📚 A-Level Mathematics Example 7.3.1 | A-Level 数学例题 7.3.1
This article walks through a typical A-Level AQA Mathematics question, Example 7.3.1, which often appears under the topic of complex numbers. We will explore the problem statement, break down the solution into clear steps, and connect it to exam-relevant techniques and common pitfalls.
本文围绕 AQA A-Level 数学常见例题 7.3.1 展开，该题通常属于复数专题。我们将梳理题目条件、分步解析解法，并联系考试高频技巧与常见易错点，帮助你稳步掌握这一类题型。
1. Problem Statement | 题目陈述
3.1 typically asks you to consider a complex number given in Cartesian form, say z = 1 + i√3, and to find its modulus and argument. In some variants, you may also be asked to express the number in polar or exponential form, or to plot it on an Argand diagram.
例题 7.3.1 通常要求你考虑一个以直角坐标（笛卡尔）形式给出的复数，例如 z = 1 + i√3，并求出它的模与辐角。在一些变体中，还要求你用极坐标形式或指数形式表示该复数，或在复平面上画出对应点。
The exact wording may vary across editions, but the core operations remain the same: compute the modulus, determine the principal argument, and use these to rewrite the complex number in an alternative form.
不同教材版本的措辞可能略有差异，但核心运算不变：计算模、确定主辐角，并利用它们将复数改写为其他形式。
Let us adopt the following concrete version for this article:
本文采用如下具体版本作为讲解对象：
Given z = 1 + i√3, find |z| and arg(z), and write z in polar form.
已知 z = 1 + i√3，求 |z| 与 arg(z)，并将 z 写成极坐标形式。
2. Prerequisite Knowledge | 前置知识回顾
tial facts about complex numbers. First, if z = a + bi, then the modulus is defined as r = |z| = √a² + b² and it represents the distance from the origin to the point (a, b) in the complex plane.
在进入解法之前，我们先回顾三个关于复数的核心事实。第一，若 z = a + bi，则模定义为 r = |z| = √a² + b²，它表示复平面上点 (a, b) 到原点的距离。
ument is simply θ = arctan(b/a), while for a < 0 we must add or subtract π as appropriate.
第二，辐角 θ 满足 tan θ = b/a，但必须根据点所在的象限来选取正确角度。当 a > 0 时，主辐角直接为 θ = arctan(b/a)；当 a < 0 时，则需要适当加减 π。
s θ + i sin θ), often abbreviated as z = r cis θ. For the exponential form, we use z = r eiθ — note that in this article we will not use LaTeX, so we only state the concept in text and Unicode.
第三，极坐标形式为 z = r(cos θ + i sin θ)，常简写为 z = r cis θ。对于指数形式，使用 z = r eiθ —— 注意本文不使用 LaTeX，因此仅以文字和 Unicode 符号说明这一概念。
3. Step-by-Step Solution | 分步解法
now solve the problem systematically. The complex number we are working with is z = 1 + i√3, where a = 1 and b = √3. Since a > 0 and b > 0, the point lies in the first quadrant, and we do not need to worry about quadrant corrections for the argument.
现在我们系统性地解答该问题。所给的复数是 z = 1 + i√3，其中 a = 1，b = √3。因为 a > 0 且 b > 0，该点位于第一象限，因此计算辐角时不需要进行象限修正。
Step 1: Compute the modulus. Using the formula r = √a² + b², we substitute a = 1 and b = √3:
步骤 1：计算模。 使用公式 r = √a² + b²，代入 a = 1 和 b = √3：
r = √(1² + (√3)²) = √1 + 3 = √4 = 2
Thus, the modulus of z is 2. This tells us that the point representing z lies on a circle of radius 2 centered at the origin.
因此，z 的模为 2。这告诉我们，表示 z 的点位于以原点为圆心、半径为 2 的圆上。
Step 2: Determine the argument. Since tan θ = b/a = √3/1 = √3, we recognise that tan(π/3) = √3. As the point is in the first quadrant, the principal argument is θ = π/3.
步骤 2：确定辐角。 因为 tan θ = b/a = √3/1 = √3，我们注意到 tan(π/3) = √3。由于该点位于第一象限，主辐角为 θ = π/3。
In A-Level exams, both radians and degrees are accepted, but radians are generally preferred in pure mathematics questions. Always check the question requirements or the mode of your calculator.
如果你更习惯角度制，则 θ = 60°。在 A-Level 考试中，弧度与角度都可接受，但在纯数学题目中通常更推荐弧度制。请务必注意题目要求以及计算器的角度模式。
Step 3: Write in polar form. he polar form is z = 2(cos(π/3) + i sin(π/3)). This can also be written as z = 2 cis(π/3).
步骤 3：写出极坐标形式。 由 r = 2 和 θ = π/3，极坐标形式为 z = 2(cos(π/3) + i sin(π/3))。也可以写作 z = 2 cis(π/3)。
If asked for the exponential form, you would simply write z = 2eiπ/3 — the modulus is the coefficient, and the argument is the exponent's imaginary part.
