1 · IB Maths
Complex loci
An equation in z traces a shape in the Argand diagram. Drag the fixed points a and b and a test point z, and watch the circle, perpendicular bisector or half-line follow.
Predict first
Start from the default settings (circle mode, r = 2.0, test point z at distance 2.24 from a, so outside the circle). If you raise the Radius, r slider to 3.0, where is z?
z is at distance |z − a| = √(2² + 1²) = 2.24 from a. A circle of radius 3.0 contains every point closer to a than 3.0, so z is now inside.
Check it in the model: Move the Radius, r slider from 2.0 to 3.0 and read the Test point line, which states whether z is inside or outside.
locus fixed point a fixed point b · drag a, b and the test point z (or use the sliders)
Point coordinates (sliders)
bisector: |z − a| = |z − b|
half-line: arg(z − a) = θ
z = x + iy, |z − a| = √((x−a₁)² + (y−a₂)²)
Try this
- In circle mode, drag z across the circle and watch the message under “Test point”. At what distance from a does it change from inside to outside?
- Switch to the perpendicular bisector. Move b so that it sits directly above a. Is the bisector horizontal or vertical? Tick “Shade” and check which side is nearer a.
- Choose the half-line, set θ = 135° and then θ = −45°. Which Cartesian form is shown and why does the x-condition change sign? Tick “Shade” to see the wedge between 0 and θ.
Exam tip: sketch the locus before doing any algebra: |z − a| = r is a circle centred at a, |z − a| = |z − b| is the perpendicular bisector of ab, and arg(z − a) = θ is a half-line from a that does not include a itself.
More on 1: Arithmetic and geometric series · Compound interest and depreciation · Proof by induction · Permutations and combinations · De Moivre's theorem · Systems of linear equations · Complex numbers on the Argand plane · Roots of a complex number · Matrix transformations · Sequences and series · Exponentials and logarithms · Binomial theorem