📚 Case Study: Complex Transformations and Matrix Representations | 案例分析：复数变换与矩阵表示实战
In AQA Year 13 Further Mathematics, mastering complex transformations and their matrix representations is essential for tackling extended analytical questions. This case study walks you through a multi-step problem that combines Möbius transformations, fixed points, geometric mapping and inverses – exactly the kind of challenge you can expect in the exam. We will solve the problem step by step, highlighting key techniques and common pitfalls.
在 AQA 13 年级进阶数学中，熟练掌握复数变换及其矩阵表示是攻克长篇分析题的关键。本案例分析将带你逐步解决一道融合莫比乌斯变换、不动点、几何映射与逆变换的综合性问题——正是你可能在考试中遇到的挑战类型。我们将分步骤解答，同时强调核心技巧与常见错误。
1. Problem Statement | 问题陈述
Consider the transformation T defined on the extended complex plane by:
考虑在扩充复平面上定义的变换 T：
w = (z − i)/(z + i)
(a) Show that T maps the real axis Im(z)=0 to the unit circle |w|=1.
(a) 证明 T 将实轴 Im(z)=0 映射到单位圆 |w|=1。
(b) Find the fixed points of T, giving your answers in the form a + bi.
(b) 求 T 的不动点，并以 a + bi 的形式给出答案。
ion of simpler transformations, and hence determine the image of the line Im(z)=1 under T.
(c) 将 T 表示为一组简单变换的复合，并由此确定直线 Im(z)=1 在 T 下的像。
(d) By writing z in terms of w, find the pre-image of the circle |w − 1| = 1 under T, and sketch or describe its locus in the z-plane.
(d) 通过将 z 用 w 表示，求出圆 |w − 1| = 1 在 T 下的原像，并在 z 平面上描绘或描述其轨迹。
2. Key Concepts: Möbius Transformations | 关键概念：莫比乌斯变换
A Mö , c, d are complex constants and ad − bc ≠ 0. These transformations map circles and lines to circles or lines in the extended complex plane, preserving angles.
莫比乌斯变换是形如 w = (az + b)/(cz + d) 的函数，其中 a, b, c, d 为复常数且 ad − bc ≠ 0。这些变换将扩充复平面上的圆和直线映射为圆或直线，并保持角度不变。
Every Mö to a non-zero scalar multiple. The composition of transformations corresponds to matrix multiplication, making matrix methods very powerful for handling multiple mappings.
每个莫比乌斯变换都可以与一个 2×2 矩阵 [[a, b], [c, d]] 相关联（允许相差一个非零标量因子）。变换的复合对应矩阵乘法，这使得矩阵方法在处理多重映射时非常有力。
1, b = −i, c = 1, d = i. The determinant is (1)(i) − (−i)(1) = 2i ≠ 0, so T is indeed a Möbius transformation.
对于我们的变换 w = (z − i)/(z + i)，有 a = 1, b = −i, c = 1, d = i。行列式为 (1)(i) − (−i)(1) = 2i ≠ 0，因此 T 确实是一个莫比乌斯变换。
3. Step (a): Image of the Real Axis | 步骤(a)：实轴的像
Set z = x + 0i with x ∈ ℝ complex conjugates of each other: (x − i) and (x + i) are negatives of the imaginary part? onjugates unless x is real. For real x, the complex conjugate of x + i is x − i. Therefore |w| = |x − i|/|x + i| dulus (the distance from (x, -1) to origin equals distance from (x, 1) to origin), we have |w| = 1 for all x ≠ −i? the unit circle, except possibly the point w = 1 which corresponds to the limit as x → ±∞.
令 z = x + 0i，其中 x ∈ ℝ。代入 T 得 w = (x − i)/(x + i)。为求像，消去 x。注意到若 x 为实数，则分母 x + i 的共轭为 x − i。因此 |w| = |x − i|/|x + i|。由于分子与分母的模相等（从 (x, -1) 到原点的距离等于从 (x, 1) 到原点的距离），对所有实 x（分母不为零）都有 |w| = 1。所以每个实 z 都映射到单位圆上的一点，仅除了可能点 w = 1，它对应 x → ±∞ 时的极限。
= 1 which is approached at both extremes. Thus the image is indeed the entire unit circle |w| = 1.
为证明整个单位圆被覆盖，注意当 x 从 −∞ 变到 ∞ 时，w 的辐角遍历除 w = 1 处角度外的所有可能角度，而 w = 1 在两端极限处被趋近。因此像确实是整个单位圆 |w| = 1。
4. Step (b): Finding the Fixed Points | 步骤(b)：求不动点
z + i: z(z + i) = z − i → z² + i z = z − i → z² + i z − z + i = 0 → z² + (i − 1)z + i = 0.
