📚 A-Level AQA Further Maths: Complex Numbers and Functions | A-Level AQA 数学：复变函数 考点精讲
Complex numbers and functions form a cornerstone of AQA Further Pure Mathematics, extending the real number system to solve equations that would otherwise have no solution and to model rotations and transformations in the plane. Understanding their algebra, geometry, and calculus is essential for high marks on Papers 1 and 2. In this revision guide, we break down every key concept you need to master, from Cartesian and polar forms to de Moivre’s theorem and complex transformations, with clear explanations and common pitfalls highlighted along the way.
复数是 AQA 进阶纯数中的核心内容，它将实数系进行了扩展，使得原本无解的方程变得可解，同时还能用来描述平面上的旋转与变换。掌握复数的代数运算、几何意义以及相关微积分知识，对于在试卷一和试卷二中取得高分至关重要。在这份复习指南中，我们梳理了从直角坐标与极坐标形式到棣莫弗定理和复变函数变换等每一个你必须掌握的关键考点，并特别指出了常见的失分陷阱。
1. Cartesian and Polar Forms | 直角坐标与极坐标形式
A complex number z can be written in Cartesian form as z = x + iy, where x = Re(z) is the real part and y = Im(z) is the imaginary part. For many AQA problems, especially those involving multiplication, division and powers, the polar form z = r(cos θ + i sin θ) is far more powerful. Here r = |z| is the modulus and θ = arg(z) is the argument, usually chosen such that –π < θ ≤ π.
复数 z 可以用直角坐标形式表示为 z = x + iy，其中 x = Re(z) 是实部，y = Im(z) 是虚部。在许多 AQA 考题中，尤其是涉及乘法、除法和幂运算时，极坐标形式 z = r(cos θ + i sin θ) 更为强大。这里的 r = |z| 是模，θ = arg(z) 是辐角，通常选取的范围是 –π < θ ≤ π。
When converting between forms, a quick sketch helps avoid errors in the quadrant of θ. Use tan θ = y/x and adjust θ depending on the signs of x and y. For example, z = 1 – i √3 gives r = 2 and tan θ = –√3, so the principal argument is –π/3, not 2π/3.
在两种形式之间转换时，画一个简图可以避免辐角所在象限的错误。利用 tan θ = y/x 再根据 x, y 的符号调整 θ。例如 z = 1 – i √3，可得 r = 2， tan θ = –√3，因此主辐角为 –π/3 而非 2π/3。
z = x + iy = r(cos θ + i sin θ), r = √(x² + y²), tan θ = y/x
2. Modulus and Argument | 模与辐角
The modulus |z| is the distance from the origin to the point (x, y) in the complex plane. It satisfies |z₁z₂| = |z₁||z₂| and |z₁ / z₂| = |z₁| / |z₂|. The argument arg(z) gives the angle, with the property arg(z₁z₂) = arg(z₁) + arg(z₂) and arg(z₁/z₂) = arg(z₁) – arg(z₂), always taken modulo 2π to stay in the principal range.
模 |z| 表示复平面上点 (x, y) 到原点的距离，满足 |z₁z₂| = |z₁||z₂| 以及 |z₁ / z₂| = |z₁| / |z₂|。辐角 arg(z) 给出了角度，且满足 arg(z₁z₂) = arg(z₁) + arg(z₂) 以及 arg(z₁/z₂) = arg(z₁) – arg(z₂)，通常都需要用 2π 的整数倍进行加减以保持在主值范围内。
Be precise when you state an argument: use radian measure and pay attention to the sign. For negative real numbers the argument is π, not –π, unless the question specifically asks for a negative principal value.
在描述辐角时要精确：使用弧度制并注意正负号。对于负实数，主辐角为 π，而不是 –π，除非题目明确要求使用负的主值。
3. Complex Conjugates | 共轭复数
The conjugate of z = x + iy is z* = x – iy (also denoted as z̄). Graphically, it reflects the point across the real axis. Key properties include z z* = |z|², (z₁ ± z₂)* = z₁* ± z₂*, (z₁z₂)* = z₁* z₂*, and (z₁ / z₂)* = z₁* / z₂*. Conjugates are extremely useful for dividing complex numbers and for proving results involving real parts.
