一、为什么需要复数：解方程 x2 + 1 = 0 | Why Complex Numbers Exist: Solving x2 + 1 = 0
在实数范围内，方程 x2 + 1 = 0 没有解，因为任何实数的平方都不可能等于 -1。为了解开这类方程，数学家引入了一个新的数 i，定义 i2 = -1。这个符号 i 被称为”虚数单位”。有了 i 之后，形如 x2 + 4 = 0 的方程就可以改写为 x2 = -4，从而得到两个解 x = 2i 和 x = -2i。这样，方程的解就从实数集推广到了复数集。
In the real number system, the equation x2 + 1 = 0 has no solution, because the square of any real number can never equal -1. To solve equations of this kind, mathematicians introduced a new number i, defined by i2 = -1. This symbol i is called the “imaginary unit”. Once i exists, an equation such as x2 + 4 = 0 can be rewritten as x2 = -4, giving the two solutions x = 2i and x = -2i. In this way the solutions of equations are extended from the real numbers to the complex numbers.
复数 z 的一般形式是 z = a + bi，其中 a 和 b 都是实数。a 叫做实部（real part），记作 Re(z)；b 叫做虚部（imaginary part），记作 Im(z)。当 b = 0 时，这个复数就是普通的实数；当 a = 0 且 b 不为 0 时，它叫做纯虚数。实数集和纯虚数集都包含在复数集之中，也就是说复数集是实数集的一个扩充。
A complex number z generally has the form z = a + bi, where a and b are both real numbers. Here a is the real part, written Re(z), and b is the imaginary part, written Im(z). When b = 0 the complex number is just an ordinary real number; when a = 0 and b is not 0 it is called a purely imaginary number. Both the set of real numbers and the set of purely imaginary numbers are contained within the set of complex numbers, which means the complex numbers form an extension of the real numbers.
复数在 Further Pure Mathematics 中占有核心地位，因为许多看似难以解决的问题，例如负数的平方根、带复系数的方程，以及三角恒等式的统一证明，都能借助复数得到简洁而优雅的解法。Edexcel 的 FP1 和 FP2 模块都要求学生熟练地处理复数。
Complex numbers occupy a central position in Further Pure Mathematics, because many problems that at first seem difficult (such as square roots of negative numbers, equations with complex coefficients, and a unified proof of trigonometric identities) can be solved concisely and elegantly using complex numbers. Edexcel’s FP1 and FP2 modules both require students to handle complex numbers fluently.
二、复数的代数形式 z = a + bi 与四则运算 | Cartesian Form z = a + bi and the Four Operations
两个复数的加减法非常直观：分别把实部与实部相加、虚部与虚部相加。例如 (3 + 2i) + (1 – 5i) = 4 – 3i，而 (3 + 2i) – (1 – 5i) = 2 + 7i。这条规则与向量的加减非常相似，为后面理解阿尔冈图埋下了伏笔。做题时建议先整理出实部和虚部，再分别合并，避免符号出错。
Addition and subtraction of two complex numbers is straightforward: add the real parts together and add the imaginary parts together. For example (3 + 2i) + (1 – 5i) = 4 – 3i, while (3 + 2i) – (1 – 5i) = 2 + 7i. This rule is very similar to the addition and subtraction of vectors, which prepares the ground for understanding the Argand diagram later. When working through questions, it is a good habit to separate the real and imaginary parts first, then combine them, so as to avoid sign errors.
乘法需要展开括号，并利用 i2 = -1 进行化简。例如 (3 + 2i)(1 – 5i) = 3 – 15i + 2i – 10i2 = 3 – 13i + 10 = 13 – 13i。请特别注意中间步骤里 i2 被替换成 -1 的那一步，这是最常出错的地方：很多同学会漏掉负号，把 -10i2 错写成 -10。
Multiplication requires expanding the brackets and then simplifying using i2 = -1. For example (3 + 2i)(1 – 5i) = 3 – 15i + 2i – 10i2 = 3 – 13i + 10 = 13 – 13i. Pay special attention to the step where i2 is replaced by -1, because this is the place where mistakes happen most often: many students miss the minus sign and incorrectly write -10i2 as -10.
