AQA AS Further Maths: Complex Numbers and the Argand Diagram — AQA AS进阶数学：复数与阿尔冈图

一、什么是虚数单位i：为什么需要它 | 1. The Imaginary Unit i: Why We Need It

在 A-Level 普通数学里，我们解方程 x² = -1 时会遇到困难，因为任何实数的平方都不可能等于负数。为了突破这个限制，数学家引入了一个全新的数，记作 i，并定义它的平方等于 -1，即 i² = -1。这个 i 被称为虚数单位，它是一切复数运算的起点。

In ordinary A-Level Mathematics, solving x² = -1 seems impossible because the square of any real number can never be negative. To break through this barrier, mathematicians introduced a brand-new number written as i, and defined its square to equal -1, that is i² = -1. This i is called the imaginary unit, and it is the starting point for all complex number work.

有了 i 之后，任何负数都可以开平方了。例如 √(-9) 可以写成 √9 × √(-1) = 3i，而 √(-4) = 2i。这意味着所有二次方程，无论判别式是正是负，现在都可以求出解。AQA AS 进阶数学的第一单元（Further Pure）正是从 i 的定义开始，逐步搭建起复数这个完整的数系。

Once i is defined, every negative number can now have a square root. For example √(-9) can be written as √9 × √(-1) = 3i, and √(-4) = 2i. This means every quadratic equation, whether its discriminant is positive or negative, can now be solved. AQA AS Further Mathematics Unit 1 (Further Pure) begins precisely with the definition of i and gradually builds up the complete system of complex numbers.

二、复数的标准形式与实部、虚部 | 2. Standard Form a + bi, Real and Imaginary Parts

一个复数通常写成标准形式 z = a + bi，其中 a 和 b 都是实数。这里的 a 叫做实部（Real Part），记作 Re(z)；b 叫做虚部（Imaginary Part），记作 Im(z)。请注意，虚部 b 本身是一个实数，它只是 i 前面的系数，而不是 bi 整体。

A complex number is usually written in standard form z = a + bi, where a and b are both real numbers. Here a is called the real part, written Re(z), and b is called the imaginary part, written Im(z). Note carefully that the imaginary part b is itself a real number; it is simply the coefficient in front of i, not the whole expression bi.

举几个例子：对于 z = 3 + 4i，实部是 3，虚部是 4；对于 z = -2i，可以看作 0 + (-2)i，所以实部是 0，虚部是 -2；对于 z = 5，可以看作 5 + 0i，实部是 5，虚部是 0。当实部为零时，我们称它为纯虚数；当虚部为零时，它就是一个普通的实数。因此，实数其实是复数的一个子集。

Consider a few examples: for z = 3 + 4i, the real part is 3 and the imaginary part is 4; for z = -2i, we can write it as 0 + (-2)i, so the real part is 0 and the imaginary part is -2; for z = 5, we can write it as 5 + 0i, so the real part is 5 and the imaginary part is 0. When the real part is zero, the number is called purely imaginary; when the imaginary part is zero, it is simply an ordinary real number. Real numbers are therefore a subset of the complex numbers.

三、复数的加法与减法 | 3. Adding and Subtracting Complex Numbers

两个复数相加或相减时，规则非常简单：实部与实部相加减，虚部与虚部相加减。也就是说 (a + bi) + (c + di) = (a + c) + (b + d)i，而 (a + bi) – (c + di) = (a – c) + (b – d)i。运算完成后，记得把结果整理回标准形式。

When adding or subtracting two complex numbers, the rule is very simple: combine the real parts together and the imaginary parts together. That is (a + bi) + (c + di) = (a + c) + (b + d)i, and (a + bi) – (c + di) = (a – c) + (b – d)i. After the calculation, remember to tidy the result back into standard form.

例如 (2 + 3i) + (5 – 7i) = (2 + 5) + (3 – 7)i = 7 – 4i，而 (2 + 3i) – (5 – 7i) = (2 – 5) + (3 – (-7))i = -3 + 10i。在 AQA 的试卷里，加法和减法通常作为大题的第一步出现，例如先合并同类项，再进行后续的乘法或除法运算。

For example (2 + 3i) + (5 – 7i) = (2 + 5) + (3 – 7)i = 7 – 4i, and (2 + 3i) – (5 – 7i) = (2 – 5) + (3 – (-7))i = -3 + 10i. In AQA exam papers, addition and subtraction usually appear as the first step of a larger question, for instance collecting like terms before moving on to multiplication or division.

