一、为什么我们需要复数？从负数的平方根说起 | Why Do We Need Complex Numbers? Starting from the Square Root of Negatives
在实数范围内，当我们尝试对一个负数取平方根时，运算会立刻失败。例如，√-1 在实数系中没有任何对应值，因为任何实数的平方都是非负数。这看起来像是一个数学上的”死胡同”，但对于16世纪的意大利数学家来说，这个问题恰恰是解开三次方程求根公式的关键钥匙。他们在推导过程中发现，即使最终答案是实数，中间步骤也必须经过”虚数”的领域。这个发现彻底改变了数学的格局。
Within the real number system, attempting to take the square root of a negative number immediately breaks down. For instance, √-1 has no counterpart in the reals, because the square of any real number is non-negative. This looks like a mathematical dead end, but for 16th-century Italian mathematicians, this very problem turned out to be the key that unlocked the general cubic formula. They discovered that even when the final answer is real, the intermediate steps must pass through the realm of “imaginary” numbers. This discovery fundamentally reshaped the landscape of mathematics.
今天的AS进阶数学课程中，复数是FM02模块的核心内容之一。复数不仅仅是书本上的抽象概念 – 它们在电子工程（交流电路分析）、量子力学（波函数）、信号处理和控制理论中都有不可替代的实际应用。理解复数，你就拥有了一把进入高等数学和工程世界的钥匙。
In today’s AS Further Mathematics course, complex numbers form one of the core topics of the FM02 module. Complex numbers are not just an abstract concept in a textbook – they have irreplaceable real-world applications in electrical engineering (AC circuit analysis), quantum mechanics (wave functions), signal processing, and control theory. Mastering complex numbers gives you the key to higher mathematics and the engineering world.
二、虚数单位 i 的定义与基本性质 | The Imaginary Unit i: Definition and Fundamental Properties
虚数单位 i 是复数理论的基石。它的定义非常简洁：i² = -1，或者说 i = √-1。这个看似简单的定义却带来了深远的影响。一旦我们接受了 i 的存在，就可以定义任何一个负数的平方根：对于任意正数 a，√-a = i√a。
of complex number theory. Its definition is elegantly simple: i² = -1, or equivalently i = √-1. This deceptively simple definition carries profound consequences. Once we accept the existence of i, we can define the square root of any negative number: for any positive number a, √-a = i√a.
与实数不同，i 的幂次呈现出周期性的规律。计算 i 的各次幂：i¹ = i，i² = -1，i³ = -i，i⁴ = 1。注意 i⁴ = 1 之后，幂次模式以4为周期循环。这意味着任何 iⁿ 都可以通过将指数 n 除以4，取余数来快速简化。例如，i²⁰²³：2023 ÷ 4 = 505 余 3，所以 i²⁰²³ = i³ = -i。这种周期性是复数运算中的一个重要捷径，AQA考试中经常考察对 i 的幂次规律的掌握。
le, i²⁰²³: 2023 ÷ 4 = 505 remainder 3, so i²⁰²³ = i³ = -i. This periodicity is an important shortcut in complex number operations, and AQA exams frequently test mastery of the power pattern of i.
三、复数的标准形式：a + bi 的完整拆解 | The Standard Form a + bi: A Complete Breakdown
一个复数 z 的标准形式写作 z = a + bi，其中 a 和 b 都是实数。a 被称为”实部”（Real Part），记作 Re(z)；b 被称为”虚部”（Imaginary Part），记作 Im(z)。注意：虚部是 b，不是 bi – 这是一个常见的考试陷阱！例如，对于复数 3 + 4i：Re(z) = 3，Im(z) = 4（不是 4i）。
dard form is written as z = a + bi, where both a and b are real numbers. a is called the “real part”, written as Re(z); b is called the “imaginary part”, written as Im(z). Pay careful attention: the imaginary part is b, not bi – this is a common exam trap! For example, for 3 + 4i: Re(z) = 3, Im(z) = 4 (not 4i).
两个复数相等，当且仅当它们的实部和虚部分别相等。也就是说，a + bi = c + di 意味着 a = c 且 b = d。这个看似平凡的性质在解含有复数的方程时极其有用 – 你可以将方程”拆分”为两个实数方程分别求解。纯实数（如 5）也可以写成复数形式 5 + 0i；纯虚数（如 3i）可以写成 0 + 3i。复数域包含了实数域作为其子集。
ingly trivial property is extremely useful when solving equations involving complex numbers – you can “split&# ginary numbers (like 3i) can be written as 0 + 3i. The complex number field contains the real number field as a subset.