记住 i 的幂次有一个循环规律：i1 = i，i2 = -1，i3 = -i，i4 = 1，之后每 4 个一循环。因此任何 in 都可以通过 n 除以 4 的余数快速求出。例如 i2025：因为 2025 除以 4 余 1，所以 i2025 = i。
Remember that the powers of i follow a cyclic pattern: i1 = i, i2 = -1, i3 = -i, i4 = 1, after which the pattern repeats every 4 steps. Any power in can therefore be found quickly by looking at the remainder when n is divided by 4. For example i2025: since 2025 divided by 4 leaves a remainder of 1, we have i2025 = i.
三、阿尔冈图：用平面表示复数 | The Argand Diagram: Representing Complex Numbers on a Plane
阿尔冈图（Argand diagram）是把复数画在平面上的方法。横轴表示实部，纵轴表示虚部，于是复数 z = a + bi 就对应平面上坐标为 (a, b) 的一个点。例如 3 + 4i 对应点 (3, 4)，-2 + i 对应点 (-2, 1)。这种几何表示让许多代数性质变得”看得见”。
The Argand diagram is a method of drawing complex numbers on a plane. The horizontal axis represents the real part and the vertical axis represents the imaginary part, so a complex number z = a + bi corresponds to a point with coordinates (a, b) on the plane. For example 3 + 4i corresponds to the point (3, 4), and -2 + i corresponds to the point (-2, 1). This geometric representation makes many algebraic properties “visible”.
在阿尔冈图上，两个复数相加相当于把它们的”位置向量”按平行四边形法则相加。正因为如此，加法和减法可以看作平面上的平移。而乘法和除法在几何上则表现为旋转和伸缩，这一点要等到极坐标形式（模-辐角形式）之后才能充分理解。
On the Argand diagram, adding two complex numbers is equivalent to adding their position vectors by the parallelogram rule. It is for this reason that addition and subtraction can be viewed as translations on the plane. Multiplication and division, on the other hand, correspond geometrically to rotation and scaling, a fact that only becomes fully clear once we meet the polar (modulus-argument) form.
考试中常要求学生把若干复数画在阿尔冈图上，或者根据图上点的位置写出对应的复数。要注意的是，实部决定点的左右位置，虚部决定上下位置，两者一定不要弄反。画图时先标出实轴 (Re) 和虚轴 (Im)，再逐个描点。
In exams students are often asked to plot several complex numbers on an Argand diagram, or to write down the complex number corresponding to a point on the diagram. Note that the real part determines the left-right position of the point and the imaginary part determines the up-down position; the two must never be swapped. When drawing, first label the real axis (Re) and the imaginary axis (Im), then plot the points one by one.
四、模与辐角：从代数形式到几何意义 | Modulus and Argument: From Algebra to Geometry
复数 z = a + bi 的模（modulus）记作 |z|，表示它在阿尔冈图上对应点到原点的距离，公式为 |z| = sqrt(a2 + b2)。例如 |3 + 4i| = sqrt(9 + 16) = 5。模一定是非负的实数，并且 |z| = 0 当且仅当 z = 0。
The modulus of a complex number z = a + bi, written |z|, is the distance from its corresponding point to the origin on the Argand diagram, given by the formula |z| = sqrt(a2 + b2). For example |3 + 4i| = sqrt(9 + 16) = 5. The modulus is always a non-negative real number, and |z| = 0 if and only if z = 0.
辐角（argument）记作 arg z，表示从正实轴方向到该点位置向量的有向角度，通常取 -π 到 π 之间的主值。例如 3 + 4i 的辐角满足 tan θ = 4/3，因此 θ 约为 0.927 弧度（约 53.1 度）。求辐角时务必先判断点在第几象限，再结合反正切的值进行修正，否则容易取到错误的角度。
The argument, written arg z, is the directed angle from the positive real axis to the position vector of the point, usually taken as a principal value between -π and π. For example the argument of 3 + 4i satisfies tan θ = 4/3, so θ is approximately 0.927 radians (about 53.1 degrees). When finding the argument, always decide which quadrant the point lies in first, then adjust the inverse-tangent value accordingly, otherwise it is easy to obtain the wrong angle.
