Loci on Argand Diagrams | Argand 图上的轨迹

📚 Loci on Argand Diagrams | Argand 图上的轨迹

An Argand diagram represents complex numbers as points in a plane, with the real part on the horizontal axis and the imaginary part on the vertical axis. Many AQA questions ask you to describe or sketch a locus: a set of complex numbers z that obey a given condition. This article covers circles, perpendicular bisectors, half-lines, regions and exam-style combinations.

Argand 图把复数表示为平面上的点，实部对应横轴，虚部对应纵轴。许多 AQA 考题要求你描述或画出轨迹，也就是满足给定条件的一组复数 z。本文涵盖圆、垂直平分线、射线、区域以及考试中常见的组合问题。

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1. What is a Locus? | 什么是轨迹

In complex numbers, a locus is the set of all points z that satisfy a rule. A rule is often written using modulus or argument. For example, the equation |z| = 3 means ‘the distance from z to the origin is 3’, so the locus is a circle centred at O with radius 3.

在复数中，轨迹就是满足某条规则的所有点 z 的集合。规则通常用模或辐角表示。例如 |z| = 3 表示 ‘z 到原点的距离为 3’，因此轨迹是以原点为圆心、半径为 3 的圆。

We normally write z = x + iy, where x, y ∈ ℝ. A locus condition can then be converted into an equation in x and y, which allows us to sketch it on ordinary coordinate axes.

我们通常设 z = x + iy，其中 x, y ∈ ℝ。轨迹条件随后可以转化为关于 x 和 y 的方程，从而可以在普通坐标轴上作图。

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2. Circles: |z − a| = r | 圆：|z − a| = r

The equation |z − a| = r, where a is a fixed complex number and r > 0, represents a circle with centre a and radius r. Geometrically, it says that every point z on the locus is exactly r units away from the fixed point a.

方程 |z − a| = r 中，a 是固定复数，r > 0，它表示以 a 为圆心、r 为半径的圆。从几何上看，它表示轨迹上每个点 z 到固定点 a 的距离都恰好为 r。

|z − a| = r

If a = α + βi and z = x + iy, then squaring both sides gives the Cartesian form:

如果 a = α + βi 且 z = x + iy，那么两边平方可得到笛卡尔形式：

(x − α)² + (y − β)² = r²

For example, |z − (2 + i)| = 3 is a circle centred at (2, 1) with radius 3. Its Cartesian equation is (x − 2)² + (y − 1)² = 9.

例如，|z − (2 + i)| = 3 表示圆心为 (2, 1)、半径为 3 的圆。它的笛卡尔方程为 (x − 2)² + (y − 1)² = 9。

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3. Perpendicular Bisectors: |z − z₁| = |z − z₂| | 垂直平分线：|z − z₁| = |z − z₂|

The locus given by |z − z₁| = |z − z₂|, where z₁ and z₂ are fixed complex numbers, is the perpendicular bisector of the line segment joining z₁ and z₂. It consists of all points that are equidistant from the two fixed points.

由 |z − z₁| = |z − z₂| 给出的轨迹是连接 z₁ 与 z₂ 的线段的垂直平分线，其中 z₁ 和 z₂ 是固定复数。该轨迹由所有到两个固定点距离相等的点组成。

|z − z₁| = |z − z₂|

To convert to Cartesian form, let z₁ = x₁ + iy₁, z₂ = x₂ + iy₂ and z = x + iy. Squaring and expanding gives a linear equation:

要转化为笛卡尔形式，设 z₁ = x₁ + iy₁，z₂ = x₂ + iy₂，z = x + iy。平方并展开后可得到线性方程：

2(x₂ − x₁)x + 2(y₂ − y₁)y = |z₂|² − |z₁|²

This linear equation is the perpendicular bisector. For instance, |z + 1| = |z − 3i| is the perpendicular bisector of the segment from −1 to 3i.

这个线性方程就是垂直平分线。例如，|z + 1| = |z − 3i| 是从 −1 到 3i 的线段的垂直平分线。

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4. Half-Lines: arg(z − a) = θ | 射线：arg(z − a) = θ

The equation arg(z − a) = θ, where θ is a fixed angle, represents a half-line starting from the point a and making an angle θ with the positive real axis. The point a itself is not included because arg 0 is undefined.