四、复数的加减乘除四则运算 | The Four Basic Operations on Complex Numbers
复数的加减法非常直观 – 只需要将实部和虚部分别相加或相减。对于 z₁ = a + bi 和 z₂ = c + di：加法 z₁ + z₂ = (a + c) + (b + d)i，减法 z₁ – z₂ = (a – c) + (b – d)i。这与向量的加法在形式上完全一致，这也是为什么我们可以把复数表示在二维平面上。
Addition and subtraction of complex numbers are very straightforward – . For z₁ = a + bi and z₂ = c + di: addition z₁ + z₂ = (a + c) + (b + d)i, subtraction z₁ – z₂ = (a – c) + (b &# vector addition, which is why we can represent complex numbers on a two-dimensional plane.
乘法稍微复杂一些，但仍然遵循代数分配律。z₁ × z₂ = (a + bi)(c + di) = ac + adi + bci + bdi²。由于 i² = -1，最后一项变为 -bd。整理后得到：z₁z₂ = (ac – bd) + (ad + bc)i。这个公式可以背诵，但更推荐的做法是每次都展开括号后用 i² = -1 替换 – 这既不容易出错，也帮助你深入理解运算过程。
di + bci + bdi². Since i² = -1, the last term becomes -bd. After rearranging: z₁z₂ = (ac &# formula, but it is better practice to expand the brackets each time and replace i² with -1 – this is less error-prone and helps you internalise the operation.
除法是最具挑战性的运算，核心思路是利用共轭复数（见下一节）将分母”实数化”。对于 z₁ ÷ z₂ = (a + bi)/(c + di)，将分子和分母同时乘以分母的共轭复数 c – di：结果的分母变为 (c + di)(c – di) = c² + d²（一个实数），分子变为 (a + bi)(c – di)。最终的商为 [(ac + bd) + (bc – ad)i]/(c² + d²)。这个方法在 AQA FM02 考试中必考，务必熟练掌握。
Division is the most challenging operation, and the core idea is to “realise&# (c + di), multiply both numerator and denominator by the conjugate of the denominator, c – di: the denominator becomes (c + di)(c – di) = c² + d² (a real number), and the numerator becomes (a + bi)(c – di). The final quotient is [(ac + bd) + (bc – ad)i]/(c² + d²). This method is guaranteed to appear in the AQA FM02 exam – make sure you master it thoroughly.
五、共轭复数及其三大核心性质 | The Complex Conjugate and Its Three Core Properties
复数 z = a + bi 的共轭复数记作 z* 或 z̄，定义为 a – bi – 只需将虚部的符号取反。共轭复数之所以重要，是因为它具有三个在解题中频繁使用的核心性质。性质一：z × z* = a² + b² = |z|²，即一个复数与其共轭的乘积等于其模的平方。这个性质是复数除法和求模运算的基础。
The complex conjugate of z = a + bi, written as z* or z̄, is defined as a – bi – e properties that are used constantly in problem-solving. Property one: z × z* = a² + b² = |z| square of its modulus. This property underpins complex division and modulus calculations.
性质二：共轭分配律。两个复数的和的共轭等于各自共轭的和：(z₁ + z₂)* = z₁* + z₂*。同样地，乘积的共轭等于各自共轭的乘积：(z₁z₂)* = z₁* × z₂*。这个性质在化简复杂表达式时非常方便。性质三：如果 z 是实数（即 Im(z) = 0），那么 z* = z。反之亦然 – 如果一个复数等于它的共轭，那么这个复数一定是实数。这一性质常用于证明题目。
sions. Property three: if z is real (i.e. Im(z) = 0), then z* = z. The converse also holds – onjugate, then the number must be real. This property is commonly used in proof questions.
