Loci on Argand Diagrams | 阿甘图上的轨迹

📚 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

Every complex 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| = √((x – 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| = r is a circle centered at a with radius 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| = 4 represents 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| is the set of points equidistant from 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

The locus arg(z – a) = θ is a half-line starting from the point a and making an angle θ with the positive 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。

The gradient of this half-line is tan θ, provided θ is not an odd multiple 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。

In the diagram, the boundary half-lines are drawn dashed 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。

When one locus is a half-line from arg(z – a) = θ, also check that the solution 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。

For a half-line arg(z – a) = θ, the closest 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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Loci on Argand Diagrams | 阿甘图上的轨迹