下表总结了四个象限里辐角的取值规律，其中 θ0 = arctan(|b/a|) 是一个 0 到 π/2 之间的锐角参考值。
The table below summarises the rules for the argument in the four quadrants, where θ0 = arctan(|b/a|) is an acute reference angle between 0 and π/2.
| 象限 / Quadrant | a 与 b 的符号 / Signs of a, b | 辐角 / Argument |
|---|---|---|
| 第一象限 / First | a > 0, b > 0 | θ = θ0 |
| 第二象限 / Second | a < 0, b > 0 | θ = π – θ0 |
| 第三象限 / Third | a < 0, b < 0 | θ = -π + θ0 |
| 第四象限 / Fourth | a > 0, b < 0 | θ = -θ0 |
五、共轭复数与复数除法 | The Complex Conjugate and Division
复数 z = a + bi 的共轭复数（complex conjugate）记作 z*（也常写作 z 上方加一横），定义为 z* = a – bi，也就是只把虚部的符号取反。共轭在阿尔冈图上表现为关于实轴的镜像对称。共轭最重要的性质是 z 乘以 z* 等于模的平方：z z* = a2 + b2 = |z|2，这是一个非负实数。
The complex conjugate of z = a + bi, written z* (also often written as z with a bar on top), is defined as z* = a – bi, which is obtained simply by changing the sign of the imaginary part. On the Argand diagram the conjugate appears as a reflection in the real axis. The most important property of the conjugate is that z times z* equals the square of the modulus: z z* = a2 + b2 = |z|2, which is a non-negative real number.
复数除法就是利用共轭来”有理化分母”。例如要计算 (3 + 2i) / (1 – i)，就把分子分母同时乘以分母的共轭 (1 + i)：原式 = (3 + 2i)(1 + i) / (1 – i)(1 + i) = (3 + 5i + 2i2) / (1 + 1) = (1 + 5i) / 2 = 1/2 + (5/2)i。这样就把结果写成了标准的 a + bi 形式。
Division of complex numbers uses the conjugate to “rationalise the denominator”. For example, to compute (3 + 2i) / (1 – i), multiply both numerator and denominator by the conjugate of the denominator, (1 + i): the expression becomes (3 + 2i)(1 + i) / (1 – i)(1 + i) = (3 + 5i + 2i2) / (1 + 1) = (1 + 5i) / 2 = 1/2 + (5/2)i. In this way the result is written in the standard a + bi form.
共轭运算还有几条常用的性质值得记住：两个数和的共轭等于各自共轭的和；(z1 z2)* = z1* z2*，即乘积的共轭等于共轭的乘积。这些性质在证明复数恒等式时非常有用，Edexcel 考试里常有”证明 |z1 z2| = |z1| |z2|”之类的题目。
The conjugate operation also has several useful properties worth remembering: the conjugate of a sum equals the sum of the conjugates, and (z1 z2)* = z1* z2*, that is, the conjugate of a product equals the product of the conjugates. These properties are extremely useful when proving complex identities, and Edexcel exams often contain questions such as “prove that |z1 z2| = |z1| |z2|”.
六、解具有复数根的二次方程与多项式方程 | Solving Quadratic and Polynomial Equations with Complex Roots
有了复数之后，任何二次方程 ax2 + bx + c = 0（其中 a、b、c 为实数）都有解。当判别式 b2 – 4ac 为负数时，方程有一对共轭复数根。例如 x2 – 2x + 5 = 0，判别式为 4 – 20 = -16，因此 x = (2 ± sqrt(-16)) / 2 = (2 ± 4i) / 2 = 1 ± 2i，两个根 1 + 2i 与 1 – 2i 互为共轭。
Once complex numbers are available, every quadratic equation ax2 + bx + c = 0 (with a, b, c real) has solutions. When the discriminant b2 – 4ac is negative, the equation has a pair of complex conjugate roots. For example x2 – 2x + 5 = 0 has discriminant 4 – 20 = -16, so x = (2 ± sqrt(-16)) / 2 = (2 ± 4i) / 2 = 1 ± 2i, and the two roots 1 + 2i and 1 – 2i are conjugates of each other.