四、复数的乘法与i的幂次循环 | 4. Multiplying Complex Numbers and the Cycle of Powers of i

两个复数相乘时，就像展开两个二项式一样使用分配律，同时牢记 i² = -1。例如 (2 + 3i)(4 – i) = 8 – 2i + 12i – 3i²，由于 -3i² = -3 × (-1) = 3，结果等于 8 + 10i + 3 = 11 + 10i。展开过程中出现的 i² 项必须替换成 -1。

To multiply two complex numbers, expand them like two binomials using the distributive law, while always remembering that i² = -1. For example (2 + 3i)(4 – i) = 8 – 2i + 12i – 3i², and since -3i² = -3 × (-1) = 3, the result is 8 + 10i + 3 = 11 + 10i. Any i² term that appears during expansion must be replaced with -1.

i 的幂次遵循一个以 4 为周期的循环，非常值得记住：i¹ = i，i² = -1，i³ = -i，i⁴ = 1，然后 i⁵ 又回到 i。这个规律可以总结为 i 的幂次每 4 个一循环。因此计算 i²⁰ 时，因为 20 是 4 的倍数，结果就是 1；而 i²¹ = i。

The powers of i follow a cycle of period 4 that is well worth memorising: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, and then i⁵ returns to i again. This pattern can be summarised by saying that powers of i repeat every four steps. To compute i²⁰, for example, since 20 is a multiple of 4, the answer is 1; and i²¹ = i.

幂次 Power	i¹	i²	i³	i⁴
结果 Result	i	-1	-i	1

五、共轭复数及其用途 | 5. The Complex Conjugate and Its Uses

复数 z = a + bi 的共轭复数记作 z*（AQA 常用 z*，有的教材写作 z̄），定义是只把虚部符号变号：z* = a – bi。例如 3 + 4i 的共轭是 3 – 4i，而 -5 – 2i 的共轭是 -5 + 2i。实数的共轭就是它本身。

The complex conjugate of z = a + bi is written z* (AQA commonly uses z*, while some textbooks write z̄), and it is defined by simply changing the sign of the imaginary part: z* = a – bi. For example the conjugate of 3 + 4i is 3 – 4i, and the conjugate of -5 – 2i is -5 + 2i. The conjugate of a real number is the number itself.

共轭复数最重要的性质是：一个复数乘以它的共轭，结果总是一个非负的实数。具体地，z × z* = (a + bi)(a – bi) = a² – (bi)² = a² + b²。这个性质是复数除法的关键工具，因为只要把分母乘以它的共轭，分母就从复数变成了实数。

The most important property of the conjugate is this: a complex number multiplied by its conjugate always gives a non-negative real number. Specifically, z × z* = (a + bi)(a – bi) = a² – (bi)² = a² + b². This property is the key tool for division, because multiplying the denominator by its conjugate turns the denominator from a complex number into a real number.

六、复数的除法 | 6. Dividing Complex Numbers

两个复数相除时，我们利用共轭复数的性质，把分母”实数化”。方法就是分子和分母同时乘以分母的共轭。例如要计算 (1 + 2i) ÷ (3 – 4i)，就在分子分母同乘 (3 + 4i)，得到 [(1 + 2i)(3 + 4i)] / [(3 – 4i)(3 + 4i)]。

When dividing two complex numbers, we use the conjugate property to make the denominator real. The method is to multiply the numerator and the denominator together by the conjugate of the denominator. For example, to compute (1 + 2i) ÷ (3 – 4i), multiply top and bottom by (3 + 4i), giving [(1 + 2i)(3 + 4i)] / [(3 – 4i)(3 + 4i)].

接着分别展开：分子 (1 + 2i)(3 + 4i) = 3 + 4i + 6i + 8i² = 3 + 10i – 8 = -5 + 10i；分母 (3 – 4i)(3 + 4i) = 3² + 4² = 25。所以结果是 (-5 + 10i) / 25 = -1/5 + 2/5 i。除法的最终答案必须写成标准形式 a + bi，实部和虚部分开表示。

Then expand each part separately: the numerator (1 + 2i)(3 + 4i) = 3 + 4i + 6i + 8i² = 3 + 10i – 8 = -5 + 10i; the denominator (3 – 4i)(3 + 4i) = 3² + 4² = 25. So the result is (-5 + 10i) / 25 = -1/5 + 2/5 i. The final answer to a division must always be written in standard form a + bi, with the real and imaginary parts separated.

