Sketch curves and regions in the complex plane defined by conditions on modulus and argument, such as circles, perpendicular bisectors, rays and half-planes

What this dot point is asking

NESA wants you to interpret conditions on a complex variable $z$ 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. The skill is reading $|z - a|$ as "the distance from $z$ to the point $a$" and $\arg(z - a)$ as "the angle of the arrow from $a$ to $z$", then drawing what those constraints force.

Modulus conditions: circles and discs

Since $|z - z_0|$ is the distance from $z$ to the fixed point $z_0$, the equation

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

To handle a condition like $|z - 1 + 2i| = 3$, rewrite the inside as $z - (1 - 2i)$, so the centre is $1 - 2i$ and the radius is $3$. Watch the sign: the centre is the value of $z_0$ that makes the bracket zero, which is the opposite sign to what appears after the minus.

Equal distances: perpendicular bisectors

The condition

says $z$ is equidistant from the points $a$ and $b$. The locus of such points is the perpendicular bisector of the segment joining $a$ and $b$. To find its Cartesian equation, write $z = x + iy$, square both moduli, and simplify; the squared terms $x^2$ and $y^2$ cancel, leaving a linear equation, confirming a straight line. Geometrically you can sketch it instantly: plot $a$ and $b$, find the midpoint, and draw the line through it at right angles to $ab$.

Argument conditions: rays

The condition

fixes the angle of the vector from $z_0$ to $z$. The locus is a ray (half-line) starting at $z_0$ and pointing in the direction $\alpha$. The starting point $z_0$ itself is excluded, because $\arg 0$ is undefined; mark it with an open circle. Only the half-line in the direction $\alpha$ counts, not the opposite direction (that would be argument $\alpha + \pi$). An inequality such as $0 \le \arg(z - z_0) \le \tfrac{\pi}{2}$ describes a wedge-shaped region between two rays.

Combining conditions: building a region

Many problems intersect two conditions, and the reliable method is to draw each one separately, decide which boundaries are solid (included) or dashed (excluded), then shade only the overlap. The figure builds the 2024 HSC region $|z| < 3$ and $0 \le \arg(z - i) \le \tfrac{\pi}{2}$ stage by stage.

Stage 1, the modulus condition. $|z| < 3$ is the open interior of the circle of radius $3$ centred at the origin. The inequality is strict, so the boundary circle is dashed and not part of the region.

Stage 2, the argument condition. $0 \le \arg(z - i) \le \tfrac{\pi}{2}$ is the quarter-plane with vertex at $i$, bounded below by the ray from $i$ in the positive real direction and on the left by the ray from $i$ in the positive imaginary direction. The inequalities are inclusive, so both rays are solid; the vertex $i$ is excluded (open circle) because $\arg 0$ is undefined.

Stage 3, shade the overlap. The answer is only where both conditions hold: the part of the quarter-plane from $i$ that lies strictly inside the circle. Its boundary is the two solid rays and the dashed arc, meeting at the open vertex $i$. Always test one interior point if unsure which side to shade; here $z = 1 + 2i$ satisfies both conditions, confirming the shaded piece.

For a sector of a disc such as $|z| \le 2$ and $0 \le \arg z \le \tfrac{\pi}{3}$ the same routine gives a "pie slice"; for an annulus $1 \le |z| \le 3$ it gives the ring between two concentric circles.

Ratio-of-distance loci (the Apollonius circle)

A condition of the form $|z - a| = k|z - b|$ with $k \ne 1$ does not give a line; it gives a circle, called an Apollonius circle. Square both sides into Cartesian form and the $x^2$ and $y^2$ terms survive (because the coefficient is $1 - k^2 \ne 0$), so completing the square reveals a centre and radius. Only the balanced case $k = 1$ collapses to the perpendicular bisector. Recognising whether a modulus condition produces a line or a circle is a common discriminating mark in HSC questions.

Conditions mixing modulus and argument

Some loci constrain both length and angle at once. The set $|z| = 2\arg z$ (with $\arg z$ in radians) is a spiral, since the radius grows in proportion to the angle. More commonly you meet a half-line crossed with a circle, or an annulus $1 \le |z| \le 3$ (the region between two concentric circles). For an annulus, shade between the two circles, drawing the inner and outer boundaries solid or dashed according to whether the inequalities are strict. Always test one convenient point (often the origin or $z = 1$) to confirm you have shaded the correct side.

A reliable sketching method

For any locus question, follow a fixed routine. First, substitute $z = x + iy$ if an algebraic equation is wanted, or interpret the modulus and argument geometrically if only a sketch is needed. Second, identify the type of object (circle, line, ray, region). Third, mark key features: centre and radius for a circle, the vertex and direction for a ray, intercepts for a line. Fourth, decide solid versus dashed boundaries from the strictness of the inequality. Finally, shade and verify with a test point.

How exam questions ask about curves and regions

• "Sketch the locus / curve given by ...": identify the object from the form ($|z - z_0| = r$ circle, $|z - a| = |z - b|$ line, $\arg(\,\cdot\,) = \alpha$ ray) and draw it with key features labelled.
• "Find the Cartesian equation of the locus ...": substitute $z = x + iy$, square any moduli, and simplify.
• "Sketch the region defined by ...": treat each inequality separately, decide solid/dashed boundaries, then shade the intersection.
• "... the locus is a straight line. Find ...": this signals an equal-distance condition, so it is a perpendicular bisector.
• "Find the point on the locus closest to / farthest from ...": usually the foot of a perpendicular (to a line) or a point on the line through the centre (for a circle).
• "Find the greatest value of $|z|$ ... / $\arg z$ ... subject to ...": an optimisation over a circle or region; sketch first, then read off the extreme point geometrically.