这一事实可以推广到实系数多项式：如果 z = a + bi 是一个实系数多项式方程 P(x) = 0 的根，那么它的共轭 z* = a – bi 也一定是根。也就是说，实系数多项式的复数根总是成对出现的。这个结论在 FP1 中经常用来在已知一个复数根的情况下，找出其余所有的根。
This fact generalises to polynomials with real coefficients: if z = a + bi is a root of a polynomial equation P(x) = 0 with real coefficients, then its conjugate z* = a – bi must also be a root. In other words, complex roots of a real polynomial always occur in conjugate pairs. This result is used frequently in FP1 to find all remaining roots once one complex root is known.
例题：已知 1 + 2i 是实系数三次方程 x3 – 3x2 + 7x – 5 = 0 的一个根，求其余两个根。由共轭根定理，1 – 2i 也是一个根。设第三个根为 r，由根与系数的关系，三个根之和等于 3（即 -(系数 x2) 的相反数），于是 (1 + 2i) + (1 – 2i) + r = 3，解得 r = 1。因此三个根为 1 + 2i、1 – 2i 和 1。
Worked example: given that 1 + 2i is a root of the cubic equation x3 – 3x2 + 7x – 5 = 0 with real coefficients, find the other two roots. By the conjugate root theorem, 1 – 2i is also a root. Let the third root be r. From the relationship between roots and coefficients, the sum of the three roots equals 3 (the negative of the coefficient of x2), so (1 + 2i) + (1 – 2i) + r = 3, giving r = 1. Hence the three roots are 1 + 2i, 1 – 2i and 1.
七、复数的极坐标形式：模-辐角形式 | Polar Form: The Modulus-Argument Form
利用模 r 和辐角 θ，任何复数都可以写成极坐标形式（也叫模-辐角形式）：z = r(cos θ + i sin θ)。其中 r = |z| 表示到原点的距离，θ = arg z 表示方向角。例如 3 + 4i 可以写成 5(cos 0.927 + i sin 0.927)。这个形式把”距离”和”方向”两个几何量清晰地分离出来。
Using the modulus r and the argument θ, any complex number can be written in polar form (also called the modulus-argument form): z = r(cos θ + i sin θ). Here r = |z| is the distance to the origin and θ = arg z is the direction angle. For example 3 + 4i can be written as 5(cos 0.927 + i sin 0.927). This form cleanly separates the two geometric quantities of “distance” and “direction”.
极坐标形式最大的威力体现在乘除法上。当两个复数相乘时，模相乘、辐角相加；当两个复数相除时，模相除、辐角相减。用公式表示：若 z1 = r1(cos θ1 + i sin θ1)，z2 = r2(cos θ2 + i sin θ2)，那么 z1 z2 = r1 r2 (cos(θ1 + θ2) + i sin(θ1 + θ2))。这解释了为什么复数乘法在阿尔冈图上表现为”旋转加伸缩”。
The greatest power of the polar form shows up in multiplication and division. When two complex numbers are multiplied, their moduli multiply and their arguments add; when they are divided, their moduli divide and their arguments subtract. In formulas: if z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2), then z1 z2 = r1 r2 (cos(θ1 + θ2) + i sin(θ1 + θ2)). This explains why complex multiplication appears on the Argand diagram as “rotation combined with scaling”.
在 Edexcel 的 FP1 考试中，要求学生能够在代数形式与极坐标形式之间自由转换。给出 z = a + bi 求极坐标形式时，先算 r = sqrt(a2 + b2)，再根据象限确定 θ；反过来，给出极坐标形式时，直接展开 a = r cos θ、b = r sin θ 即可得到代数形式。
In Edexcel FP1 exams, students are required to convert freely between the Cartesian form and the polar form. Given z = a + bi and asked for the polar form, first compute r = sqrt(a2 + b2), then determine θ according to the quadrant; conversely, given the polar form, simply expand a = r cos θ and b = r sin θ to obtain the Cartesian form.
八、棣莫弗定理与单位根 | De Moivre’s Theorem and Roots of Unity
棣莫弗定理（De Moivre’s theorem）是极坐标形式的直接延伸：对任意整数 n，(cos θ + i sin θ)n = cos(nθ) + i sin(nθ)。也就是说，求一个复数的幂，只需把辐角乘以 n，模再作相应次方。这个定理极大地简化了高次幂的计算，也常常用来推导倍角的三角恒等式。
De Moivre’s theorem is a direct extension of the polar form: for any integer n, (cos θ + i sin θ)n = cos(nθ) + i sin(nθ). In other words, to raise a complex number to a power, simply multiply its argument by n and raise its modulus to the corresponding power. This theorem greatly simplifies the computation of high powers, and is also used to derive double-angle and multiple-angle trigonometric identities.