不动点满足 z = (z − i)/(z + i)。两边乘以 z + i：z(z + i) = z − i → z² + i z = z − i → z² + i z − z + i = 0 → z² + (i − 1)z + i = 0。
nant is Δ = (i − 1)² − 4i = (i² − 2i + 1) − 4i = (−1 − 2i + 1) − 4i = −6i. Write −6i = 6 e−iπ/2 = 6 ei 3π/2. The square roots are √6 e−iπ/4 and √6 ei 3π/4.
使用二次方程求根公式，其中 a = 1, b = i − 1, c = i。判别式为 Δ = (i − 1)² − 4i = (i² − 2i + 1) − 4i = (−1 − 2i + 1) − 4i = −6i。将 −6i 写为 6 e−iπ/2 或 6 ei 3π/2。其平方根为 √6 e−iπ/4 和 √6 ei 3π/4。
Now √6 e−iπ/4 = √6 (cos π/4 − i sin π/4) = √6 (1/√2 − i/√2) = √3 (1 − i). The second root is √6 ei 3π/4 = √6 (−1/√2 + i/√2) = √3 (−1 + i) = −√3 (1 − i). So the two fixed points are:
现在 √6 e−iπ/4 = √6 (cos π/4 − i sin π/4) = √6 (1/√2 − i/√2) = √3 (1 − i)。第二个根为 √6 ei 3π/4 = √6 (−1/√2 + i/√2) = √3 (−1 + i) = −√3 (1 − i)。因此两个不动点为：
z = [(1 − i) ± √3 (1 − i)] / 2 = (1 − i)(1 ± √3)/2.
Thus the fixed points are z₁ = (1+√3)/2 − i (1+√3)/2 and z₂ = (1−√3)/2 − i (1−√3)/2.
从而不动点为 z₁ = (1+√3)/2 − i (1+√3)/2 和 z₂ = (1−√3)/2 − i (1−√3)/2。
5. Geometric Interpretation of the Transformation | 变换的几何解释
ro), we can understand T as a composition of translation, inversion, rotation and scaling that sends the upper half-plane to the unit disc. This is a typical Cayley-type map.
由于 T 将实轴映射到单位圆，且点 z = i 映到 w = 0（因为 (i − i)/(i + i)=0），而 z = −i 映到无穷远点（分母为零），我们可以将 T 理解为一个由平移、反演、旋转和伸缩构成的复合映射，将上半平面映到单位圆盘。这是一个典型的凯莱型映射。
followed by reflection in the real axis), then z₃ = −2i z₂ (rotation by −π/2 and scaling by 2), and finally w = 1 + z₃ (translation by 1).
事实上，T 可以分解为：w = (z − i)/(z + i) = 1 − 2i/(z + i)。从 z₁ = z + i（上移 i），然后 z₂ = 1/z₁（单位圆反演再关于实轴反射），然后 z₃ = −2i z₂（旋转 −π/2 并伸缩 2），最后 w = 1 + z₃（平移 1）。
Recognising such decompositions helps when finding images of standard lines or circles without heavy algebra.
在进行标准直线或圆的像求解时，辨识这类分解有助于减少复杂的代数运算。
6. Matrix Representation of T | T 的矩阵表示
i], [1, i]] up to a non-zero scalar multiple. The inverse transformation corresponds to the inverse matrix, and compositions with other Möbius maps correspond to matrix products.
表示 T 的矩阵为 M = [[1, -i], [1, i]]（可相差一个非零标量因子）。逆变换对应逆矩阵，与其他莫比乌斯映射的复合则对应矩阵乘积。
ad−bc = 2i, so the inverse matrix is (1/2i) [[i, i], [-1, 1]] = [[1/2, 1/2], [i/2, -i/2]]? Let' ]] = [[1/2, 1/2], [-1/(2i), 1/(2i)]]. Multiply numerator and denominator for the third row? i/2. So the inverse matrix is [[1/2, 1/2], [i/2, -i/2]]. This corresponds to w = ( (1/2)z' + (1/2) ) / ( (i/2)z' - i/2 )? Not needed now, but we can solve directly for z.
为求逆变换，计算 M 的逆。对于 2×2 矩阵 [[a, b], [c, d]]，逆为 (1/(ad−bc)) [[d, -b], [-c, a]]。这里 ad−bc = 2i，故逆矩阵为 (1/2i) [[i, i], [-1, 1]] = [[1/2, 1/2], [i/2, -i/2]]。（计算细节：-1/(2i) = i/2。）该矩阵对应一个莫比乌斯变换，但我们可以直接由 w 解出 z，后续步骤将使用。
7. Step (c): Image of the Line Im(z)=1 | 步骤(c)：直线 Im(z)=1 的像
Let z = x + i, x ∈ ℝ (x²+4). Let u = x²/(x²+4) and v = −2x/(x²+4). Eliminate x. Notice that u > 0, and u + iv satisfies (u − 1/2)² + v² = 1/4. Indeed, from v = −2x/(x²+4), we can express x in terms of v, etc., leading to the circle equation. Alternatively, use the property that Mö rcle. By examining three points, e.g., x=0 gives w=0, x=2 gives w=2/(2+2i)=1/(1+i)= (1−i)/2, and as x→∞, w→1. So the circle passes through 0, (1−i)/2, and 1. The circle that includes these points is indeed |w − 1/2| = 1/2.
