Leave lesson

Pure · Complex numbers

1 / 11

Loci in the Argand diagram

Reading a condition on z as a set of points in the Argand diagram: |z − a| = r is a circle centre a radius r, |z − a| = |z − b| is the perpendicular bisector of a and b, and arg(z − a) = θ is a half-line from a. Describing and sketching these loci, turning them into Cartesian equations, and shading the regions given by the matching inequalities.

Pure · Complex numbers

Loci in the Argand diagram

Reading a condition on z as a set of points in the Argand diagram: |z − a| = r is a circle centre a radius r, |z − a| = |z − b| is the perpendicular bisector of a and b, and arg(z − a) = θ is a half-line from a. Describing and sketching these loci, turning them into Cartesian equations, and shading the regions given by the matching inequalities.

Why it works

Every locus here comes from one idea from the previous section: ∣z1−z2∣|z_1 - z_2| is the distance between the points representing z1z_1 and z2z_2 on the Argand diagram, and arg⁡(z1−z2)\arg(z_1 - z_2) is the direction from one to the other. Read the condition as a sentence about distances or directions and the shape draws itself.

A circle: ∣z−a∣=r|z - a| = r. Write aa for the point it represents. Then ∣z−a∣|z - a| is the distance from the moving point zz to the fixed point aa. "That distance is always rr" is the definition of a circle, centre aa, radius rr. The one thing to watch is the sign: ∣z−a∣|z - a| has centre +a+a, so ∣z+3−2i∣=∣z−(−3+2i)∣|z + 3 - 2i| = |z - (-3 + 2i)| has centre −3+2i-3 + 2i, i.e. (−3,2)(-3, 2). In Cartesian form, putting z=x+iyz = x + iy and a=p+qia = p + qi, ∣z−a∣=r  ⟺  (x−p)2+(y−q)2=r2.|z - a| = r \iff (x - p)^2 + (y - q)^2 = r^2.

A perpendicular bisector: ∣z−a∣=∣z−b∣|z - a| = |z - b|. This says the point zz is the same distance from aa as from bb. The set of points equidistant from two fixed points is the perpendicular bisector of the line segment joining them. To get its Cartesian equation, square both sides — the x2x^2 and y2y^2 terms cancel, leaving a straight line. For example ∣z−1∣=∣z+i∣|z - 1| = |z + i| (points (1,0)(1,0) and (0,−1)(0,-1)): squaring gives (x−1)2+y2=x2+(y+1)2(x-1)^2 + y^2 = x^2 + (y+1)^2, so −2x+1=2y+1-2x + 1 = 2y + 1, i.e. y=−xy = -x.

A half-line: arg⁡(z−a)=θ\arg(z - a) = \theta. Here z−az - a is the vector from aa to zz, and its argument is fixed at θ\theta. So zz lies on the ray (half-line) starting at aa and pointing in the direction θ\theta. It is a half-line, not a full line, because only one direction has argument θ\theta (the opposite direction has argument θ±π\theta \pm \pi). The starting point aa itself is excluded — arg⁡0\arg 0 is undefined. For arg⁡(z−2)=π4\arg(z - 2) = \frac{\pi}{4} the ray starts at (2,0)(2, 0) and goes up at 45∘45^\circ: for x>2x > 2, y=x−2y = x - 2.

Regions from inequalities. Swap == for ≤\le or ≥\ge and you shade a region instead of a curve: ∣z−a∣≤r|z - a| \le r is the inside of the circle (the closed disc), ∣z−a∣≥∣z−b∣|z - a| \ge |z - b| is the half-plane on the bb-side of the bisector, and 0≤arg⁡(z−a)≤π20 \le \arg(z - a) \le \frac{\pi}{2} is a wedge between two half-lines.