一个经典应用是计算单位根（roots of unity），也就是方程 zn = 1 的解。这个方程恰好有 n 个不同的复数解，它们在阿尔冈图上均匀地分布在一个以原点为圆心、半径为 1 的单位圆上，相邻两个根之间的夹角为 2π/n。例如方程 z3 = 1 有三个根：1、cos(2π/3) + i sin(2π/3) 和 cos(4π/3) + i sin(4π/3)。
A classic application is computing the roots of unity, that is, the solutions of the equation zn = 1. This equation has exactly n distinct complex solutions, which lie evenly spaced on the unit circle centred at the origin with radius 1, with an angle of 2π/n between any two neighbouring roots. For example the equation z3 = 1 has three roots: 1, cos(2π/3) + i sin(2π/3) and cos(4π/3) + i sin(4π/3).
利用棣莫弗定理还可以推导出许多重要的三角恒等式。例如考虑 (cos θ + i sin θ)3 的两种写法：一方面按棣莫弗定理它等于 cos 3θ + i sin 3θ；另一方面用二项式展开并利用 i2 = -1，再令两边虚部相等，就能得到 sin 3θ = 3 sin θ – 4 sin3 θ。这种”实部虚部分别比较”的技巧在 FP2 中非常常见。
De Moivre’s theorem can also be used to derive many important trigonometric identities. For example, consider the two ways of writing (cos θ + i sin θ)3: on one hand, by De Moivre’s theorem it equals cos 3θ + i sin 3θ; on the other hand, expanding by the binomial theorem and using i2 = -1, then equating the imaginary parts of both sides, gives sin 3θ = 3 sin θ – 4 sin3 θ. This technique of “comparing real and imaginary parts separately” is very common in FP2.
九、常见易错点与考试技巧 | Common Pitfalls and Exam Techniques
第一个高频错误是化简乘法时漏掉 i2 = -1 里的负号，例如把 -10i2 直接当成 -10 而不是 +10。第二个错误是求辐角时不判断象限，直接把 arctan(b/a) 当作最终答案，导致落在第二、三象限的角取错。第三个错误是极坐标形式与代数形式转换时把 cos 和 sin 记混，或者忘了先求出模 r。
The first frequent mistake is dropping the minus sign in i2 = -1 when simplifying a product, for example treating -10i2 directly as -10 instead of +10. The second mistake is failing to check the quadrant when finding the argument, and taking arctan(b/a) directly as the final answer, which gives the wrong angle for points in the second or third quadrant. The third mistake is mixing up cos and sin when converting between polar and Cartesian forms, or forgetting to compute the modulus r first.
考试中还有一类”证明性质”的题目，例如证明 |z1 z2| = |z1| |z2| 或 arg(z1 z2) = arg z1 + arg z2。处理这类题的通用思路是先把两个复数都写成极坐标形式，再利用乘法规则直接得出结论，比在代数形式下展开要干净得多。答题时记得把”取模”和”取辐角”分两步写清楚。
Exams also contain a class of “prove a property” questions, such as proving |z1 z2| = |z1| |z2| or arg(z1 z2) = arg z1 + arg z2. The general approach to these questions is to write both complex numbers in polar form first, then use the multiplication rule to reach the conclusion directly; this is much cleaner than expanding in Cartesian form. When answering, remember to write the “take modulus” and “take argument” steps separately and clearly.
最后建议：答题时始终把结果整理成标准的 a + bi 形式，除非题目明确要求极坐标形式。涉及多个复数运算时，每一步都检查实部和虚部是否分离正确。对于单位根和棣莫弗定理的题目，先把复数写成 cos + i sin 的形式再套定理，可以避免大量繁琐的展开。
A final piece of advice: always tidy up the result into the standard a + bi form, unless the question explicitly asks for the polar form. When a question involves several operations on complex numbers, check at every step that the real and imaginary parts are separated correctly. For questions on roots of unity and De Moivre’s theorem, first write the complex number in cos + i sin form before applying the theorem, which avoids a great deal of tedious expansion.
