Regions and curves in the complex plane (WACE Year 12 Mathematics Specialist Unit 3)

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 $|z - a|$ is the distance from $z$ to the fixed point $a$, the equation $|z - a| = r$ describes all points at distance $r$ from $a$, a circle of radius $r$ centred at $a$. The inequality $|z - a| \le r$ shades the closed disc; $|z - a| > r$ shades the outside.

The condition $|z - a| = |z - b|$ says $z$ is equidistant from $a$ and $b$, which is the perpendicular bisector of the segment joining them. The inequality $|z - a| < |z - b|$ shades the half-plane nearer to $a$.

Modulus of a quotient or product condition

Some conditions involve the modulus of a quotient. Because $\left|\dfrac{z - a}{z - b}\right| = k$ can be rewritten as $|z - a| = k\,|z - b|$, it describes the set of points whose distance to $a$ is a fixed multiple $k$ of the distance to $b$. When $k = 1$ this is the perpendicular bisector, but for $k \neq 1$ it is a circle, known as a circle of Apollonius. Squaring $|z - a|^2 = k^2 |z - b|^2$ and expanding with $z = x + iy$ 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 $\arg(z - a) = \alpha$ describes all points $z$ for which the displacement from $a$ points in the fixed direction $\alpha$. This is a ray (half-line) starting at $a$ (open, since $z = a$ has no argument) at angle $\alpha$ to the horizontal. A condition like $\tfrac{\pi}{6} \le \arg(z - a) \le \tfrac{\pi}{3}$ 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 $|z - a| + |z - b| = k$ (a constant sum of distances to two fixed points) is an ellipse with foci $a$ and $b$, provided $k$ exceeds the distance between them. A condition $\operatorname{Re}(z) = c$ is the vertical line $x = c$, and $\operatorname{Im}(z) = c$ is the horizontal line $y = c$. 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 $\arg(z - a) = \alpha$ gives the relationship $\dfrac{y - a_2}{x - a_1} = \tan\alpha$, but only on the correct half-line: the ray points in the direction $\alpha$, so the signs of $x - a_1$ and $y - a_2$ must match that direction. Writing the Cartesian relation as a line and then restricting to the correct ray (and excluding the start point $a$) 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).