复数 z = x + iy 的共轭复数是 z* = x – iy（也记作 z̄）。在几何上，它表示该点关于实轴的反射。重要性质包括 z z* = |z|²，(z₁ ± z₂)* = z₁* ± z₂*，(z₁z₂)* = z₁* z₂* 以及 (z₁ / z₂)* = z₁* / z₂*。共轭复数在复数除法以及涉及实部的证明中极为有用。
To divide two complex numbers, multiply numerator and denominator by the conjugate of the denominator, turning the denominator into a real number. This is a routine skill that AQA examiners expect to be performed quickly and accurately.
进行复数除法时，将分子和分母同时乘以分母的共轭复数，分母就变成了实数。这是一项基础技能，AQA 考官希望考生能够迅速且准确地完成。
(a + ib) / (c + id) = (a + ib)(c – id) / (c² + d²)
4. Operations in Polar Form | 极坐标形式的运算
Multiplication and division in polar form reveal the geometric beauty of complex numbers: rotate and scale. If z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), then z₁z₂ = r₁r₂(cos(θ₁+θ₂) + i sin(θ₁+θ₂)) and z₁/z₂ = r₁/r₂(cos(θ₁–θ₂) + i sin(θ₁–θ₂)). This is the basis for de Moivre’s theorem and for understanding complex transformations.
极坐标形式下的乘除法揭示了复数的几何美：旋转与缩放。若 z₁ = r₁(cos θ₁ + i sin θ₁)，z₂ = r₂(cos θ₂ + i sin θ₂)，则 z₁z₂ = r₁r₂(cos(θ₁+θ₂) + i sin(θ₁+θ₂))，z₁/z₂ = r₁/r₂(cos(θ₁–θ₂) + i sin(θ₁–θ₂))。这是棣莫弗定理以及理解复变函数变换的基础。
A common exam question gives a complex number in polar form and asks you to find z₁z₂, then interpret the result geometrically. You should be able to describe the rotation (by adding arguments) and the scaling (by multiplying moduli) clearly.
常见的考题是给出极坐标形式的复数，要求计算 z₁z₂，并解释其几何意义。你需要能够清晰地描述旋转（辐角相加）和缩放（模相乘）的效果。
5. De Moivre’s Theorem | 棣莫弗定理
De Moivre’s theorem states that for any integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). This extends to rational exponents when finding roots, though careful treatment of multiple values is needed. The theorem is the engine behind deriving trigonometric identities and finding powers of complex numbers quickly.
棣莫弗定理指出，对于任意整数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。该定理也可以扩展到有理指数来求根，但需要小心处理多值情况。这一理论是推导三角恒等式以及快速计算复数幂次的核心工具。
For a complex number z = r(cos θ + i sin θ), zⁿ = rⁿ(cos(nθ) + i sin(nθ)). AQA often asks you to express cos 3θ or sin 4θ in terms of powers of cos θ and sin θ using the binomial expansion together with de Moivre. Practice setting cos nθ = Re(cos θ + i sin θ)ⁿ and equating real and imaginary parts.
对于 z = r(cos θ + i sin θ)，有 zⁿ = rⁿ(cos(nθ) + i sin(nθ))。AQA 经常要求利用二项式展开与棣莫弗定理，将 cos 3θ 或 sin 4θ 表示成 cos θ 和 sin θ 幂次的组合。务必练习设 cos nθ = Re(cos θ + i sin θ)ⁿ，然后分别令实部和虚部相等。
zⁿ = rⁿ (cos(nθ) + i sin(nθ))
6. Roots of Complex Numbers | 复数的根
To solve zⁿ = w, write w in polar form w = R(cos φ + i sin φ). Then the n distinct n-th roots are given by zₖ = R^(1/n) [cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)] for k = 0, 1, …, n – 1. Geometrically, the roots lie on a circle of radius R^(1/n) and are equally spaced by an angle of 2π/n.
要解方程 zⁿ = w，先把 w 写成极坐标形式 w = R(cos φ + i sin φ)，那么 n 个不同的 n 次方根由 zₖ = R^(1/n) [cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)] 给出，其中 k = 0, 1, …, n – 1。几何上，这些根均匀分布在半径为 R^(1/n) 的圆上，相邻两根的辐角差为 2π/n。
When finding roots, always list all k values explicitly and, if asked, sketch the positions on the Argand diagram. Examiners look for correct symmetry: if a polynomial has real coefficients, roots occur in conjugate pairs.
