📚 Loci on Argand Diagrams | 阿甘图上的轨迹
This article explains how to describe and sketch loci in the Argand plane, a key topic in AQA A-Level Mathematics. A locus is a set of points satisfying a given condition, and in complex numbers these conditions usually involve modulus or argument.
本文讲解如何在阿甘平面中描述和绘制轨迹，这是 AQA A-Level 数学中的一个关键考点。轨迹是满足给定条件的一组点，在复数中，这些条件通常涉及模或辐角。
1. The Argand Plane and Distance | 阿甘平面与距离
plex number z = x + yi is represented by the point (x, y) in the Argand plane. The modulus |z - a| measures the distance between the point z and the point a.
每个复数 z = x + yi 都对应阿甘平面上的点 (x, y)。模 |z - a| 表示点 z 与点 a 之间的距离。
If a = p + qi, then |z - a| = |(x - p) + (y - q)i| - p)² + (y - q)²). This distance interpretation is the foundation of every locus problem.
若 a = p + qi，则 |z - a| = |(x - p) + (y - q)i| = √((x - p)² + (y - q)²)。这一距离解释是所有轨迹问题的基础。
2. Circle: |z - a| = r | 圆：|z - a| = r
The locus |z - a| adius r. The centre is the complex number a, and the radius is the positive real number r.
轨迹 |z - a| = r 是以 a 为圆心、r 为半径的圆。圆心是复数 a，半径是正实数 r。
z = x + yi, a = p + qi ⇒ (x - p)² + (y - q)² = r²
For example, |z - 3 + 2i| nts a circle with centre 3 - 2i and radius 4. In Cartesian form: (x - 3)² + (y + 2)² = 16.
例如，|z - 3 + 2i| = 4 表示圆心为 3 - 2i、半径为 4 的圆。直角坐标形式为：(x - 3)² + (y + 2)² = 16。
3. Interior and Exterior of a Circle | 圆的内部与外部
The inequality |z - a| < r describes the interior of the circle, excluding the boundary. The inequality |z - a| ≤ r includes the boundary as a solid line in a sketch.
不等式 |z - a| < r 描述圆的内部（不含边界）。不等式 |z - a| ≤ r 包含边界，画图时用实线表示。
Similarly, |z - a| > r is the exterior of the circle, with the boundary drawn as a dashed line because it is not included.
类似地，|z - a| > r 是圆的外部，边界用虚线表示，因为边界不包含在区域内。
These conventions are essential for correct region questions in the exam.
这些约定在考试的区域类问题中至关重要。
4. Perpendicular Bisector: |z - a| = |z - b| | 垂直平分线：|z - a| = |z - b|
The locus |z - a| = |z - b| m a and b. This is exactly the perpendicular bisector of the line segment joining a and b.
轨迹 |z - a| = |z - b| 是到 a 和 b 距离相等的点集。这正是连接 a 和 b 的线段的垂直平分线。
|z - (x₁ + y₁i)| = |z - (x₂ + y₂i)|
To find its Cartesian equation, square both sides and expand:
要求其直角坐标方程，可两边平方并展开：
(x - x₁)² + (y - y₁)² = (x - x₂)² + (y - y₂)²
After simplification the x² and y² terms cancel, leaving a linear equation in x and y.
化简后 x² 和 y² 项相消，得到关于 x 和 y 的一次方程。
5. Half-Line: arg(z - a) = θ | 射线：arg(z - a) = θ
tive real axis. The endpoint a is not included, because z - a = 0 has no defined argument.
轨迹 arg(z - a) = θ 是从点 a 出发、与正实轴成角 θ 的射线。端点 a 不包含在内，因为 z - a = 0 的辐角无定义。
e of π/2. For example, arg(z - i) = π/4 is the half-line starting at i with direction 45°.
这条射线的斜率为 tan θ，前提是 θ 不是 π/2 的奇数倍。例如，arg(z - i) = π/4 是起点为 i、方向为 45° 的射线。
6. Sector Regions from Arguments | 辐角形成的扇形区域
The inequality α < arg(z - a) < β describes an infinite sector with vertex at a, bounded by the two half-lines along α and β.
不等式 α < arg(z - a) < β 表示以 a 为顶点、由沿 α 和 β 的两条射线围成的无限扇形区域。
when the inequalities are strict. If ≤ or ≥ is used, the corresponding boundary is solid.