六、阿甘图：复数在平面上的几何表示 | The Argand Diagram: Geometric Representation of Complex Numbers on a Plane
阿甘图（Argand Diagram）是理解复数的一个革命性工具。它以法国数学家Jean-Robert Argand命名，将复数 z = a + bi 映射到一个二维平面上：横轴（x轴）表示实部 a，纵轴（y轴）表示虚部 b。这样，每个复数都对应平面上的一个唯一点 (a, b)，而复数的加减法恰好对应了向量的加减法。阿甘图将抽象的复数概念可视化，让我们能够从几何角度理解复数运算。
ensional plane: the horizontal axis (x-axis) represents the real part a, and the vertical axis (y-axis) represents the imaginary part b. In this way, each complex number corresponds to a unique point (a, b) on the plane, and complex addition/subtraction correspond exactly to vector addition/subtraction. The Argand diagram visualises abstract complex number concepts, allowing us to understand complex operations from a geometric perspective.
在阿甘图上，共轭 z* = a – bi 就是 z 关于实轴的镜像反射。而乘以 i 的效果则是将复平面上的点逆时针旋转 90°。例如，1（点 (1,0)）乘以 i 变成 i（点 (0,1)），再乘 i 变成 -1（点 (-1,0)），再乘 i 变成 -i（点 (0,-1)） – 这直观地解释了为什么 i⁴ = 1。AQA考试中经常要求考生在阿甘图上作图或解释几何变换，因此掌握阿甘图上的运算是得分的关键。
On the Argand diagram, the conjugate z* = a &# ckwise. For example, 1 (point (1,0)) multiplied by i becomes i (point (0,1)), multiplied by i again becomes -1 (point (-1,0)), and once more becomes -i (point (0,-1)) – ric transformations, so mastering operations on the Argand diagram is key to scoring well.
七、模与辐角：从直角坐标到极坐标的桥梁 | Modulus and Argument: The Bridge from Cartesian to Polar Form
复数 z = a + bi 的模（Modulus），记作 |z| 或 r，定义为从原点到点 (a,b) 的距离：|z| = √a² + b²。这恰好等于 √z × z*，与上一节中提到的性质一完美呼应。模始终是非负实数，代表复数在阿甘图上的”大小”或”长度”。
The modulus of a complex number z = a + bi, written |z| or r, is defined as the distance from the origin to the point (a, b): |z| = √a² + b². This is exactly equal to √z × z*, perfectly echoing property one from the previous section. The modulus is always a non-negative real number, representing the “size” or “length” of the complex number on the Argand diagram.
辐角（Argument），记作 arg(z) 或 θ，是从正实轴到连接原点与点 (a,b) 的线段所成的角，通常以弧度为单位，取主值范围 (-π, π] 或 [0, 2π)。计算辐角使用 θ = arctan(b/a)，但必须注意象限！使用 atan2(b, a) 函数可以自动处理象限问题。例如，z = -1 + i 位于第二象限：|z| = √((-1)² + 1²) = √2，arg(z) = 3π/4（而不是 arctan(1/(-1)) = -π/4，后者对应第四象限）。象限判断错误是AS考试中最常见的失分点之一。
, 2π). To compute the argument, use θ = arctan(b/a), but you must account for the quadrant! andles quadrant issues automatically. For example, z = -1 + i lies in the second quadrant: |z| = √((-1)² + 1²) = √2 -π/4, which corresponds to the fourth quadrant). Quadrant misjudgement is one of the most common marks lost in AS exams.
八、二次方程与复数根：判别式为负时发生了什么？| Quadratic Equations with Complex Roots: What Happens When the Discriminant Is Negative?
在GCSE阶段，当你遇到判别式 Δ = b² – 4ac < 0 的二次方程时，答案总是"无实数解"。到了AS进阶数学，这个答案被拓展了：方程仍然有两个解，它们是共轭复数对。例如，x² + 4x + 13 = 0：判别式 Δ = 16 - 52 = -36 < 0。使用求根公式：x = [-4 ± √-36]/2 = [-4 ± 6i]/2 = -2 ± 3i。所以两个根是 -2 + 3i 和 -2 - 3i，它们互为共轭。
At GCSE level, when you encounter a quadratic equation with discriminant Δ = b² – 4ac < 0, the answer is always "no real solutions" + 4x + 13 = 0: discriminant Δ = 16 - 52 = -36 < 0. Using the quadratic formula: x = [-4 ± √-36 2 = -2 ± 3i. So the two roots are -2 + 3i and -2 - 3i, which are conjugates of each other.