求解根时，要明确列出所有 k 值对应的根，并要求时在阿尔冈图上画出它们的位置。考官会关注对称性是否正确：如果多项式系数为实数，则根会以共轭对的形式出现。
7. Euler’s Formula | 欧拉公式
Euler’s formula e^(iθ) = cos θ + i sin θ provides a compact exponential notation for polar form: z = r e^(iθ). Multiplication and division become straightforward: z₁z₂ = r₁r₂ e^(i(θ₁+θ₂)). De Moivre’s theorem is simply (e^(iθ))ⁿ = e^(inθ). This notation is used heavily in AQA for advanced manipulations, especially in FM2 with differential equations.
欧拉公式 e^(iθ) = cos θ + i sin θ 为极坐标形式提供了一个简洁的指数记法：z = r e^(iθ)。此时乘除法变得异常简单：z₁z₂ = r₁r₂ e^(i(θ₁+θ₂))，棣莫弗定理也仅仅是 (e^(iθ))ⁿ = e^(inθ)。AQA 在进阶运算中大量运用此记法，尤其是在 FM2 微分方程部分。
You must be comfortable switching between the three forms: x + iy, r(cos θ + i sin θ), and r e^(iθ). Many questions link these by asking you to express a complex number in exponential form and then use it to solve equations or evaluate integrals.
你必须熟练掌握三种形式之间的转换：x + iy，r(cos θ + i sin θ) 以及 r e^(iθ)。很多题目会把它们串联起来，要求先用指数形式表示一个复数，再用于解方程或计算积分。
8. Loci and Geometry | 轨迹与几何
Argand diagrams bring complex equations to life. Common loci include: |z – a| = r, a circle centred at a; |z – a| = |z – b|, the perpendicular bisector of the segment joining a and b; arg(z – a) = θ, a half-line from a at angle θ. Intersection of loci often yields the roots of an equation geometrically.
阿尔冈图让复数方程变得具象化。常见的轨迹包括：|z – a| = r 代表以 a 为圆心、r 为半径的圆；|z – a| = |z – b| 表示连接 a 和 b 线段的垂直平分线；arg(z – a) = θ 表示从 a 出发、与正实轴夹角为 θ 的射线。轨迹的交点往往就对应方程根的几何位置。
AQA frequently asks you to sketch or identify the locus given a condition, or to find the Cartesian equation of the locus. Use substitution z = x + iy and simplify using modulus and argument properties. Checking boundary behaviour and whether the locus includes the point at infinity is often part of the mark scheme.
AQA 经常要求根据条件画出或识别出轨迹，或者求出轨迹的笛卡尔方程。此时可代入 z = x + iy，并利用模和辐角的性质进行化简。注意边界情况以及轨迹是否包含无穷远点，这些细节通常在评分标准中有体现。
9. Transformations by Complex Functions | 复变函数的变换
Mappings of the form w = f(z), such as w = z + a (translation), w = az (rotation and enlargement), or w = 1/z (inversion), appear in AQA Further Maths syllabuses. Understanding how lines and circles transform under these elementary mappings is a crucial skill. For instance, w = 1/z maps lines not passing through the origin to circles through the origin, and vice versa.
形如 w = f(z) 的映射，例如 w = z + a（平移）、w = az（旋转与缩放）或 w = 1/z（反演），是 AQA 进阶数学考纲中的重要内容。理解直线和圆在这些初等映射下如何变换是一项关键技能。例如，w = 1/z 将不经过原点的直线映射为过原点的圆，反之亦然。
When a transformation is given, substitute z = x + iy, then express w = u + iv. Equate real and imaginary parts to find relations between u, v and x, y. Eliminate x, y to obtain the equation of the image locus in the w-plane. Practice with T: z ↦ w = (z – i)/(z + i) and the famous Möbius transformations.
给定一个变换后，可代入 z = x + iy，然后写出 w = u + iv。令实部和虚部分别相等，建立 u、v 与 x、y 的关系式，再消去 x、y 得到像在 w 平面上的轨迹方程。多练习 T: z ↦ w = (z – i)/(z + i) 以及经典的莫比乌斯变换。
10. Solving Equations with Complex Roots | 求解含复根方程
Polynomial equations with real coefficients always have roots that occur in conjugate pairs. If you find one complex root a + ib, then a – ib is also a root. Use this fact to factorise cubic or quartic equations quickly. For example, given that 2 + i is a root of z³ – z² – 7z + 15 = 0, the other two roots are its conjugate 2 – i and a real root found by comparing the product of the two complex roots.