七、阿尔冈图：在平面上表示复数 | 7. The Argand Diagram: Representing Complex Numbers on a Plane

复数可以直观地画在平面上，这个平面叫做阿尔冈图（Argand diagram）。它的横轴（x 轴）表示实部，纵轴（y 轴）表示虚部。于是复数 z = a + bi 就对应平面上的一个点 (a, b)。例如 3 + 4i 对应点 (3, 4)，-2 + i 对应点 (-2, 1)。

Complex numbers can be drawn visually on a plane called the Argand diagram. Its horizontal axis (the x-axis) represents the real part, and its vertical axis (the y-axis) represents the imaginary part. A complex number z = a + bi therefore corresponds to a point (a, b) on the plane. For example 3 + 4i corresponds to the point (3, 4), and -2 + i corresponds to the point (-2, 1).

在阿尔冈图上，共轭复数表现为关于实轴的镜像对称：z = a + bi 在实轴上方，z* = a – bi 就在实轴下方，两点关于 x 轴完全对称。这个几何图像能帮助你理解为什么 z × z* 是实数，也能帮助你快速判断一个复数落在哪个象限。

On the Argand diagram, a number and its conjugate are mirror images across the real axis: z = a + bi lies above the real axis while z* = a – bi lies below it, the two points being perfectly symmetric about the x-axis. This geometric picture helps you understand why z × z* is real, and also helps you quickly judge which quadrant a complex number lies in.

八、模与幅角：复数的极坐标 | 8. Modulus and Argument: Polar Coordinates of a Complex Number

除了用实部和虚部描述一个复数，我们还可以用”距离和方向”来描述它。复数 z = a + bi 的模（modulus）记作 |z|，定义为它到原点的距离，公式是 |z| = √(a² + b²)。例如 3 + 4i 的模是 √(3² + 4²) = 5。模永远是非负的实数。

Besides describing a complex number by its real and imaginary parts, we can also describe it by its distance and direction. The modulus of z = a + bi, written |z|, is defined as its distance from the origin, with the formula |z| = √(a² + b²). For example the modulus of 3 + 4i is √(3² + 4²) = 5. The modulus is always a non-negative real number.

复数 z 的幅角（argument）记作 arg(z)，是从正实轴逆时针转到该复数所在方向的角。它通常以弧度表示，取值范围（主值）是 -π < θ ≤ π。例如 1 + i 的幅角是 π/4，因为它在第一象限与两个坐标轴成 45 度角。计算幅角时要用到反正切，同时必须根据复数所在的象限对结果进行调整。

The argument of z, written arg(z), is the angle measured anticlockwise from the positive real axis to the direction of the complex number. It is usually expressed in radians, and its principal value lies in the range -π < θ ≤ π. For example the argument of 1 + i is π/4, because it makes a 45-degree angle with both axes in the first quadrant. To compute the argument you use the inverse tangent, but you must adjust the result according to the quadrant in which the complex number lies.

复数 z	模 |z|	幅角 arg(z)
1 + i	√2	π/4
-1 + i	√2	3π/4
0 – 3i	3	-π/2

九、模-幅角形式 z = r(cosθ + i sinθ) | 9. Modulus-Argument Form

如果一个复数 z = a + bi 的模是 r、幅角是 θ，那么它的实部 a = r cosθ，虚部 b = r sinθ。于是 z 可以写成模-幅角形式：z = r(cosθ + i sinθ)。这种形式把复数的”距离”和”方向”信息直接写了出来，在乘法和除法中特别有用。

If a complex number z = a + bi has modulus r and argument θ, then its real part is a = r cosθ and its imaginary part is b = r sinθ. We can therefore write z in modulus-argument form: z = r(cosθ + i sinθ). This form writes out the distance and direction information directly, and it is especially useful for multiplication and division.

例如复数 1 + i 的模是 √2、幅角是 π/4，所以它的模-幅角形式是 √2(cos π/4 + i sin π/4)。反过来，如果题目给出模-幅角形式 2(cos π/3 + i sin π/3)，你可以立刻算出 cos π/3 = 1/2、sin π/3 = √3/2，从而还原成标准形式 1 + √3 i。这两种形式之间的互相转换是 AQA 考试的常见考点。

For example the complex number 1 + i has modulus √2 and argument π/4, so its modulus-argument form is √2(cos π/4 + i sin π/4). Conversely, if a question gives the modulus-argument form 2(cos π/3 + i sin π/3), you can immediately evaluate cos π/3 = 1/2 and sin π/3 = √3/2 to recover the standard form 1 + √3 i. Converting between these two forms is a common exam topic in AQA papers.

