📚 Introduction to Complex Numbers | 复数入门
al solution because no real number squared can give −1. Complex numbers extend the number system so that every polynomial equation has a solution, giving us a powerful tool for algebra, geometry and applied mathematics.
在 A Level 数学中，很多方程在实数范围内有解，但有些简单的方程没有实数解。例如，x² + 1 = 0 没有实数解，因为任何实数的平方都不可能等于 −1。复数扩展了数系，使每一个多项式方程都有解，为代数、几何和应用数学提供了强大工具。
1. From Real Numbers to Complex Numbers | 从实数到复数
The real number system includes rational and irrational numbers, such as 3, ½, √2 and π. However, the square root of a negative number is not defined as a real number. Complex numbers are introduced by defining a new number whose square is negative.
实数系包括有理数和无理数，例如 3、½、√2 和 π。但是，负数的平方根在实数范围内没有定义。复数通过定义一种平方为负数的新数来引入。
number. Although the name suggests it is not real, imaginary and complex numbers are legitimate mathematical objects with rigorous definitions and many real-world applications in physics, engineering and signal processing.
历史上，数学家把 x² = −1 的解称为虚数。虽然这个名字听起来不像真实存在，但虚数和复数是严格的数学对象，在物理、工程和信号处理中有许多实际应用。
2. The Imaginary Unit i | 虚数单位 i
The imaginary unit is denoted by i and is defined by the equation:
虚数单位记作 i，其定义为：
i² = −1
ns we use the defining property i² = −1 rather than treating i as an ordinary square root.
因此 i 是 −1 的一个平方根。我们常写作 √−1 = i，但在计算中应使用定义性质 i² = −1，而不是把 i 当作普通平方根来处理。
The powers of i repeat in a cycle of four. The key values are:
i 的幂以四为周期循环，关键值为：
| i¹ = i | i² = −1 |
| i³ = −i | i⁴ = 1 |
i, i⁶ = −1, i⁷ = −i and i⁸ = 1. This cycle is useful when simplifying higher powers of i.
i⁴ 之后模式重复，因此 i⁵ = i，i⁶ = −1，i⁷ = −i，i⁸ = 1。这个循环在化简 i 的高次幂时非常有用。
3. Standard Form and Terminology | 标准形式与术语
A complex number is written in standard form as:
复数写成标准形式为：
z = a + bi
ry part of z is Im(z) = b. Note that the imaginary part is the real coefficient b, not bi.
其中 a 和 b 是实数。z 的实部为 Re(z) = a，虚部为 Im(z) = b。注意虚部是实数系数 b，而不是 bi。
is 3 and the imaginary part is −4. A real number such as 5 can be written as 5 + 0i, and a pure imaginary number such as 2i can be written as 0 + 2i.
例如，在 z = 3 − 4i 中，实部是 3，虚部是 −4。实数 5 可以写成 5 + 0i，纯虚数 2i 可以写成 0 + 2i。
4. Equality of Complex Numbers | 复数相等
are equal. In symbols, if z₁ = a + bi and z₂ = c + di, then z₁ = z₂ means 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。用符号表示，如果 z₁ = a + bi 且 z₂ = c + di，那么 z₁ = z₂ 就意味着 a = c 且 b = d。
real numbers in complex equations. For example, if x + yi = 4 + 7i, then x = 4 and y = 7.
这条规则在求解复数方程中的未知实数时特别有用。例如，如果 x + yi = 4 + 7i，那么 x = 4 且 y = 7。
When equating real and imaginary parts, always make sure both sides are expressed in standard form a + bi first.
在比较实部和虚部时，务必先把两边都写成标准形式 a + bi。
5. Adding and Subtracting Complex Numbers | 复数的加法与减法
parts together and add the imaginary parts together. If z₁ = a + bi and z₂ = c + di, then:
复数相加时，实部与实部相加，虚部与虚部相加。如果 z₁ = a + bi 且 z₂ = c + di，那么：
z₁ + z₂ = (a + c) + (b + d)i
Subtraction works in the same way by subtracting the real parts and subtracting the imaginary parts:
减法同理，实部减实部，虚部减虚部：
z₁ − z₂ = (a − c) + (b − d)i
(4 + 6i) − (1 + 2i) = 3 + 4i. Addition and subtraction do not require the use of i² = −1.
例如，(5 + 2i) + (3 − 7i) = 8 − 5i，(4 + 6i) − (1 + 2i) = 3 + 4i。加法和减法不需要使用 i² = −1。
6. Multiplying Complex Numbers | 复数的乘法
ing brackets normally and then replacing i² with −1. If z₁ = a + bi and z₂ = c + di, then:
复数相乘时，像普通代数式一样展开括号，然后将 i² 替换为 −1。如果 z₁ = a + bi 且 z₂ = c + di，那么：
z₁z₂ = (ac − bd) + (ad + bc)i
pand directly. For example, (2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i² = 2 − 5i + 12 = 14 − 5i.
这个公式很有用，但你也可以直接展开。例如，(2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i² = 2 − 5i + 12 = 14 − 5i。
nd remember that the imaginary unit i is not a variable: its defining property is i² = −1.