方程 arg(z − a) = θ 中 θ 是固定角度，它表示从点 a 出发、与正实轴夹角为 θ 的射线。由于 arg 0 没有定义，点 a 本身不包含在轨迹内。

arg(z − a) = θ

If a = α + βi, then z − a = (x − α) + i(y − β). The condition arg(z − a) = θ means there exists r > 0 such that:

若 a = α + βi，则 z − a = (x − α) + i(y − β)。条件 arg(z − a) = θ 表示存在 r > 0 使得：

x − α = r cos θ,   y − β = r sin θ

When sketching, draw an open circle at a to show that a is excluded, and draw a ray from a in the direction θ. AQA uses the principal argument, usually −π < θ ≤ π, so make sure your angle lies in that range.

作图时，在 a 处画一个空心圆以表示 a 不包含在内，然后从 a 沿 θ 方向画射线。AQA 使用辐角主值，通常为 −π < θ ≤ π，因此要确保角度在该范围内。

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5. Regions and Inequalities | 区域与不等式

Inequalities involving modulus or argument describe regions rather than single curves. For example, |z − a| < r is the interior of a circle, while |z − a| > r is the exterior. The boundary is included when the inequality is ≤ or ≥.

涉及模或辐角的不等式描述的是区域而不是单条曲线。例如，|z − a| < r 表示圆的内部，|z − a| > r 表示圆的外部。当不等式为 ≤ 或 ≥ 时，边界包含在内。

An argument inequality such as 0 < arg z < π/2 describes an infinite sector in the first quadrant, bounded by the positive real axis and the positive imaginary axis. Strict inequalities mean the boundary rays are not included.

诸如 0 < arg z < π/2 的辐角不等式描述的是第一象限内的无限扇形区域，由正实轴和正虚轴围成。严格不等式表示边界射线不包含在内。

When combining modulus and argument conditions, you may need to shade the intersection of regions. Always test a point to confirm the correct side of a boundary.

当同时出现模和辐角条件时，你可能需要对区域的交集进行阴影标注。始终要检验一个点，以确认边界的正确一侧。

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6. Combining Loci and Intersections | 组合轨迹与交点

Exam questions often ask for the points that satisfy two loci simultaneously. Geometrically, this means finding the intersection of the two curves. A circle and a line may meet in zero, one or two points; a circle and a half-line may meet in at most one point because a half-line only extends in one direction.

考试题经常要求找出同时满足两个轨迹的点。从几何上讲，这意味着求两条曲线的交点。圆与直线可能没有交点、有一个交点或有两个交点；圆与射线最多只有一个交点，因为射线只向一个方向延伸。

Algebraically, substitute one locus condition into the other. For example, if |z − (1 + 2i)| = 2 and arg(z − 1) = π/4, write z on the half-line as z = 1 + r(cos π/4 + i sin π/4), then substitute into the circle equation to solve for r > 0.

代数上，将一个轨迹条件代入另一个。例如，若 |z − (1 + 2i)| = 2 且 arg(z − 1) = π/4，可把射线写为 z = 1 + r(cos π/4 + i sin π/4)，然后代入圆的方程，解出 r > 0。

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7. Algebraic Conversion to Cartesian Form | 代数转化为笛卡尔形式

Converting a complex locus to an equation in x and y is a key skill. For moduli, square both sides and use the identity |z − a|² = (x − α)² + (y − β)². For arguments, use trigonometry or the relationship y − β = (tan θ)(x − α) where the half-line is in the correct quadrant.

将复数轨迹转化为 x 和 y 的方程是一项关键技能。对于模，将两边平方并使用恒等式 |z − a|² = (x − α)² + (y − β)²。对于辐角，可使用三角关系或 y − β = (tan θ)(x − α)，同时注意射线所在的象限。

Be careful with argument equations: tan θ alone does not determine the quadrant. You must check the signs of x − α and y − β, or use r > 0 with cos θ and sin θ explicitly.

处理辐角方程时要小心：仅用 tan θ 无法确定象限。必须检查 x − α 和 y − β 的符号，或者明确使用 r > 0 以及 cos θ 和 sin θ。

For example, arg(z − 2) = 3π/4 gives y = −(x − 2) with x < 2 and y > 0, but writing y = tan(3π/4)(x − 2) only gives the line y = −x + 2, not the half-line restriction.

