Loci in the Complex Plane
Why This Matters
# Loci in the Complex Plane - A-Level Further Mathematics Summary This lesson explores geometric representations of complex number equations and inequalities as loci (paths or regions) in the Argand diagram. Students learn to interpret conditions like |z - a| = r as circles, arg(z - a) = θ as half-lines from point a, and |z - a| = |z - b| as perpendicular bisectors, alongside more sophisticated loci involving combinations of modulus and argument. These concepts are fundamental for A-Level Further Mathematics examinations, regularly appearing as multi-mark questions requiring both algebraic manipulation and geometric interpretation, often combined with transformations, regions satisfying multiple conditions, and applications to solving complex number problems graphically.
Key Words to Know
Core Concepts & Theory
A locus in the complex plane is the set of all points (complex numbers) satisfying a given condition. Understanding loci requires mastery of the modulus-argument form and geometric interpretation.
Key Definitions:
- Modulus |z| represents the distance from the origin to point z
- |z - a| represents the distance from point z to the fixed point a
- arg(z - a) represents the angle from the positive real axis to the line joining a to z
Fundamental Locus Types:
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Circle: |z - a| = r describes a circle with centre a and radius r. The condition |z - a| < r gives the interior, while |z - a| > r gives the exterior.
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Perpendicular Bisector: |z - a| = |z - b| represents all points equidistant from a and b, forming the perpendicular bisector of the line segment joining them.
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Half-line (Ray): arg(z - a) = θ represents a half-line starting at point a, making angle θ with the positive real axis.
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Sector: α ≤ arg(z - a) ≤ β describes the region between two half-lines from point a.
The Apollonius Circle: |z - a|/|z - b| = k (where k ≠ 1) represents a circle. When k = 1, it degenerates to the perpendicular bisector.
Mnemonic - CRAB: Circles (modulus equations), Rays (argument equations), Apollonius (ratio of distances), Bisectors (equal distances).
Cambridge Key Point: Always sketch the locus first. The Argand diagram is your most powerful tool for visualizing complex number relationships.
Detailed Explanation with Real-World Examples
Complex loci have profound applications in engineering, physics, and navigation systems.
GPS Positioning Analogy: Consider how GPS determines your location. When your device receives a signal from a satellite at known position a, the time delay tells you |z - a| = r₁ (you're somewhere on a circle around that satellite). A second satellite at position b gives |z - b| = r₂ (another circle). Your position is where these circles intersect—exactly how we solve locus problems!
Radio Broadcasting: The condition |z - a| = |z - b| models the boundary where signals from two radio towers (at positions a and b) have equal strength. This perpendicular bisector helps engineers plan coverage areas and identify interference zones.
Radar Systems: Military radar uses arg(z - a) = θ to describe the direction of detected objects. The condition α ≤ arg(z - a) ≤ β represents a scanning sector—the region the radar actively monitors.
Robotics Path Planning: The Apollonius circle |z - a|/|z - b| = k models scenarios where a robot must maintain a specific ratio of distances from two obstacles. For k = 2, the robot stays twice as far from point b as from point a, enabling safe navigation.
Visual Thinking: Imagine standing in a field with two trees. Walking so you're always equidistant from both traces the perpendicular bisector. Walking so you're always twice as far from one tree as the other traces an Apollonius circle.
Real-world insight: Every locus condition translates to a geometric constraint. Train yourself to see |z - a| as distance from a and arg(z - a) as direction from a.
Worked Examples & Step-by-Step Solutions
Example 1: Sketch and describe the locus |z - 3 - 4i| = 5.
Solution:
- Let z = x + yi, so |z - (3 + 4i)| = 5
- This is |(x - 3) + (y - 4)i| = 5
- √[(x - 3)² + (y - 4)²] = 5
- (x - 3)² + (y - 4)² = 25
Examiner note: Circle with centre (3, 4) and radius 5. Mark the centre clearly on your diagram.
Example 2: Find the Cartesian equation of |z - 2i| = |z - 4|.
Solution:
- Let z = x + yi
- |x + (y - 2)i| = |(x - 4) + yi|
- √[x² + (y - 2)²] = √[(x - 4)² + y²]
- Square both sides: x² + y² - 4y + 4 = x² - 8x + 16 + y²
- Simplify: -4y + 4 = -8x + 16
- 2x - y + 3 = 0 (perpendicular bisector)
Examiner note: This line passes through the midpoint (2, 1) and is perpendicular to the line joining 2i and 4.
Example 3: Describe the region arg(z - 1 - i) = π/4.
Solution:
- Starting point: 1 + i
- Angle π/4 from positive real axis
- Half-line from (1, 1) at 45° to the horizontal
- Excludes the point 1 + i itself
Examiner note: Use a ruler for rays. Show clearly with an arrow that the line extends to infinity.
Common Exam Mistakes & How to Avoid Them
Mistake 1: Confusing |z - a| with |z| - a Why it happens: Misunderstanding modulus notation. Fix: Remember |z - ...
Cambridge Exam Technique & Mark Scheme Tips
Command Word Mastery:
- "Sketch" (3-4 marks): Draw the locus with labelled axes, key points (centres, intersecti...
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Exam Tips
- 1.Always start by identifying the type of locus (circle, perpendicular bisector, half-line) based on the form of the complex equation.
- 2.For modulus conditions, remember that |z - a| is the distance from z to a. For argument conditions, arg(z - a) is the angle of the vector from a to z.
- 3.When sketching, clearly label key points (centers, starting points of half-lines, points 'a' and 'b') and indicate the direction of half-lines or the shaded region for inequalities.
- 4.If in doubt, substitute z = x + iy and convert the complex equation into a Cartesian equation to verify the shape. This is particularly useful for Apollonius's circles or complex argument conditions.
- 5.Pay close attention to strict inequalities (e.g., < or >) vs. non-strict inequalities (e.g., <= or >=) as they determine whether boundaries are included (solid lines) or excluded (dashed lines).