令 z = x + i, x ∈ ℝ。代入 T：w = (x + i − i)/(x + i + i) = x/(x + 2i)。为求轨迹，分离实部和虚部：w = x(x − 2i)/(x² + 4) = (x²)/(x²+4) − i (2x)/(x²+4)。令 u = x²/(x²+4)，v = −2x/(x²+4)。消去 x，注意到 u > 0，且 (u − 1/2)² + v² = 1/4。事实上，由 v 可反解 x，得到该圆方程。也可用性质：不过极点的直线在莫比乌斯变换下映为圆。直线 Im(z)=1 不过极点 z=−i，故像为圆。取三点：x=0 得 w=0，x=2 得 w=2/(2+2i)=1/(1+i)= (1−i)/2，当 x→∞ 时 w→1。过这三点的圆正是 |w − 1/2| = 1/2。
Thus the image is the circle with centre 1/2 and radius 1/2 in the w-plane.
因此像为 w 平面上圆心在 1/2、半径为 1/2 的圆。
8. Step (d): Pre-image of the Circle |w − 1|=1 | 步骤(d)：圆 |w−1|=1 的原像
We need the set of z such that |w − 1| → z − w z = w i + i → z(1 − w) = i(w + 1) → z = i(w + 1)/(1 − w). This is valid for w ≠ 1.
我们需要满足 |w − 1| = 1 的 z 的集合。首先反解变换：由 w = (z − i)/(z + i)，两边乘出： z − i → w z + w i = z − i → z − w z = w i + i → z(1 − w) = i(w + 1) → z = i(w + 1)/(1 − w)，其中 w ≠ 1。
Now impose |w − 1| = 1. Let w = 1 + eiθ (since |w−1|=1). Then w + 1 = 2 + eiθ and 1 − w = −eiθ. So z = i (2 + eiθ) / (−eiθ) = −i (2 e−iθ ). Then w − 1 = (z − i)/(z + i) − 1 = (z − i − z − i)/(z + i) = −2i/(z + i). The condition |w−1|=1 becomes |−2i/(z + i)| = 1 → 2/|z + i| = 1 → |z + i| = 2.
现在施加条件 |w − 1| = 1。令 w = 1 + eiθ（因为 |w−1|=1）。则 w + 1 = 2 + eiθ，1 − w = −eiθ。于是 z = i (2 + eiθ) / (−eiθ) = −i (2 e−iθ + 1)。这是参数形式。要得到直角坐标轨迹，用 w 的表达式：w − 1 = (z − i)/(z + i) − 1 = (z − i − z − i)/(z + i) = −2i/(z + i)。条件 |w−1|=1 化为 |−2i/(z + i)| = 1 → 2/|z + i| = 1 → |z + i| = 2。
circle, so it is included in the pre-image as the point at infinity in the extended plane.
因此原像是 z 平面上圆心在 −i、半径为 2 的圆。注意 z = ∞ 对应 w = 1，而该点在原始圆上，因此在扩充平面的原像中包含无穷远点。
9. Combining with Other Transformations | 与其他变换的复合
Suppose we have another transformation S defined by w' = (2w)/(w + 1). The combined transformation S ∘ T sends z directly to w'. Using matrices, the matrix for T is MT = [[1, -i], [1, i]], and for S we have a=2, b=0, c=1, d=1, so MS = [[2, 0], [1, 1]]. The product MS MT is [[2, 0], [1, 1]] [[1, -i], [1, i]] = [[2, -2i], [2, 1−i]]. This gives the combined Möbius map w' = (2z − 2i) / (2z + (1−i)). This approach saves time and reduces algebra errors.
假设有另一个变换 S 定义为 w' = (2w)/(w + 1)。复合变换 S ∘ T 直接将 z 映射到 w'。利用矩阵，T 的矩阵为 MT = [[1, -i], [1, i]]，S 对应 a=2, b=0, c=1, d=1，故 MS = [[2, 0], [1, 1]]。乘积 MS MT = [[2, 0], [1, 1]] [[1, -i], [1, i]] = [[2, -2i], [2, 1−i]]。从而得到复合映射 w' = (2z − 2i) / (2z + (1
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