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

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

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 $z_1$.

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.

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.

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.

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 $2$.