十、复数的几何轨迹：圆与射线 | Geometric Loci in the Complex Plane: Circles and Rays
复数还有一个重要的几何应用：描述平面上的轨迹（locus）。形如 |z – a| = r 的方程表示以 a 对应的点为圆心、r 为半径的圆；形如 arg(z – a) = θ 的方程表示从点 a 出发、与正实轴成角 θ 的一条射线（半直线）。这类问题把复数与几何轨迹直接联系起来，是 Edexcel FP2 的重要考点。
Complex numbers have another important geometric application: describing loci on the plane. An equation of the form |z – a| = r represents a circle centred at the point corresponding to a, with radius r; an equation of the form arg(z – a) = θ represents a ray (half-line) starting from the point a and making an angle θ with the positive real axis. This kind of problem connects complex numbers directly with geometric loci, and is an important topic in Edexcel FP2.
理解这些轨迹的关键在于记住 |z – a| 的几何意义：它是 z 对应的点与 a 对应的点之间的距离。因此 |z – a| = r 就是”到定点 a 的距离恒等于 r 的所有点”，这正是圆的定义。同理，arg(z – a) 是向量 (z – a) 的方向角，令它等于固定角度 θ，就得到一条射线。
The key to understanding these loci is to remember the geometric meaning of |z – a|: it is the distance between the point representing z and the point representing a. Therefore |z – a| = r means “all points whose distance to the fixed point a is exactly r”, which is precisely the definition of a circle. Similarly, arg(z – a) is the direction angle of the vector (z – a), and setting it equal to a fixed angle θ gives a ray.
例题：在阿尔冈图上画出满足 |z – (2 + i)| = 3 的点的轨迹，并说明它是什么图形。由于 |z – (2 + i)| 表示 z 到点 (2, 1) 的距离，该方程表示以 (2, 1) 为圆心、半径为 3 的圆。而满足 arg(z – i) = π/4 的点，则构成从点 (0, 1) 出发、方向角为 π/4（即 45 度）的一条射线，且这条射线不包括起点本身。
Worked example: on an Argand diagram, sketch the locus of points satisfying |z – (2 + i)| = 3 and state what shape it is. Since |z – (2 + i)| is the distance from z to the point (2, 1), the equation represents a circle with centre (2, 1) and radius 3. Meanwhile, the points satisfying arg(z – i) = π/4 form a ray starting from the point (0, 1) with direction angle π/4 (that is, 45 degrees), and this ray does not include its starting point itself.
Summary | 总结
复数是 Further Pure Mathematics 的基石：虚数单位 i 满足 i2 = -1，使得 x2 + 1 = 0 这类方程第一次有了解。复数的代数形式 z = a + bi 支持加减乘除四则运算，阿尔冈图则把复数变成平面上的点，让模（到原点的距离）和辐角（方向角）都有了直观的几何意义。共轭复数把除法转化为乘法，并揭示了实系数多项式复数根成对出现的规律。
Complex numbers are the cornerstone of Further Pure Mathematics: the imaginary unit i satisfies i2 = -1, giving equations such as x2 + 1 = 0 a solution for the first time. The Cartesian form z = a + bi supports the four arithmetic operations, while the Argand diagram turns complex numbers into points on a plane, giving the modulus (distance to the origin) and the argument (direction angle) a clear geometric meaning. The complex conjugate turns division into multiplication, and reveals the rule that complex roots of a real polynomial occur in conjugate pairs.
极坐标形式 z = r(cos θ + i sin θ) 让乘除法的几何本质（模相乘除、辐角相加减）一目了然，进而引出棣莫弗定理 (cos θ + i sin θ)n = cos(nθ) + i sin(nθ) 和单位根 zn = 1 的均匀分布解。掌握这些内容，就能从容应对 Edexcel FP1 和 FP2 中绝大多数关于复数的题目。
The polar form z = r(cos θ + i sin θ) makes the geometric nature of multiplication and division (moduli multiply or divide, arguments add or subtract) immediately clear, and in turn leads to De Moivre’s theorem (cos θ + i sin θ)n = cos(nθ) + i sin(nθ) and the evenly spaced solutions of the roots of unity zn = 1. Once these ideas are mastered, you can handle the vast majority of complex-number questions in Edexcel FP1 and FP2 with confidence.
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