4.2 Kinematics of motion in a straight line

|z−a|=r describes a circle, arg(z−a)=θ a ray, and |z−a|=|z−b| the perpendicular bisector of AB. Each condition is a set of allowed points.

Sketch reference points first, then apply equality or inequality signs. For multiple conditions, keep only their intersection and note excluded endpoints or rays.

|z−(1+i)|=2 is a circle centred at (1,1) with radius 2.

The modulus condition gives distance, while an argument condition gives direction; confusing them changes the locus completely.

|z−a|<r selects the interior of a circle and |z−a|>r its exterior. An argument inequality selects a wedge, with boundary rays included or excluded according to the sign.

Test one convenient point to decide which side of each boundary is wanted, then combine conditions. State whether boundaries are included.

|z|≤2 and arg z between 0 and π/2 gives a closed quarter-disk in the first quadrant.

Sketching only the boundary is incomplete when an inequality describes a region.

A condition such as |z−a|=|z−b| gives a perpendicular bisector, while |z−a|=r gives a circle and arg(z−a)=θ gives a ray. Multiple conditions mean an intersection of sets.

Translate each condition separately, sketch boundaries, and test a point to select the correct side for inequalities.

|z−1|=|z+1| is the imaginary axis: points equidistant from 1 and −1 have real part zero.

The equal-distance locus is not the line joining the two points; it is perpendicular to that line.

To find where a line meets a locus, parameterise the line, substitute its coordinates into the modulus or argument condition, and solve for permitted parameter values.

Check the parameter interval and retain only points satisfying every original condition. A quadratic parameter may represent two, one or no intersections.

Substituting z=1+iλ into |z|=√2 gives 1+λ²=2, so λ=±1 and two points on the circle.

Solving the squared modulus equation can introduce a sign or domain issue; verify in the unsquared condition.