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NSWMaths Extension 2Syllabus dot point

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.

Generated by Claude Opus 4.76 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Modulus conditions: circles and discs
  3. Equal distances: perpendicular bisectors
  4. Argument conditions: rays
  5. Combining conditions

What this dot point is asking

NESA wants you to interpret conditions on a complex variable zz 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 zz0|z - z_0| is the distance from zz to the fixed point z0z_0, the equation

zz0=r|z - z_0| = r

is the set of points at distance rr from z0z_0, a circle of centre z0z_0 and radius rr. The corresponding inequalities give regions: zz0<r|z - z_0| < r is the open interior (boundary dashed and excluded), zz0r|z - z_0| \le r is the closed disc, and zz0>r|z - z_0| > r is the exterior.

To handle a condition like z1+2i=3|z - 1 + 2i| = 3, rewrite the inside as z(12i)z - (1 - 2i), so the centre is 12i1 - 2i and the radius is 33.

Equal distances: perpendicular bisectors

The condition

za=zb|z - a| = |z - b|

says zz is equidistant from the points aa and bb. The locus of such points is the perpendicular bisector of the segment joining aa and bb. To find its Cartesian equation, write z=x+iyz = x + iy, square both moduli, and simplify; the squared terms x2x^2 and y2y^2 cancel, leaving a linear equation, confirming a straight line.

Argument conditions: rays

The condition

arg(zz0)=α\arg(z - z_0) = \alpha

fixes the angle of the vector from z0z_0 to zz. The locus is a ray (half-line) starting at z0z_0 and pointing in the direction α\alpha. The starting point z0z_0 itself is excluded, because arg0\arg 0 is undefined; mark it with an open circle. An inequality such as 0arg(zz0)π20 \le \arg(z - z_0) \le \tfrac{\pi}{2} describes a wedge-shaped region between two rays.

Combining conditions

Many problems intersect two conditions. For example, z2|z| \le 2 and 0argzπ30 \le \arg z \le \tfrac{\pi}{3} together describe a sector of the disc of radius 22: 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.