, associative, and distributive laws.
例如，(3 + 2i) + (1 – 4i) = 4 – 2i，(3 + 2i) – (1 – 4i) = 2 + 6i。这些运算遵循通常的交换律、结合律和分配律。
5. Multiplication by a Real Scalar | 实数倍复数
Multiplying a complex number by a real scalar k multiplies both the real and imaginary parts by k:
复数乘以实数标量 k 时，实部和虚部都乘以 k：
k(a + bi) = ka + kbi
For instance, 3(2 – 5i) = 6 – 15i. This operation is closely related to geometric scaling in the complex plane.
例如，3(2 – 5i) = 6 – 15i。这个运算与复平面中的几何缩放密切相关。
6. Multiplication of Complex Numbers | 复数乘法
To multiply two complex numbers, use the distributive law and the fact i² = -1:
两个复数相乘时，使用分配律并利用 i² = -1：
(a + bi)(c + di) = (ac – bd) + (ad + bc)i
Derivation: (a + bi)(c + di) = ac + adi + bci + bdi² = ac + (ad + bc)i – bd = (ac – bd) + (ad + bc)i.
推导：(a + bi)(c + di) = ac + adi + bci + bdi² = ac + (ad + bc)i – bd = (ac – bd) + (ad + bc)i。
Example: (2 + 3i)(4 – i) = 8 – 2i + 12i – 3i² = 8 + 10i + 3 = 11 + 10i.
例：(2 + 3i)(4 – i) = 8 – 2i + 12i – 3i² = 8 + 10i + 3 = 11 + 10i。
7. Complex Conjugate | 共轭复数
The conjugate of z = a + bi is denoted z̄ and is defined as z̄ = a – bi. The conjugate has the effect of reflecting the point z across the real axis in the complex plane.
复数 z = a + bi 的共轭记作 z̄，定义为 z̄ = a – bi。共轭的作用是将复平面中的点 z 关于实轴作镜像反射。
Key properties include: z + z̄ = 2a (a real number), z – z̄ = 2bi (purely imaginary), and z·z̄ = a² + b² (a non-negative real number). The product z·z̄ equals the square of the modulus of z.
关键性质包括：z + z̄ = 2a（实数），z – z̄ = 2bi（纯虚数），以及 z·z̄ = a² + b²（非负实数）。乘积 z·z̄ 等于 z 的模的平方。
Conjugation also distributes over addition, subtraction, multiplication, and division: the conjugate of a sum (or product) is the sum (or product) of the conjugates.
共轭运算还满足对加法、减法、乘法和除法的分配性：和（或积）的共轭等于共轭的和（或积）。
8. Division of Complex Numbers | 复数的除法
To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator. This converts the denominator into a real number:
复数相除时，将分子和分母同时乘以分母的共轭，从而把分母化为实数：
(a + bi) / (c + di) = ((a + bi)(c – di)) / ((c + di)(c – di))
Since (c + di)(c – di) = c² + d², the result is:
由于 (c + di)(c – di) = c² + d²，结果为：
(a + bi) / (c + di) = (ac + bd)/(c² + d²) + (bc – ad)/(c² + d²) i
Example: (5 + i)/(2 – i) = ((5 + i)(2 + i))/((2 – i)(2 + i)) = (10 + 5i + 2i + i²)/(4 + 1) = (9 + 7i)/5 = 9/5 + 7/5 i.
例：(5 + i)/(2 – i) = ((5 + i)(2 + i))/((2 – i)(2 + i)) = (10 + 5i + 2i + i²)/(4 + 1) = (9 + 7i)/5 = 9/5 + (7/5)i。
9. Modulus and Argument | 模与辐角
The modulus of z = a + bi is defined as |z| = √(a² + b²), representing the distance from the origin to the point (a, b) in the complex plane. The argument, arg(z), is the angle θ that the line from the origin to z makes with the positive real axis.