始终分别合并实部项和虚部项，并记住虚数单位 i 不是变量：它的定义性质是 i² = −1。
7. Complex Conjugates | 共轭复数
The complex conjugate of z = a + bi is denoted by z* and is defined as:
复数 z = a + bi 的共轭复数记作 z*，其定义为：
z* = a − bi
ties. The sum z + z* = 2a is real, and the product zz* = a² + b² is real and non-negative.
换句话说，共轭复数就是把虚部的符号改变。共轭复数有两个重要性质：和 z + z* = 2a 是实数，积 zz* = a² + b² 是实数且非负。
5. The product of a complex number and its conjugate is especially important for division.
例如，如果 z = 3 + 4i，那么 z* = 3 − 4i，z + z* = 6，zz* = 9 + 16 = 25。复数与其共轭的乘积对除法尤其重要。
8. Dividing Complex Numbers | 复数的除法
To divide one complex number by another, multiply both the numerator and the denominator by the conjugate of the denominator. This converts the denominator into a real number.
要将一个复数除以另一个复数，将分子和分母同时乘以分母的共轭复数。这样可以把分母化为实数。
For example, to calculate (3 + 2i) ÷ (1 − i), multiply by the conjugate of 1 − i, which is 1 + i:
例如，计算 (3 + 2i) ÷ (1 − i)，乘以 1 − i 的共轭 1 + i：
(3 + 2i)(1 + i) ÷ (1 − i)(1 + i) = (1 + 5i) ÷ 2 = ½ + (5/2)i
In general, if z₁ = a + bi and z₂ = c + di, then:
一般来说，如果 z₁ = a + bi 且 z₂ = c + di，那么：
z₁ ÷ z₂ = (ac + bd)/(c² + d²) + (bc − ad)/(c² + d²) i
You do not need to memorise this formula if you can multiply by the conjugate confidently.
只要你能熟练地乘以共轭复数，就不需要记忆这个公式。
9. Solving Quadratic Equations | 解二次方程
negative discriminants to have solutions. For ax² + bx + c = 0, the quadratic formula is:
复数使得判别式为负的二次方程也有解。对于 ax² + bx + c = 0，二次公式为：
x = (−b ± √b² − 4ac) ÷ 2a
If the discriminant b² − 4ac is negative, the square root becomes an imaginary number written in terms of i.
如果判别式 b² − 4ac 为负，平方根就变成用 i 表示的虚数。
example, solve x² + 4x + 13 = 0. The discriminant is 4² − 4 × 1 × 13 = 16 − 52 = −36, so:
例如，解方程 x² + 4x + 13 = 0。判别式为 4² − 4 × 1 × 13 = 16 − 52 = −36，因此：
x = (−4 ± √−36) ÷ 2 = (−4 ± 6i) ÷ 2 = −2 ± 3i
When a polynomial has real coefficients, any complex roots always occur in conjugate pairs. Here the roots are −2 + 3i and −2 − 3i.
当多项式具有实系数时，任何复数根总是成共轭对出现。这里根为 −2 + 3i 和 −2 − 3i。
10. Argand Diagrams and Modulus-Argument Form | 阿尔冈图与模-辐角形式
axis is the imaginary axis. The complex number z = a + bi is plotted at the point (a, b).
复数可以在阿尔冈图上用几何方式表示。横轴是实轴，纵轴是虚轴。复数 z = a + bi 对应点 (a, b)。
The modulus of z is the distance from the origin to the point, given by:
z 的模是原点到该点的距离，公式为：
|z| = √a² + b²
e origin to the point. It satisfies tan θ = b/a, but you must choose the correct quadrant.
z 的辐角记作 arg z，是从正实轴到原点与该点连线的角 θ。它满足 tan θ = b/a，但必须选择正确的象限。
The modulus-argument form is:
模-辐角形式为：
z = r(cos θ + i sin θ)
For example, z = 1 + i√3 has modulus |z| = √1 + 3 = 2 and argument θ = π/3, so z = 2(cos π/3 + i sin π/3).
例如，z = 1 + i√3 的模为 |z| = √1 + 3 = 2，辐角为 θ = π/3，所以 z = 2(cos π/3 + i sin π/3)。
11. Summary of Key Skills | 关键技能总结
roducts, use the conjugate to divide, and represent solutions geometrically on an Argand diagram when needed.
处理复数时，始终先确定实部和虚部。利用 i² = −1 化简乘积，利用共轭进行除法，并在需要时用阿尔冈图几何表示解。
Key skills include simplifying powers of i, adding, subtracting, multiplying and dividing in standard form, solving quadratic equations with negative discriminants, and converting between standard form and modulus-argument form.
关键技能包括化简 i 的幂，以标准形式进行加减乘除，解判别式为负的二次方程，以及在标准形式和模-辐角形式之间转换。
With these foundations, complex numbers become a natural extension of the real number system and prepare you for more advanced topics such as de Moivre's theorem and complex roots of unity.
有了这些基础，复数就自然成为实数系的扩展，并为进一步学习棣莫弗定理和单位复数根等更高级主题做好准备。
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