如果有需要写出指数形式，直接写 z = 2eiπ/3 —— 模是系数，辐角是指数的虚部。
4. Visualising the Solution | 图形理解
An Argand diagram helps us interpret these results geometrically. The point (1, √3) is plotted with the real part on the horizontal axis and the imaginary part on the vertical axis. The line segment from the origin to this point has length 2, and the angle it makes with the positive real axis is 60°.
复平面（阿甘图）有助于我们从几何上理解这些结果。点 (1, √3) 的实部在水平轴、虚部在垂直轴上。连接原点到该点的线段长度为 2，它与正实轴之间的夹角为 60°。
This visual interpretation is crucial in A-Level exams. A question might provide a diagram and ask you to identify the modulus and argument, or conversely, to plot a complex number from its polar form. Quick sketching can prevent quadrant errors.
这种几何解释在 A-Level 考试中至关重要。题目可能提供图形，要求你识别模和辐角；或者反过来，让你根据极坐标形式绘制复数点。快速画草图可以有效避免象限错误。
For z = 1 + i√3, the '1' tells us how far right we move, and the '√3' tells us how far up. The right triangle formed has legs of length 1 and √3, and the hypotenuse is exactly 2 — a perfect 30-60-90 triangle.
对于 z = 1 + i√3，'1' 决定向右移动的距离，'√3' 决定向上移动的距离。形成的直角三角形两直角边为 1 和 √3，斜边恰好为 2 —— 这是一个完美的 30-60-90 三角形。
A good habit is to always mark the quadrant on your diagram before computing the argument. In this example, the point is clearly in the first quadrant, so no adjustment is necessary. For second or third quadrant points, you would need to add π or -π as appropriate.
一个好习惯是在计算辐角之前，先在图中标出点所在的象限。本例中，点明显在第一象限，因此不需要做任何调整。对于第二或第三象限的点，则需相应加上 π 或 -π。
5. Table of Key Values | 关键数值表
The following table summarises the intermediate computations in the solution. It is useful for revision and quick reference — similar tables appear in many A-Level revision guides.
下表总结了求解过程中的各步骤计算值，方便复习和快速查阅 —— 类似的表格在许多 A-Level 复习指南中都会出现。
| Quantity | Value | Explanation |
| a | 1 | Real part of z |
| b | √3 | Imaginary part of z |
| r = |z| | 2 | √1 + 3 |
| tan θ | √3 | b/a = √3/1 |
| θ = arg(z) | π/3 | First quadrant, tan⁻¹(√3) |
n consistently, so an answer of 7π/3 would be marked wrong for a principal argument, even though it is a valid co-terminal angle.
注意辐角取的是主辐角，即满足 -π < θ ≤ π。A-Level AQA 一致采用这一约定，所以虽然 7π/3 是合法的同终边角，但如果题目要求主辐角，写成 7π/3 就会被判错。
6. Common Errors and Variations | 常见错误与变式
t frequent mistakes students make is forgetting to check the quadrant. Consider z = -1 + i√3. The modulus is still 2, but the argument is no longer π/3. In fact, the point lies in the second quadrant, so the correct principal argument is 2π/3.
学生最容易犯的错误之一是忘记检查象限。考虑 z = -1 + i√3，模仍然是 2，但辐角不再是 π/3。实际上该点位于第二象限，正确的主辐角是 2π/3。
Another common error is to write the argument in degrees mixed with radian expressions, or to forget the factor of 2 when converting from Cartesian to polar form. Always verify that r equals the distance from the origin to the point, not the sum of the components.
另一个常见错误是将角度制与弧度制混用，或者在从直角坐标转化为极坐标时漏掉因子 2。请务必验证 r 等于原点到点的距离，而不是两个分量的简单加和。
Some textbook variants of Example 7.3.1 use z = √3 + i. In that case, the modulus is still 2, but the argument is π/6 because tan θ = 1/√3. This is a subtle difference: the real and imaginary parts are swapped, so the angle changes from 60° to 30°.
部分教材的例题 7.3.1 变式使用 z = √3 + i。此时模仍为 2，但辐角为 π/6，因为 tan θ = 1/√3。这是一个细微差别：实部和虚部互换，角度从 60° 变为 30°。
Another variation might ask you to verify that z² = -2 + 2√3 i. Using De Moivre' /3 = 2π/3. Converting back to Cartesian form gives 4(cos(2π/3) + i sin(2π/3)) = 4(-1/2 + i√3/2) = -2 + 2√3 i, which matches.