十、解含复数根的二次方程 | 10. Solving Quadratic Equations with Complex Roots

引入复数之后，任何一个二次方程 ax² + bx + c = 0 现在都有两个解（可能相同）。当判别式 Δ = b² – 4ac 为负数时，方程的解就是一对共轭复数。求根公式仍然是 x = [-b ± √(b² – 4ac)] / 2a，只是根号下的负数要用 i 来处理。

With complex numbers introduced, every quadratic equation ax² + bx + c = 0 now has two solutions (possibly equal). When the discriminant Δ = b² – 4ac is negative, the solutions are a pair of complex conjugates. The quadratic formula is still x = [-b ± √(b² – 4ac)] / 2a, except that the negative number under the square root is handled using i.

例如解方程 x² – 2x + 5 = 0，判别式 Δ = 4 – 20 = -16，所以 √(-16) = 4i，于是 x = [2 ± 4i] / 2 = 1 ± 2i。可以看到两个根 1 + 2i 和 1 – 2i 正好互为共轭。这是一个普遍规律：实系数二次方程若有复数根，它们一定成对共轭出现。

For example, to solve x² – 2x + 5 = 0, the discriminant is Δ = 4 – 20 = -16, so √(-16) = 4i, giving x = [2 ± 4i] / 2 = 1 ± 2i. Notice that the two roots 1 + 2i and 1 – 2i are exactly conjugates of each other. This is a general rule: if a quadratic equation with real coefficients has complex roots, they always occur as a conjugate pair.

十一、AQA考试常见题型与答题技巧 | 11. Common AQA Exam Question Types and Technique

AQA AS 进阶数学关于复数的题目通常按固定的模式设计。常见的第一问是给出两个复数 z₁ 和 z₂，要求计算 z₁ + z₂、z₁z₂ 或 z₁/z₂；第二问往往要求把它们画在阿尔冈图上；第三问则常常要求求模或幅角，并把结果写成模-幅角形式。

AQA AS Further Mathematics questions on complex numbers usually follow a fixed pattern. A common first part gives two complex numbers z₁ and z₂ and asks for z₁ + z₂, z₁z₂, or z₁/z₂; a second part often asks you to plot them on an Argand diagram; a third part frequently asks for the modulus or argument, or for the answer written in modulus-argument form.

答题时有三条技巧值得牢记。第一，每一步都保持标准形式 a + bi，不要在中间步骤混用多种形式。第二，除法务必”同乘共轭”，并清楚写出分母如何变成实数。第三，求幅角时画一个草图，先判断象限再写答案，因为反正切函数本身无法区分相差 π 的角。

Three techniques are worth remembering when answering. First, keep every step in standard form a + bi, and do not mix several different forms within your working. Second, for division always multiply by the conjugate and show clearly how the denominator becomes real. Third, when finding the argument, draw a sketch and decide the quadrant before writing the answer, because the inverse tangent function cannot by itself distinguish angles that differ by π.

此外，纯虚数、实轴上的点、以及共轭点的对称关系，都是 AQA 喜欢用来考察理解的细节。把 i 的幂次循环背熟，能让你在化简形如 i²⁰²⁵ 的式子时节省大量时间。平时练习时建议把”计算、画图、求模与幅角”这三步连成一套完整的流程反复训练。

In addition, purely imaginary numbers, points on the real axis, and the symmetry of conjugate points are all details that AQA likes to use to test understanding. Memorising the cycle of powers of i will save you a great deal of time when simplifying expressions such as i²⁰²⁵. When practising, it is a good idea to link the three steps of calculation, plotting, and finding modulus and argument into one complete routine and rehearse it repeatedly.

十二、用实部与虚部分别相等来解方程 | 12. Equating Real and Imaginary Parts to Solve Equations

复数相等有一个严格的判据：两个复数相等，当且仅当它们的实部相等、虚部也相等。这个看似简单的性质，是 AQA 进阶数学里解”求未知实数”类题目的核心工具。如果题目给出 a + bi = c + di，那么立刻可以得到 a = c 且 b = d 两个方程。

Equality of complex numbers has a strict criterion: two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This seemingly simple property is the core tool for the “find the unknown real numbers” type of question in AQA Further Mathematics. If a question gives a + bi = c + di, then you immediately obtain the two equations a = c and b = d.