对于实系数二次方程 ax² + bx + c = 0，如果 Δ ≥ 0，两根是实数；如果 Δ < 0，两根构成复共轭对，且两根之和 = -b/a，两根之积 = c/a - 这些韦达定理在复数域中完全成立。AQA常常考察利用根的和与积来构造二次方程，这类题型需要灵活运用韦达定理。
ots form a complex conjugate pair, and the sum of roots = -b/a, the product of roots = c/a - Vieta' iven sum and product of roots, a question type that requires flexible application of Vieta's formulas.
九、共轭根定理：多项式复数根的对称性规律 | The Conjugate Root Theorem: The Symmetry of Complex Roots in Polynomials
共轭根定理是AS进阶数学中一个优美而强大的结论：如果一个实系数多项式（所有系数都是实数）有一个复数根 z = a + bi，那么它的共轭 z* = a – bi 也一定是该多项式的根。这意味着，实系数多项式的复数根总是成对出现。这个定理可以推广到任意次数的实系数多项式。
coefficients are real numbers) has a complex root z = a + bi, then its conjugate z* = a &# n conjugate pairs. This theorem generalises to real-coefficient polynomials of any degree.
一个典型的AQA考题是：已知 2 + i 是三次多项式 x³ + px² + qx + 10 = 0 的一个根，求实数 p 和 q。根据共轭根定理，2 – i 也是根。设第三个根为 α（必为实数，因为奇数次实系数多项式至少有一个实根）。利用因式分解 (x – (2+i))(x – (2-i))(x – α) = x³ + px² + qx + 10，展开后比较系数即可求出 p, q 和 α 的值。这个解题流程融合了共轭根定理、因式分解和比较系数法，是FM02的核心考点。
x³ + px² + qx + 10 = 0, find the real numbers p and q. By the Conjugate Root Theorem, 2 &# d-degree real-coefficient polynomial has at least one real root). Using factorisation (x – (2+i))(x – (2-i))(x &# lution flow combines the Conjugate Root Theorem, factorisation, and comparing coefficients – it is a core examination topic in FM02.
十、考试策略与常见失分陷阱：如何稳拿FM02复数题满分 | Exam Strategy and Common Pitfalls: How to Secure Full Marks on FM02 Complex Number Questions
在AQA AS进阶数学FM02考试中，复数相关题目通常占试卷总分的15%-20%。以下是最常见的四种失分陷阱及应对策略。陷阱一：混淆虚部与带i的项。题目要求写出 Im(z) 时，答案必须是实数 b，而不是 bi。陷阱二：除法运算中忘记乘以分母的共轭，或者乘了分子却忘了乘分母。建议在草稿纸上先写出完整的 (a+bi)(c-di)/[(c+di)(c-di)] 形式，再分步计算。
In the AQA AS Further Mathematics FM02 exam, complex number questions typically account for 15%-20% of the total marks. Here are the four most common pitfalls and counter-strategies. Pitfall one: confusing the imaginary part with the i-containing term. When the question asks for Im(z), the answer must be the real number b, not bi. Pitfall two: forgetting to multiply by the denominator’s conjugate during division, or multiplying the numerator but forgetting the denominator. It is advisable to write out the full form (a+bi)(c-di)/[(c+di)(c-di)] on scratch paper first, then compute step by step.
陷阱三：计算辐角时忽略象限。记住黄金法则 – 先画阿甘图，确定复数所在的象限，再用 arctan 求参考角，最后根据象限调整到正确的辐角。陷阱四：i的幂次简化错误。当指数较大时，不要试图硬算 i 的每一次幂；直接用 n mod 4 来确定结果。n ≡ 0 → 1，n ≡ 1 → i，n ≡ 2 → -1，n ≡ 3 → -i。考试中时间紧迫，这个口诀可以帮你节省宝贵的两分钟。
Pitfall three: ignoring the quadrant when computing the argument. Remember the golden rule – o the correct argument based on the quadrant. Pitfall four: errors in simplifying powers of i. When the exponent is large, do not attempt to compute each power of i individually; use n mod 4 directly to determine the result. n ≡ 0 → 1, n ≡ 1 → i, n ≡ 2 → -1, n ≡ 3 → -i. Time is tight in exams, and this mantra can save you two precious minutes.