另一种变式可能要求你验证 z² = -2 + 2√3 i。使用棣莫弗定理，z² 的模为 2² = 4，辐角为 2 × π/3 = 2π/3。转化为直角坐标形式得 4(cos(2π/3) + i sin(2π/3)) = 4(-1/2 + i√3/2) = -2 + 2√3 i，与题目一致。
7. Connection to De Moivre's Theorem | 联系棣莫弗定理
and argument, it is often a building block for more advanced questions involving De Moivre's theorem. Once you have z = 2 cis(π/3), you can compute powers easily:
虽然例题 7.3.1 的核心是求模和辐角，但它通常是涉及棣莫弗定理的高阶问题的铺垫。一旦你得到 z = 2 cis(π/3)，就可以轻松计算幂：
z³ = (2 cis(π/3))³ = 8 cis(π) = 8(cos π + i sin π) = -8
Indeed, if you cube the original expression, you get (-2 + 2√3 i)(1 + i√3), which is a computational mess compared to the elegant polar method. This illustrates the power of converting to polar form early.
事实上，如果你直接对原表达式做立方：(−2 + 2√3 i)(1 + i√3)，计算过程会很繁琐，远不如极坐标方法简洁。这就体现出尽早将复数化为极坐标形式的优势。
Similarly, roots of complex numbers are much easier with polar form. For example, to find the cube roots of z, you take the cube root of the modulus (21/3) and divide the argument by 3, then add 2πk/3 for k = 0, 1, 2.
同样地，求复数的根在极坐标形式下也更简单。例如求 z 的立方根，只需将模开三次方（21/3），将辐角除以 3，再对 k = 0, 1, 2 依次加上 2πk/3 即可。
The transition from a single worked example to these broader applications is exactly what examiners expect. Mastering the basic steps of Example 7.3.1 ensures you have the foundation for more complex manipulations.
从单一例题向这些更广应用过渡，正是考官所期待的能力。掌握例题 7.3.1 的基本步骤，确保你拥有处理更复杂运算的基础。
8. Practice Question | 配套练习
To solidify your understanding, try the following question by yourself before checking the answer below. The style is chosen to mirror common exam questions.
为巩固理解，请先独立尝试下面的练习，再对照下方答案。本题风格贴合常见考题。
Question: Given w = -1 - i, find |w| and arg(w), and write w in polar form.
练习题：已知 w = -1 - i，求 |w| 与 arg(w)，并写出 w 的极坐标形式。
walk through the solution. First, a = -1 and b = -1. The modulus is r = √((-1)² + (-1)²) = √2. The point lies in the third quadrant since both coordinates are negative.
我们逐步看解答。首先，a = -1，b = -1。模为 r = √((-1)² + (-1)²) = √2。由于两个坐标均为负，该点在第三象限。
alently 5π/4, but the principal value convention requires -π < θ ≤ π, so we choose -3π/4).
对于辐角，先计算参考角：tan θ_ref = 1，因此 θ_ref = π/4。由于点在第三象限，主辐角为 -3π/4（等价于 5π/4，但主值约定要求 -π < θ ≤ π，因此我们选择 -3π/4）。
Therefore, the polar form is w = √2 π/4), or expanded as w = √2(cos(-3π/4) + i sin(-3π/4)). As a quick check, cos(-3π/4) = -1/√2 and sin(-3π/4) = -1/√2, which gives w = √2(-1/√2 - i/√2) = -1 - i — correct.
因此，极坐标形式为 w = √2 cis(-3π/4)，展开为 w = √2(cos(-3π/4) + i sin(-3π/4))。快速验证：cos(-3π/4) = -1/√2，sin(-3π/4) = -1/√2，于是 w = √2(-1/√2 - i/√2) = -1 - i —— 正确。
9. Exam Tips Summary | 考试技巧总结
Here is a concise checklist that summarises the key takeaways from Example 7.3.1. Refer to this list the night before the exam or whenever you practise similar complex number questions.
下面是例题 7.3.1 的精简要点列表。大考前夜或每次练习类似复数题目时，都可以参照此清单进行回顾。
- Always compute the modulus with r = √a² + b², never r = a + b.
- Determine the quadrant from the signs of a and b before finding the argument.
- Use radians for pure mathematics unless the question explicitly requests degrees.
- Write the final answer in a clear, standard format such as 2 cis(π/3).
- Check your answer by converting polar form back to Cartesian form. It only takes a few seconds and catches many errors.
- 计算模时务必使用 r = √a² + b²，千万不要写成 r = a + b。
- 在求辐角之前，先根据 a 与 b 的符号判断象限。
- 纯数学中默认使用弧度制，除非题目明确要求角度制。
- 将最终答案写成规范清晰的形式，例如 2 cis(π/3)。
- 通过将极坐标还原为直角坐标来验算。这只需要几秒钟，却能发现许多错误。
By internalising these steps, you can solve Example 7.3.1 and its many variants quickly and accurately, freeing up time for more complex exam questions.
通过内化这些步骤，你可以快速准确地解答例题 7.3.1 及其众多变式，为考试中更复杂的问题节省出宝贵时间。
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