Regions and curves in the complex plane in QCE Specialist Mathematics Unit 3

What this dot point is asking

QCAA wants you to translate a condition on a complex number $z$ into a curve or region on the Argand plane and sketch it accurately. The conditions use modulus (distance), argument (angle) and inequalities. Recognising that $|z - z_0|$ is a distance and $\arg(z - z_0)$ is a direction lets you convert algebra into geometry quickly.

The answer

Modulus as distance

For $z = x + iy$, the expression $|z - z_0|$ is the distance from the point $z$ to the fixed point $z_0$ on the plane. So

is a circle of radius $r$ centred at $z_0$. The inequality $|z - z_0| \leq r$ is the closed disc (interior plus boundary), and $|z - z_0| > r$ is the exterior.

Perpendicular bisector from equal distances

The condition $|z - z_1| = |z - z_2|$ says $z$ is equidistant from $z_1$ and $z_2$. The set of such points is the perpendicular bisector of the segment joining $z_1$ and $z_2$, a straight line. The inequality $|z - z_1| < |z - z_2|$ is the half-plane closer to $z_1$.

Argument conditions give rays

The condition $\arg(z - z_0) = \alpha$ describes all points whose direction from $z_0$ is the fixed angle $\alpha$. This is a ray (half-line) starting at $z_0$ (open endpoint, since $z = z_0$ has no defined argument) at angle $\alpha$ to the positive real direction. A condition like $0 \leq \arg(z - z_0) \leq \tfrac{\pi}{3}$ is the wedge-shaped region between two rays.

Converting to Cartesian form

To confirm a sketch, substitute $z = x + iy$. For example $|z - 2| = 3$ becomes $\sqrt{(x-2)^2 + y^2} = 3$, so $(x - 2)^2 + y^2 = 9$: a circle, centre $(2, 0)$, radius $3$. Squaring both sides removes the surd, turning the modulus condition into a recognisable Cartesian equation.

Boundaries: solid or dashed

Strict inequalities ($<$, $>$) exclude the boundary, drawn as a dashed curve. Inclusive inequalities ($\leq$, $\geq$) include the boundary, drawn solid. The marker checks both the correct shape and the correct boundary style, and whether the interior or exterior is shaded.

Combining conditions

Two or more conditions joined by "and" give the intersection of the regions, the overlap. Sketch each region, then shade only the common part. This is where careful boundary handling matters most, since the answer is the set satisfying every condition at once.

Circles from a general modulus equation

Not every circle is presented as $|z - z_0| = r$. A relation such as $|z - 1| = 2|z + 2|$ also gives a circle (an Apollonius circle), found by squaring. With $z = x + iy$, the left side squared is $(x - 1)^2 + y^2$ and the right is $4[(x + 2)^2 + y^2]$. Expanding and collecting gives $3x^2 + 3y^2 + 18x + 15 = 0$, that is $x^2 + y^2 + 6x + 5 = 0$, which completes the square to $(x + 3)^2 + y^2 = 4$: a circle of centre $(-3, 0)$ and radius $2$. The lesson is that any equation linear in $|z - a|^2$ terms reduces to a circle or a line, and squaring is the reliable route to its standard form.

Lines and half-planes from real and imaginary parts

Conditions on $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ are the simplest: $\operatorname{Re}(z) = k$ is the vertical line $x = k$, and $\operatorname{Im}(z) \geq k$ is the horizontal half-plane $y \geq k$. More subtly, a condition like $\operatorname{Re}(z + \bar z)$ or one mixing $z$ and $\bar z$ should be expanded into $x$ and $y$ first, after which the geometry is read directly. Because $z + \bar z = 2x$ and $z - \bar z = 2iy$, these substitutions convert any conjugate expression into Cartesian coordinates cleanly.

Why the start point of a ray is excluded

For $\arg(z - z_0) = \alpha$, the point $z = z_0$ is excluded because $\arg(0)$ is undefined: a zero vector has no direction. Mark the endpoint with an open circle. This is a small but regularly rewarded detail, and it is the geometric reason a ray is a half-line missing its tip rather than a full segment.