在图中，当不等式为严格不等时，边界射线画虚线。若使用 ≤ 或 ≥，则相应边界画实线。
Always measure the argument anticlockwise from the positive real axis. A negative angle means a clockwise rotation.
始终从正实轴逆时针方向测量辐角。负角表示顺时针旋转。
7. Cartesian Equations from Loci | 由轨迹求直角坐标方程
To convert a locus condition into a Cartesian equation, write z = x + yi and use |x + yi - a| = √((x - p)² + (y - q)²). Squaring often removes the square root.
要把轨迹条件转化为直角坐标方程，令 z = x + yi，并利用 |x + yi - a| = √((x - p)² + (y - q)²)。平方通常能去掉根号。
| Complex form | Cartesian form | Locus |
| |z - a| = r | (x - p)² + (y - q)² = r² | Circle |
| |z - a| = |z - b| | Linear equation in x, y | Perpendicular bisector |
| arg(z - a) = θ | y - q = tan θ (x - p) | Half-line |
For the half-line, the equation is valid only on the correct side of a, so state the domain restriction where necessary.
对于射线，方程只在 a 的正确一侧成立，因此必要时需注明定义域限制。
8. Intersections of Loci | 轨迹的交点
To find where two loci intersect, solve their Cartesian equations simultaneously. For example, a circle and a perpendicular bisector meet in 0, 1, or 2 points.
要找出两条轨迹的交点，联立它们的直角坐标方程即可。例如，圆与垂直平分线可能相交于 0、1 或 2 个点。
n lies on the correct side of a. Sometimes the Cartesian line has a second intersection on the opposite half-line, which must be rejected.
当一条轨迹是 arg(z - a) = θ 形成的射线时，还需检查解是否位于 a 的正确一侧。有时直线方程会在另一侧产生第二个交点，此时必须舍去。
In region problems, intersections of boundary curves are often vertices of the shaded region. Mark them clearly on your sketch.
在区域问题中，边界曲线的交点通常是阴影区域的顶点。请在草图中清楚地标出它们。
9. Working with Shaded Regions | 处理阴影区域
A typical AQA question gives two or three conditions, such as |z - 4| ≤ 3 and 0 ≤ arg(z) ≤ π/4, and asks you to shade the region satisfying all of them.
一个典型的 AQA 题目会给出两三个条件，如 |z - 4| ≤ 3 和 0 ≤ arg(z) ≤ π/4，并要求你画出满足所有条件的阴影区域。
You should:
你应该：
- Sketch each locus boundary carefully first.
- 先仔细画出每条轨迹边界。
- Use solid lines if the boundary is included, dashed if not.
- 若边界包含则用实线，不包含则用虚线。
- Test a point such as z = 0 to decide which side of each boundary to shade.
- 取一个测试点（如 z = 0）来判断每条边界的哪一侧需要阴影。
Always label the final region with the letter R if the question asks for it.
如果题目要求，请用字母 R 标注最终区域。
10. Finding Maximum and Minimum Values | 求最大值与最小值
Questions often ask for the greatest or least value of |z| or arg(z) subject to a locus condition. Geometrically, |z| is the distance from the origin to a point on the locus; arg(z) is the angle the point makes with the positive real axis.
题目常常要求在某轨迹条件下求 |z| 或 arg(z) 的最大值与最小值。几何上，|z| 是原点到轨迹上点的距离；arg(z) 是该点与正实轴所成的角度。
For a circle, the maximum distance from the origin is |centre| + r and the minimum is ||centre| - r|. The corresponding points lie on the line joining the origin and the centre.
对于圆，原点到圆的最大距离为 |圆心| + r，最小距离为 ||圆心| - r|。相应的点位于原点和圆心连线上。
est point to the origin is the foot of the perpendicular from the origin to the half-line. If this foot lies behind a, then the nearest point on the locus is a itself.
对于射线 arg(z - a) = θ，离原点最近的点是从原点到射线的垂足。若垂足落在 a 的后面，则轨迹上离原点最近的点就是 a。
Always sketch the situation before attempting calculation. A good diagram reveals which point is closest or furthest.
在计算前务必先画图。好的图形能直接看出哪个点最近或最远。
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