例如，arg(z − 2) = 3π/4 给出 y = −(x − 2) 且 x < 2、y > 0；但如果只写 y = tan(3π/4)(x − 2)，得到的只是直线 y = −x + 2，而不是受限的射线。

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8. Sketching Strategy and Common Mistakes | 作图策略与常见错误

When sketching a locus, first identify its type: circle, perpendicular bisector, half-line, or region. Mark fixed points such as centres or endpoints. Use a dashed line or open circle for excluded boundaries, and solid lines for included boundaries.

作轨迹图时，首先判断类型：圆、垂直平分线、射线或区域。标出固定点，如圆心或端点。被排除的边界用虚线或空心圆表示，包含的边界用实线表示。

Common mistakes include confusing |z − a| with z − a itself, forgetting to exclude the endpoint of a half-line, and using degrees instead of radians. Another frequent error is drawing a full line when the locus should be only a half-line.

常见错误包括混淆 |z − a| 与 z − a 本身、忘记排除射线的端点、使用角度制而不是弧度制。另一个常见错误是当轨迹应为射线时却画成了整条直线。

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9. Exam-Style Examples | 考试风格例题

Example 1: Describe and sketch the locus of z satisfying |z − (3 − 2i)| = 4.

例 1：描述并画出满足 |z − (3 − 2i)| = 4 的 z 的轨迹。

This is a circle with centre 3 − 2i, so at (3, −2), and radius 4. The Cartesian equation is (x − 3)² + (y + 2)² = 16.

这是一个圆，圆心为 3 − 2i，即 (3, −2)，半径为 4。笛卡尔方程为 (x − 3)² + (y + 2)² = 16。

Example 2: Find the locus of z if |z − i| = |z + 3| and describe it.

例 2：若 |z − i| = |z + 3|，求 z 的轨迹并加以描述。

This is the perpendicular bisector of the segment joining i and −3. Letting z = x + iy, squaring gives x² + (y − 1)² = (x + 3)² + y², which simplifies to 2y + 1 = 6x + 9, or 6x − 2y + 8 = 0, equivalently 3x − y + 4 = 0.

这是连接 i 与 −3 的线段的垂直平分线。设 z = x + iy，平方后得到 x² + (y − 1)² = (x + 3)² + y²，化简得 2y + 1 = 6x + 9，即 6x − 2y + 8 = 0，等价于 3x − y + 4 = 0。

Example 3: Shade the region satisfying |z − 1| ≤ 2 and 0 ≤ arg(z − 1) ≤ π/4.

例 3：画出满足 |z − 1| ≤ 2 且 0 ≤ arg(z − 1) ≤ π/4 的区域。

The first condition is a closed disk centred at 1 with radius 2. The second condition is a sector from the point 1, bounded by rays at angles 0 and π/4. Their intersection is a wedge-shaped part of the disk, including the two boundary rays and the circle arc.

第一个条件是以 1 为圆心、半径为 2 的闭圆盘。第二个条件是以点 1 为顶点的扇形，由角度 0 和 π/4 的两条射线围成。两者的交集是圆盘中楔形的一部分，包含两条边界射线和圆弧。

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10. Summary and Key Formulae | 总结与关键公式

The table below summarises the main loci you need to recognise for AQA questions on Argand diagrams.

下表总结了你在 AQA Argand 图题目中需要识别的主要轨迹。

Locus	Geometric meaning	中文解释
|z − a| = r	Circle centre a, radius r	圆，圆心 a，半径 r
|z − z₁| = |z − z₂|	Perpendicular bisector of segment z₁z₂	线段 z₁z₂ 的垂直平分线
arg(z − a) = θ	Half-line from a at angle θ	从 a 出发、角度为 θ 的射线
|z − a| < r	Interior of a circle	圆的内部
|z − a| > r	Exterior of a circle	圆的外部
θ₁ < arg(z − a) < θ₂	Sector between two half-lines	两条射线之间的扇形区域

Remember to convert to x and y form when an exact Cartesian equation is requested, and always check whether boundaries are included before shading a region.

当题目要求精确的笛卡尔方程时，请记得转化为 x 和 y 的形式；在给区域上色之前，务必检查边界是否包含在内。

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