十一、阿甘图上的轨迹问题：用复数方程描述几何图形 | Loci on the Argand Diagram: Describing Geometric Shapes with Complex Equations
在AQA FM02考试中，”轨迹”（locus）题型是一个高频考点，要求考生用复数方程描述阿甘图上的几何路径。最常见的轨迹类型有三种。第一种：|z – w| = r。这个方程表示以复数 w 对应的点为圆心、半径为 r 的圆。例如，|z – (3 + 4i)| = 5 表示以点 (3,4) 为圆心、半径为5的圆。这个方程本质上就是平面上所有与点 (3,4) 距离为5的点的集合。
In the AQA FM02 exam, “locus&# y topic, requiring candidates to describe geometric paths on the Argand diagram using complex equations. There are three most common locus types. Type one: |z – w| le centred at the point corresponding to the complex number w, with radius r. For example, |z – (3 + 4i)| 4) with radius 5. This equation is essentially the set of all points on the plane whose distance from (3,4) is 5.
第二种：|z – w₁| = |z – w₂|。这个方程表示到两个定点 w₁ 和 w₂ 距离相等的点的轨迹 – 即连接 w₁ 和 w₂ 的线段的垂直平分线（perpendicular bisector）。当考试题给出这种形式时，你不需要展开复杂的代数推导，直接识别出它是垂直平分线，然后找到中点坐标和线段斜率，就可以写出直线的方程。第三种：arg(z – w) = θ。这个方程表示从点 w 出发、与正实轴成角 θ 的一条射线（half-line），不包括 w 点本身。
Type two: |z – w₁| = |z – w₂|. This equation represents the locus of points equidistant from two fixed points w₁ and w₂ – the slope of the segment, and you can write the equation of the line. Type three: arg(z &# the point w, making an angle θ with the positive real axis, excluding the point w itself.
一道典型的综合题：在同一张阿甘图上，画出满足 |z – 4| = 3 和 arg(z) = π/4 的所有点 z，并找出它们的交点。第一个条件是圆心在 (4,0)、半径3的圆；第二个条件是从原点出发、角度为 π/4 的射线。交点可以通过解圆方程 x² + y² = (到原点的距离)² 和直线 y = x 的方程组来找到。具体计算：设 z = x + ix，代入 |z – 4| = 3 得到 |(x-4) + ix| = √((x-4)² + x²) = 3。两边平方：(x-4)² + x² = 9，展开：x² – 8x + 16 + x² = 9，整理：2x² – 8x + 7 = 0，解得 x = (8 ± √64 – 56)/4 = (8 ± √8)/4 = 2 ± √2/2。因为 arg(z) = π/4，x 和 y 均为正，所以取 x = 2 + √2/2。这种将几何与代数结合起来的多步骤题目是AQA最爱的出题方式。
A typical integrated question: on the same Argand diagram, sketch all points z satisfying |z – 4| gin)² together with the line equation y = x. Specifically: let z = x + ix, substitute into |z – 4| = 3 to get |(x-4) + ix| = √((x-4)² + x²) = 3. Square both sides: (x-4)² + x² = 9, expand: x² – 8x + 16 + x² = 9, simplify: 2x² – 8x + 7 = 0, solve: x = (8 ± √64 – 56)/4 = (8 ± √8)/4 = 2 ± √2/2. Since arg(z) = π/4, both x and y are positive, so take x = 2 + √2/2. This kind of multi-step question that combines geometry and algebra is AQA’s favourite way to test this topic.
十二、复数在解方程组中的妙用 | Using Complex Numbers to Solve Systems of Equations
复数的一个巧妙应用是帮助求解某些实数方程组。当我们面对看似”不对称”的方程组时，有时可以通过引入复数将问题转化为更优雅的形式。考虑方程组：如果题目给出 z + 1/z = 2cosθ，要求找出 z 的值。将等式两边乘以 z：z² – 2z cosθ + 1 = 0。使用求根公式：z = cosθ ± √cos²θ – 1 = cosθ ± √-sin²θ = cosθ ± i sinθ。这正是复数极坐标形式的雏形 – z = eiθ 或 z = e-iθ，而 z + 1/z = eiθ + e-iθ = 2cosθ，验证无误。
numbers is helping to solve certain systems of real equations. When faced with seemingly “asymmetric&# uation: if a question gives z + 1/z = 2cosθ and asks for z. Multiply both sides by z: z² – 2z cosθ + 1 = 0. Using the quadratic formula: z = cosθ ± √cos²θ – 1 = cosθ ± √-sin²θ = cosθ ± i sinθ. This is precisely the embryonic form of complex polar form – z = eiθ or z = e-iθ, and indeed z + 1/z = eiθ + e-iθ = 2cosθ, verified.
