How do equations and inequalities in the complex variable describe curves and regions in the Argand plane?
Sketch curves and regions in the complex plane defined by conditions on modulus and argument, such as circles, perpendicular bisectors, rays and half-planes
A focused answer to the HSC Maths Extension 2 dot point on curves and regions in the Argand plane. Loci from modulus conditions, perpendicular bisectors, argument rays, and regions from inequalities, with verified worked examples.
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What this dot point is asking
NESA wants you to interpret conditions on a complex variable as geometric objects in the Argand plane. A modulus condition typically gives a circle or its interior; an argument condition gives a ray; an equality of two distances gives a perpendicular bisector. You must sketch the curve or region accurately, including whether boundaries are included.
Modulus conditions: circles and discs
Since is the distance from to the fixed point , the equation
is the set of points at distance from , a circle of centre and radius . The corresponding inequalities give regions: is the open interior (boundary dashed and excluded), is the closed disc, and is the exterior.
To handle a condition like , rewrite the inside as , so the centre is and the radius is .
Equal distances: perpendicular bisectors
The condition
says is equidistant from the points and . The locus of such points is the perpendicular bisector of the segment joining and . To find its Cartesian equation, write , square both moduli, and simplify; the squared terms and cancel, leaving a linear equation, confirming a straight line.
Argument conditions: rays
The condition
fixes the angle of the vector from to . The locus is a ray (half-line) starting at and pointing in the direction . The starting point itself is excluded, because is undefined; mark it with an open circle. An inequality such as describes a wedge-shaped region between two rays.
Combining conditions
Many problems intersect two conditions. For example, and together describe a sector of the disc of radius : a pie slice. Sketch each condition separately, then shade only the overlap. Be precise about which boundary lines or arcs are solid (included) and which are dashed (excluded).
Exam-style practice questions
Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 HSC3 marksSketch the region defined by |z| < 3 and 0 <= arg(z - i) <= pi/2.Show worked answer →
Build the region as the intersection of two conditions.
Condition 1, |z| < 3: the open interior of the circle of radius 3 centred at the origin O (boundary not included, so draw it dashed).
Condition 2, 0 <= arg(z - i) <= pi/2: the argument of (z - i) is the angle of the vector from the point i (that is, 0 + 1i) to z. Requiring this angle to lie between 0 and pi/2 inclusive selects the closed quarter-plane with vertex at i, bounded below by the ray from i in the positive x direction and on the left by the ray from i in the positive y direction. Both bounding rays are included.
The answer is the overlap: the part of that quarter-plane from vertex i that also lies strictly inside the circle |z| = 3.
Mark notes: 1 mark for the correct interior of |z| < 3, 1 mark for the correct wedge from i, 1 mark for shading only the intersection with the correct open and closed boundaries.
2021 HSC3 marksSketch the region of the complex plane defined by Re(z) >= Arg(z), where Arg(z) is the principal argument of z.Show worked answer →
Write z = x + iy, so Re(z) = x and Arg(z) = theta is the principal argument in (-pi, pi].
The boundary is the curve x = theta, that is, the real part equals the angle. In polar terms a point at angle theta is included when x = r cos(theta) >= theta.
Features to capture on the sketch:
- Near the positive x-axis theta is close to 0 and x > 0, so the condition holds: the region includes the positive real axis and a band around it.
- As theta increases towards pi (upper half-plane) the requirement x >= theta becomes harder to satisfy, because x is bounded while theta grows, so the region narrows and is cut off by the boundary curve x = theta.
- For theta < 0 (lower half-plane) the right-hand side theta is negative, so x >= theta holds over a much larger area; essentially the whole lower half-plane to the right of the boundary curve is included.
Shade the side of the boundary curve x = Arg(z) on which Re(z) is the larger quantity, including the boundary itself (since the relation is >=).
Mark notes: 1 mark for identifying the boundary Re(z) = Arg(z), 1 mark for correct behaviour in the upper versus lower half-plane, 1 mark for shading the correct closed region.