例如，已知 (x + yi)² = -5 + 12i，要求实数 x 和 y。先展开左边得到 (x² – y²) + 2xyi，再让实部等于 -5、虚部等于 12，得到方程组 x² – y² = -5 和 2xy = 12。解这个方程组就能求出 x 和 y 的值。这类题目把复数运算与联立方程结合起来，是考试中区分度较高的一类题。

For example, suppose (x + yi)² = -5 + 12i and you are asked to find the real numbers x and y. First expand the left side to get (x² – y²) + 2xyi, then set the real parts equal to -5 and the imaginary parts equal to 12, giving the system of equations x² – y² = -5 and 2xy = 12. Solving this system yields the values of x and y. This type of question combines complex arithmetic with simultaneous equations and is one of the more discriminating question types in the exam.

十三、乘以i的几何意义：旋转90度 | 13. Multiplying by i: A 90-Degree Rotation

在阿尔冈图上，一个复数乘以 i 有一个非常优美的几何解释：它会绕着原点逆时针旋转 90 度。例如 2 + 0i（实轴上的点 2）乘以 i 得到 2i（虚轴上的点），恰好是逆时针转了 90 度；再乘一次 i 得到 -2，又转了 90 度；再乘 i 得到 -2i，继续旋转。

On the Argand diagram, multiplying a complex number by i has a very elegant geometric interpretation: it rotates the point 90 degrees anticlockwise about the origin. For example 2 + 0i (the point 2 on the real axis) multiplied by i gives 2i (a point on the imaginary axis), exactly a 90-degree anticlockwise turn; multiplying by i again gives -2, another 90 degrees; multiplying by i once more gives -2i, continuing the rotation.

这个几何图像解释了为什么 i 的幂次每 4 个一循环：连续乘 4 次 i 就是旋转 360 度，回到原来的位置，所以 i⁴ = 1。理解这个旋转关系，能帮助你在阿尔冈图上快速心算乘法结果，也是 AQA 考察几何理解时的常见角度。

This geometric picture explains why powers of i repeat every four steps: multiplying by i four times in a row rotates through 360 degrees and returns to the starting position, so i⁴ = 1. Understanding this rotation relationship helps you quickly compute multiplication results mentally on the Argand diagram, and it is a common angle AQA uses to test geometric understanding.

十四、由已知复数根构造二次方程 | 14. Constructing a Quadratic Equation from a Given Complex Root

如果已知一个二次方程的一个根是复数，那么它的共轭一定是另一个根，因为实系数二次方程的复数根总是成对出现。利用这一点，我们可以”反向”构造出方程。设一根为 α = p + qi，则另一根为 β = p – qi。

If we know that one root of a quadratic equation is a complex number, then its conjugate must be the other root, because complex roots of a quadratic equation with real coefficients always occur in pairs. Using this fact, we can construct the equation in reverse. Let one root be α = p + qi, so the other root is β = p – qi.

由根与系数的关系，两根之和 S = α + β = 2p，两根之积 P = αβ = p² + q²。于是这个二次方程可以写成 x² – Sx + P = 0，也就是 x² – 2px + (p² + q²) = 0。例如根是 3 + 4i 时，S = 6、P = 25，方程就是 x² – 6x + 25 = 0。你可以用判别式验证：Δ = 36 – 100 = -64，确实有复数根。

From the relationships between roots and coefficients, the sum of the roots is S = α + β = 2p and the product is P = αβ = p² + q². The quadratic equation can therefore be written as x² – Sx + P = 0, that is x² – 2px + (p² + q²) = 0. For example, if the root is 3 + 4i, then S = 6 and P = 25, and the equation is x² – 6x + 25 = 0. You can verify this with the discriminant: Δ = 36 – 100 = -64, which is indeed negative, confirming complex roots.

十五、阿尔冈图上的轨迹：圆与射线 | 15. Loci on the Argand Diagram: Circles and Half-Lines

轨迹（locus）是 AQA 进阶数学里关于复数的进阶考点。最常见的轨迹有两种。第一种是 |z – a| = r，它表示”到定点 a 的距离恒等于 r 的所有点”，在阿尔冈图上是一个以 a 为圆心、r 为半径的圆。例如 |z – 3| = 2 表示圆心在 3（即点 (3,0)）、半径为 2 的圆。

Loci are an advanced topic on complex numbers in AQA Further Mathematics. The two most common loci are the following. The first is |z – a| = r, which represents all points whose distance from the fixed point a is always equal to r; on the Argand diagram this is a circle with centre a and radius r. For example |z – 3| = 2 describes a circle centred at 3 (the point (3,0)) with radius 2.