复数 z = a + bi 的模定义为 |z| = √(a² + b²)，表示复平面中原点到点 (a, b) 的距离。辐角 arg(z) 是原点与 z 的连线与正实轴之间的夹角 θ。
For a given z, we have a = |z|cos θ and b = |z|sin θ, so z can be written as |z|(cos θ + i sin θ). This is the polar form, which is closely linked to the algebraic form and is essential for solving problems involving powers and roots.
对于给定的 z，有 a = |z|cos θ，b = |z|sin θ，因此 z 可写成 |z|(cos θ + i sin θ)。这就是极坐标形式，与代数形式紧密相连，在求解幂和根的问题中至关重要。
10. The Complex Plane and Geometric Interpretation | 复平面与几何意义
In the complex plane, the algebraic form z = a + bi corresponds to the point (a, b). The real part a is the horizontal coordinate, and the imaginary part b is the vertical coordinate. Addition and subtraction of complex numbers correspond to vector addition and subtraction.
在复平面中，代数形式 z = a + bi 对应于点 (a, b)。实部 a 是横坐标，虚部 b 是纵坐标。复数的加法和减法对应于向量的加法和减法。
Moreover, multiplication by i corresponds to a 90° counterclockwise rotation about the origin. For example, multiplying z = 1 + i by i gives iz = i + i² = -1 + i, which is indeed z rotated 90° counterclockwise.
此外，乘以 i 等价于绕原点逆时针旋转 90°。例如，z = 1 + i 乘以 i 得到 iz = i + i² = -1 + i，这确实是 z 逆时针旋转 90° 后的结果。
| Algebraic Form | Point in Complex Plane |
| z = 3 + 4i | (3, 4) |
| z = 3 – 4i (conjugate) | (3, -4) |
| z = -3 + 4i | (-3, 4) |
11. Solving Quadratic Equations | 解二次方程
The algebraic form of complex numbers enables us to solve quadratic equations with negative discriminants. For the equation ax² + bx + c = 0, the solutions are given by x = (-b ± √(b² – 4ac))/(2a). If b² – 4ac < 0, then the square root is imaginary, and the two solutions are complex conjugates.
复数的代数形式使我们能够求解判别式为负的二次方程。对于方程 ax² + bx + c = 0，解为 x = (-b ± √(b² – 4ac))/(2a)。若 b² – 4ac < 0，则平方根为虚数，两个解是一对共轭复数。
Example: Solve x² + 4x + 5 = 0. Using the quadratic formula, x = (-4 ± √(16 – 20))/2 = (-4 ± √(-4))/2 = (-4 ± 2i)/2 = -2 ± i. The solutions are -2 + i and -2 – i.
例：解 x² + 4x + 5 = 0。由求根公式，x = (-4 ± √(16 – 20))/2 = (-4 ± √(-4))/2 = (-4 ± 2i)/2 = -2 ± i。解为 -2 + i 和 -2 – i。
12. Applications and Exam Tips | 应用与考试要点
In IB examinations, complex numbers in algebraic form are frequently tested through operations, conjugation, and modulus calculations. Common pitfalls include forgetting that Im(z) is the real coefficient of i, misapplying i² = -1 during multiplication, and failing to rationalise the denominator correctly in division.
在 IB 考试中，代数形式的复数经常通过运算、共轭和模的计算进行考查。常见错误包括忘记 Im(z) 是 i 的实数系数、在乘法中误用 i² = -1，以及在除法中未能正确地将分母有理化。
It is also important to recognise that the modulus satisfies |z₁z₂| = |z₁||z₂| and |z₁/z₂| = |z₁|/|z₂| for z₂ ≠ 0. These properties are useful for simplifying expressions without fully expanding them.
还要注意模满足 |z₁z₂| = |z₁||z₂| 和 |z₁/z₂| = |z₁|/|z₂|（其中 z₂ ≠ 0）。这些性质有助于在不完全展开的情况下化简表达式。
Practice converting between algebraic and polar forms, as this skill underpins many IB questions on complex numbers. Always check whether your final answer is in the form requested by the question.
要练习在代数形式和极坐标形式之间转换，这一技能是 IB 复数题目的基础。始终检查最终答案是否符合题目要求的形式。
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