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

$|z-a|$ is the distance from the point $z$ to the point $a$ in the Argand diagram. So $|z-a|=r$ is the set of points at distance $r$ from $a$: a circle with centre $a$ and radius $r$ (here $r>0$). Read the centre carefully: $|z+2-3\mathrm{i}|=|z-(-2+3\mathrm{i})|$, so the centre is $-2+3\mathrm{i}$. To get the Cartesian equation put $z=x+y\mathrm{i}$ and square: $|z-(a+b\mathrm{i})|=r$ becomes $(x-a)^2+(y-b)^2=r^2$.

$|z-a|<r$ is the inside of the circle and $|z-a|>r$ is the outside. A strict inequality leaves the boundary out (drawn dashed); $\le$ or $\ge$ includes it (drawn solid). A region such as $1\le|z-2|\le3$ is the annulus between two circles with the same centre. Test one point to check which side you shade: for $|z-2|>3$, $z=0$ gives $|{-2}|=2$, which is not greater than $3$, so $0$ is not in the region.

$|z-a|=|z-b|$ means $z$ is the same distance from $a$ as from $b$. The locus is the perpendicular bisector of the line segment joining $a$ and $b$, a straight line. Example: $|z-2|=|z-2\mathrm{i}|$. Squaring with $z=x+y\mathrm{i}$: $(x-2)^2+y^2=x^2+(y-2)^2$, which simplifies to $y=x$. $|z-a|<|z-b|$ is the half-plane on the side of the bisector containing $a$.

$\arg(z-a)=\theta$ is the set of points for which the line from $a$ to $z$ makes the angle $\theta$ with the positive real direction: a half-line starting at $a$ (not including $a$ itself, as $\arg 0$ is undefined). The principal argument lies in $(-\pi,\pi]$. Example: $\arg(z+2-\mathrm{i})=\frac{\pi}{4}$ starts at $-2+\mathrm{i}$ and goes up and to the right. With $z=x+y\mathrm{i}$: $\frac{y-1}{x+2}=\tan\frac{\pi}{4}=1$, so $y=x+3$ with $x>-2$. $\alpha<\arg(z-a)<\beta$ is the wedge between two half-lines from $a$.

Many questions give two conditions together, for example $|z-3\mathrm{i}|\le3$ and $\arg z\ge\frac{\pi}{4}$. Sketch each boundary, shade the intersection, and use Cartesian forms to find where boundaries meet. Worked example: where does $\arg z=\frac{\pi}{4}$ meet $|z-3\mathrm{i}|=3$? The half-line is $y=x$ with $x>0$, and the circle is $x^2+(y-3)^2=9$. Substituting, $2x^2-6x=0$, so $x=0$ or $3$. $x=0$ is the origin, which is not on the half-line, so the point is $3+3\mathrm{i}$. Areas of regions come from sectors and segments, using radians: sector $\frac12r^2\theta$, segment $\frac12r^2(\theta-\sin\theta)$.