第二种常见的轨迹是 arg(z – a) = θ，它表示”从定点 a 出发、方向为 θ 的所有点”，在阿尔冈图上是一条以 a 为起点、沿方向 θ 延伸的半直线（射线）。把这两种轨迹与前面的模、幅角定义联系起来，你就能用几何的方法快速判断一个复数满足的条件对应的图形。

The second common locus is arg(z – a) = θ, which represents all points lying in direction θ from the fixed point a; on the Argand diagram this is a half-line (a ray) starting at a and extending in the direction θ. By linking these two loci back to the definitions of modulus and argument, you can quickly identify the geometric figure corresponding to the condition that a complex number satisfies.

十六、常见错误与避坑清单 | 16. Common Mistakes and a Checklist to Avoid Them

复习复数时，有几类错误在 AQA 考试里反复出现。第一类是把 i 写成实数并参与错误运算，例如忘记 i² = -1，直接把 i² 当成 i 或 1。第二类是除法时只乘分母、忘记分子也要同乘共轭，导致答案整体出错。第三类是写答案时把实部和虚部混在一起，没有整理成标准形式 a + bi。

When revising complex numbers, several kinds of mistake recur in AQA exams. The first is treating i as a real number and using it incorrectly, for example forgetting that i² = -1 and treating i² as i or 1. The second is, during division, multiplying only the denominator by the conjugate and forgetting that the numerator must be multiplied as well, which makes the whole answer wrong. The third is mixing the real and imaginary parts together in the final answer instead of tidying it into standard form a + bi.

第四类是求幅角时直接套用 arctan 而不看象限，例如把 -1 + i 的幅角错写成 -π/4，而正确的主值是 3π/4。第五类是在模-幅角形式与标准形式之间转换时，把 sin 和 cos 的位置或符号写反。对照下面这份清单逐条检查，能帮你大幅减少不必要的失分。

The fourth is finding the argument by applying arctan without checking the quadrant, for example writing the argument of -1 + i as -π/4 when the correct principal value is 3π/4. The fifth is swapping or mis-signing sin and cos when converting between modulus-argument form and standard form. Checking against the following list one item at a time will help you greatly reduce unnecessary marks lost.

考试前请确认你已经能做到：化简任何 i 的幂次；用共轭完成除法并把结果写成标准形式；在阿尔冈图上正确标出复数及其共轭；由实部虚部求出模与幅角，并注意幅角的主值范围；把给定根反向构造出二次方程。把这些基础动作练熟，复数这一章就能稳拿分数。

Before the exam, make sure you can do all of the following: simplify any power of i; perform division using the conjugate and write the result in standard form; plot a complex number and its conjugate correctly on the Argand diagram; find the modulus and argument from the real and imaginary parts while observing the principal range of the argument; and construct a quadratic equation from a given root. Once these basic moves are fluent, this chapter will reliably earn marks.

Summary | 总结

复数是 AQA AS 进阶数学第一单元的核心内容。本文从虚数单位 i（满足 i² = -1）出发，依次介绍了复数的标准形式、加减乘除四则运算、共轭复数的性质，以及阿尔冈图和模-幅角这两个几何工具，最后说明了如何解含复数根的二次方程。

Complex numbers are the core of AQA AS Further Mathematics Unit 1. This article started from the imaginary unit i (satisfying i² = -1), then covered the standard form, the four arithmetic operations, the properties of the conjugate, and the two geometric tools of the Argand diagram and the modulus-argument form, and finally showed how to solve quadratic equations with complex roots.

掌握复数的关键在于两点：一是牢记 i² = -1 以及 i 的幂次每 4 个一循环；二是熟练运用”同乘共轭”来完成除法。只要把代数运算与阿尔冈图上的几何图像对应起来，复数这一章就能学得扎实而轻松，为后续的矩阵、根与系数关系等进阶内容打好基础。

The key to mastering complex numbers lies in two points: first, remember that i² = -1 and that powers of i repeat every four steps; second, be fluent in using multiplication by the conjugate to perform division. Once you connect the algebra with the geometric picture on the Argand diagram, this chapter becomes solid and manageable, laying a strong foundation for later topics such as matrices and relationships between roots and coefficients.

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