How do equations and inequalities in z carve out lines, circles and regions in the Argand plane?
Describe and sketch subsets of the complex plane defined by equations and inequalities in modulus and argument
WACE Specialist Unit 3 loci in the complex plane: circles from modulus conditions, perpendicular bisectors from equal distances, rays from argument conditions, and shaded regions from inequalities, translated to Cartesian form, with a worked example.
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What this dot point is asking
SCSA wants you to interpret modulus and argument conditions geometrically, sketch the resulting curves and regions, and where needed convert to Cartesian equations.
Modulus conditions give circles and distances
Since is the distance from to the fixed point , the equation describes all points at distance from , a circle of radius centred at . The inequality shades the closed disc; shades the outside.
The condition says is equidistant from and , which is the perpendicular bisector of the segment joining them. The inequality shades the half-plane nearer to .
Modulus of a quotient or product condition
Some conditions involve the modulus of a quotient. Because can be rewritten as , it describes the set of points whose distance to is a fixed multiple of the distance to . When this is the perpendicular bisector, but for it is a circle, known as a circle of Apollonius. Squaring and expanding with produces a Cartesian equation that, after completing the square, reveals the centre and radius. Recognising the family the condition belongs to before grinding through the algebra is the efficient approach.
Argument conditions give rays
The condition describes all points for which the displacement from points in the fixed direction . This is a ray (half-line) starting at (open, since has no argument) at angle to the horizontal. A condition like shades a wedge between two rays.
More loci: ratios and sums of distances
Beyond circles, bisectors and rays, SCSA can set a few richer conditions. A condition of the form (a constant sum of distances to two fixed points) is an ellipse with foci and , provided exceeds the distance between them. A condition is the vertical line , and is the horizontal line . Recognising the geometric object before converting to Cartesian form saves time and avoids algebra errors, because the shape tells you what equation to expect.
Converting an argument condition carefully
An argument condition needs more care than a modulus condition when converted to Cartesian form. The equation gives the relationship , but only on the correct half-line: the ray points in the direction , so the signs of and must match that direction. Writing the Cartesian relation as a line and then restricting to the correct ray (and excluding the start point ) is the rigorous way to handle it.
Combining conditions
Many SCSA questions intersect two conditions, for example a disc and a wedge. Sketch each boundary, decide which side each inequality shades, and the answer is the overlap. Always state whether boundaries are included (solid) or excluded (dashed).
Exam-style practice questions
Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
WACE 20226 marksCalculator-assumed. On an Argand diagram sketch the region .Show worked answer →
An intersection of a disc and a wedge.
Rewrite , so both conditions are centred at the point , i.e. .
The first condition is the closed disc of radius centred at . The second is the wedge of directions from with angle between (the positive real direction) and (the positive imaginary direction), i.e. the first-quadrant wedge measured from .
The region is the quarter-disc: that part of the disc lying up and to the right of the centre, bounded by the two radii along the positive horizontal and vertical directions and the arc. Boundaries are solid (inequalities are weak). Markers reward the correct centre , the disc, the quarter-wedge, and the overlap shaded with solid boundaries.
WACE 20245 marksCalculator-free. The locus of satisfies . Find its Cartesian equation and describe it geometrically.Show worked answer →
A perpendicular-bisector locus.
The condition says is equidistant from (the point ) and (the point ), so the locus is the perpendicular bisector of the segment joining them.
Algebraically, put . Then and . Setting equal: , so , giving , hence , that is .
Markers reward recognising the perpendicular bisector, the squared-modulus expansion, and the line .
