Loci in the Argand diagramEdexcel A-Level Further Maths: Revision notes

Because $|z-a|$ is the distance from $z$ to the point $a$, the locus $|z-a|=r$ is a circle with centre $a$ and radius $r$. Read the centre as the number subtracted, reversing signs: $|z-3+2i|=|z-(3-2i)|$ has centre $3-2i$.

• $|z-a|\le r$: the circle and its inside (a closed disc).
• $|z-a|>r$: the outside of the circle, boundary excluded. Cartesian form: with $z=x+iy$ and $a=p+qi$, $|z-a|=r$ becomes $(x-p)^2+(y-q)^2=r^2$.

This describes the points equidistant from $a$ and $b$: the perpendicular bisector of the line joining them. For $|z-1|=|z-i|$, squaring gives $(x-1)^2+y^2=x^2+(y-1)^2$, so $y=x$. For the region $|z-a|\le|z-b|$, the points at least as close to $a$ as to $b$ lie on one side of the bisector, the side containing $a$. Test a point (such as $a$ itself) to decide which side, rather than guessing.

$\arg(z-a)=\theta$ is a half-line starting at the point $a$ (not included), making an angle $\theta$ with the positive real direction. It is not a full line. Example: $\arg(z-2i)=\frac{\pi}{4}$ with $z=x+iy$ gives $\tan\frac{\pi}{4}=\frac{y-2}{x}$ with $x>0$, so $y=x+2$ for $x>0$. The region $\alpha<\arg(z-a)<\beta$ is the wedge between two half-lines from $a$ at angles $\alpha$ and $\beta$, with the half-lines and $a$ excluded for strict inequalities.

Regions combine conditions, such as $|z-3-4i|\le5$ together with $|z|\le|z-6|$. Draw each boundary, shade or test each region, then take the overlap. Example: $|z|\le|z-6|$ gives $x\le3$. The circle $|z-3-4i|=5$ has centre $(3,4)$, so the line $x=3$ passes through the centre and $S_1\cap S_2$ is a semicircle of area $\frac{25\pi}{2}$. The boundaries meet where $x=3$ and $(y-4)^2=25$, at $3+9i$ and $3-i$. To find intersections, substitute one Cartesian equation into the other. For a half-line, reject solutions outside its allowed range, for example when $x<0$ on a locus with $x>0$.