Curves and regions in the complex plane - TCE Mathematics Specialised (Tasmania)

What this dot point is asking

This dot point asks you to turn algebraic statements about a complex variable $z$ into pictures, and to sketch the set of all points satisfying them. The key translation, developed on the Argand diagram, is that $|z - a|$ is the distance from $z$ to the fixed point $a$, and $\arg(z - a)$ is the direction from $a$ to $z$.

Circles from a modulus condition

The equation $|z - a| = r$ says the distance from $z$ to the fixed point $a$ is constant and equal to $r$. That is exactly the definition of a circle of radius $r$ centred at $a$. If $a = p + iq$, writing $z = x + iy$ gives

the familiar Cartesian circle. The corresponding region $|z - a| \le r$ is the closed disc, and $|z - a| > r$ is everything outside the circle.

Perpendicular bisectors from equal distances

The equation $|z - a| = |z - b|$ says $z$ is equidistant from the two fixed points $a$ and $b$. The set of such points is the perpendicular bisector of the segment joining $a$ and $b$. The inequality $|z - a| < |z - b|$ is the half-plane of points closer to $a$.

Rays from an argument condition

The equation $\arg(z - a) = \theta$ fixes the direction from the point $a$ to $z$. The locus is a ray (half-line) starting at $a$, pointing at angle $\theta$ to the positive real axis. The starting point $a$ itself is excluded, because $\arg(0)$ is undefined. A condition like $\alpha \le \arg(z - a) \le \beta$ describes a wedge-shaped region between two rays.

Combining conditions

Many questions intersect two loci, for example a circle and a ray, or shade the overlap of two regions. Sketch each condition separately in light pencil, then identify the common set. Note carefully whether each boundary is included (use a solid line for $\le$ or $\ge$) or excluded (a dashed line for strict inequality), and whether endpoints of rays are open circles.

The weighted-distance circle (an Apollonius circle)

A condition like $|z + 1| = \sqrt{2}\,|z - i|$, where the two distances are in a fixed ratio other than $1$, is not a perpendicular bisector. It is still a circle, called an Apollonius circle, and the past question above shows the standard method: write $z = x + iy$, square both sides to clear the moduli, and complete the square. The ratio $\sqrt{2}$ shifts the centre off the midpoint and sets a finite radius. Recognising that an unequal ratio of moduli gives a circle (while an equal ratio gives a line) saves you from expecting the wrong shape.

Why this matters

Loci questions reward the student who thinks geometrically first and reaches for algebra only to confirm. Knowing the core shapes (circle, perpendicular bisector, ray, Apollonius circle) on sight lets you sketch quickly and spend your time on the intersections and boundary details that earn the final marks.