在AS阶段，这类问题通常不会直接要求使用 de Moivre 定理（那是A2的内容），但理解 z + 1/z 与三角函数之间的内在联系，可以大大简化某些代数方程的求解过程。考试中如果遇到 z + k/z = c（其中k > 0, c为实数）这类形式的方程，记得尝试将其转化为关于 z 的二次方程，然后利用判别式和复数根的知识来解决。
At AS level, such problems typically do not directly require de Moivre&# and c is real) in the exam, remember to try converting them into quadratic equations in z, then use discriminant and complex root knowledge to solve them.
十三、精选练习题与详细解答 | Selected Practice Questions with Detailed Solutions
以下精选了三道AQA风格的FM02复数题目，每题附有详细的分步解答，供你检验自己对上述所有知识点的掌握程度。
Below are three carefully selected AQA-style FM02 complex number questions, each with detailed step-by-step solutions, to help you test your mastery of all the topics covered above.
题目一 | Question 1：已知 z₁ = 3 + 2i 和 z₂ = 1 – 5i。计算 (a) z₁z₂，(b) z₁/z₂，将结果写成 a + bi 的形式。
解答 | Solution：(a) z₁z₂ = (3+2i)(1-5i) = 3 – 15i + 2i – 10i² = 3 – 13i + 10 = 13 – 13i。
(b) z₁/z₂ = (3+2i)/(1-5i)。分子分母同乘分母的共轭 (1+5i)：= (3+2i)(1+5i)/[(1-5i)(1+5i)] = (3 + 15i + 2i + 10i²)/(1 + 25) = (3 + 17i – 10)/26 = (-7 + 17i)/26 = -7/26 + (17/26)i。
题目二 | Question 2：证明复数 z = (1+i)/(1-i) 的模为1，并求其辐角。
解答 | Solution：分子分母同乘 (1+i)：z = (1+i)²/[(1-i)(1+i)] = (1 + 2i + i²)/(1 + 1) = (1 + 2i – 1)/2 = 2i/2 = i。因此 |z| = |i| = 1，arg(z) = arg(i) = π/2。更简短的方法：注意到 |1+i| = √2 且 |1-i| = √2，所以 |z| = |1+i|/|1-i| = √2/√2 = 1。
题目三 | Question 3：已知 1+2i 是实系数二次方程 x² + px + q = 0 的一个根，求实数 p 和 q 的值。
解答 | Solution：根据共轭根定理，另一个根为 1-2i。两根之和 = (1+2i) + (1-2i) = 2 = -p，所以 p = -2。两根之积 = (1+2i)(1-2i) = 1 – (2i)² = 1 + 4 = 5 = q。因此 p = -2, q = 5。验证：方程为 x² – 2x + 5 = 0，判别式 = 4 – 20 = -16，确实有复数根。
Summary | 总结
复数作为AS AQA进阶数学FM02模块的核心主题，将实数域扩展到了一个更广阔的代数结构。从虚数单位 i 的基本定义（i² = -1），到标准形式 a + bi，再到四则运算、共轭性质、阿甘图、模与辐角，每一步的构建都逻辑严密。在实际解题中，复共轭是处理除法和分母有理化的核心工具，而阿甘图则提供了直观的几何视角。二次方程的复数根和共轭根定理共同构成了多项式理论在复数域中的基石。掌握这些内容不仅是为了应对考试，更是为进一步学习复数极坐标形式、欧拉公式 eiθ = cosθ + i sinθ 以及 de Moivre 定理打下坚实的基础。
dard form a + bi, to the four basic operations, conjugate properties, the Argand diagram, and modulus and argument – geometric perspective. Complex roots of quadratics and the Conjugate Root Theorem together form the foundation of polynomial theory in the complex domain. Mastering these topics is not just about exam preparation – it lays the groundwork for further study of complex polar form, Euler’s formula eiθ = cosθ + i sinθ, and de Moivre’s theorem.
更多咨询请联系16621398022（同微信）
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图