具有实系数的多项式方程，其根总是成对共轭出现。一旦找到一个复根 a + ib，则 a – ib 也是一个根。利用这个事实可以快速对三次或四次方程进行因式分解。例如，若已知 2 + i 是 z³ – z² – 7z + 15 = 0 的一个根，则另外两个根为其共轭 2 – i 以及一个实根（通过比较两个复根的乘积求得）。
AQA may also ask you to solve equations like zⁿ = a complex number, or to use substitution to reduce an equation to a simpler polynomial in z or w. Always check the general solution and account for all branches.
AQA 还可能要求你求解 zⁿ = 某个复数 这样的方程，或利用代换将方程化为关于 z 或 w 的更简单多项式。务必牢记通解形式，并列出所有分支。
11. Proofs and Trigonometric Applications | 证明与三角应用
A significant proportion of AQA marks comes from proving identities using complex numbers. Typical tasks: prove that cos 5θ = 16 cos⁵θ – 20 cos³θ + 5 cos θ by expanding (cos θ + i sin θ)⁵ and equating real parts. Also, you may need to sum series like ∑ cos kθ using geometric series with e^(iθ).
AQA 试卷中相当比例的分值来自于利用复数证明恒等式。典型任务是：将 (cos θ + i sin θ)⁵ 展开并取实部，从而证明 cos 5θ = 16 cos⁵θ – 20 cos³θ + 5 cos θ。此外，你可能还需要借助 e^(iθ) 的等比数列来求和，例如 ∑ cos kθ。
Another classic proof: for any complex numbers z and w, |z + w| ≤ |z| + |w| (triangle inequality) and its consequence in geometric problems. Be ready to justify each algebraic step, especially when manipulating arguments and moduli.
另一个经典证明是：对于任意复数 z 和 w，有 |z + w| ≤ |z| + |w|（三角不等式）及其在几何问题中的推论。每一步代数变换都要有明确的理由，尤其是在处理辐角和模时。
cos nθ = Re (e^(inθ)), sin nθ = Im (e^(inθ))
12. Common Exam Pitfalls and Tips | 常见失分陷阱与应考技巧
Pitfall 1: Using degrees instead of radians for arguments. AQA demands radian measure unless specified otherwise. Always set your calculator to radian mode for the Further Maths papers.
陷阱一：在表示辐角时使用角度而非弧度。除非特别说明，AQA 要求使用弧度制。在进阶数学考试中，务必将计算器设置为弧度模式。
Pitfall 2: Forgetting that |z| is always non-negative. When solving |z – a| = 2, students sometimes write z = a ± 2 without considering that this describes two points, not a circle. Carefully distinguish between equations in real variables and complex equations.
陷阱二：忘记 |z| 始终是非负的。解 |z – a| = 2 时，有的同学习惯性地写成 z = a ± 2，却没有意识到这描述的是两个点，而不是一个圆。要仔细区分实数方程与复数方程。
Pitfall 3: Losing a root when taking the nth root. For z³ = 8, the roots are not just z = 2, but also the two complex cube roots of unity times 2. Write out all k values.
陷阱三：开 n 次方根时丢掉根。对于 z³ = 8，根不仅仅是 z = 2，还有两个复数立方根乘以 2。务必写出所有 k 值对应的根。
Pitfall 4: Misapplying de Moivre for non-integer n. When n is rational, you must add 2kπ before dividing by the denominator to obtain all roots. Direct application (cos θ + i sin θ)^(1/n) = cos(θ/n) + i sin(θ/n) only gives one root and will lose marks.
陷阱四：对非整数 n 错误使用棣莫弗定理。当 n 为有理数时，必须在除以分母之前加上 2kπ，才能得到所有根。直接使用 (cos θ + i sin θ)^(1/n) = cos(θ/n) + i sin(θ/n) 只能得到一个根，会严重失分。
Tip: Always double-check whether a locus includes boundaries or endpoints. A half-line arg(z – a) = θ does not include point a itself; indicate this with an open circle on your sketch.
应考技巧：始终检查轨迹是否包含边界或端点。射线 arg(z – a) = θ 不包含点 a 本身，在作图时需要用空心圆标示。
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