Further loci and regions in the Argand diagramEdexcel A-Level Further Maths: Revision notes

In the Argand diagram, $z=x+iy$ and $|z-a|$ is the distance from $z$ to the point $a$.

• $|z-a|=r$: a circle, centre $a$, radius $r$ (and $|z-a|\leq r$ is the disc).
• $|z-a|=|z-b|$: the perpendicular bisector of the line joining $a$ and $b$.
• $\arg(z-a)=\theta$: a half-line starting at $a$ (excluding $a$) at angle $\theta$ to the positive real direction. Inequalities describe regions: $|z-a|<r$ is inside the circle; $\arg(z-a)$ between two values is the wedge between two half-lines; $p\leq\mathrm{Re}(z)\leq q$ is a vertical strip and $p\leq\mathrm{Im}(z)\leq q$ a horizontal strip.

If $k=1$ the locus is the perpendicular bisector. If $k\neq1$ it is a circle (the circle of Apollonius). Method: let $z=x+iy$, square both sides, and complete the square. Example: $|z-1|=2|z-4|$: $(x-1)^2+y^2=4\left[(x-4)^2+y^2\right]\;\Rightarrow\;x^2+y^2-10x+21=0\;\Rightarrow\;(x-5)^2+y^2=4.$ Centre $5$, radius $2$. It meets the real axis at $3$ and $7$, as a check: $|3-1|=2=2|3-4|$ and $|7-1|=6=2|7-4|$. Greatest $|z|$ on the circle is $5+2=7$.

Since $\arg\frac{z-a}{z-b}=\arg(z-a)-\arg(z-b)$, this is the angle at $z$ between the lines to $b$ and $a$: the locus is an arc of a circle through $a$ and $b$ (the end points are excluded). If $\beta=\frac{\pi}{2}$ it is a semicircle (angle in a semicircle); if $\beta<\frac{\pi}{2}$ a major arc; if $\beta>\frac{\pi}{2}$ a minor arc. $\beta=0$ or $\pi$ gives a line. For $\arg\frac{z-2}{z+2}=\frac{\pi}{2}$ multiply by the conjugate of the denominator: $\frac{z-2}{z+2}=\frac{x^2+y^2-4+4iy}{|z+2|^2}.$ Real part $0$ gives $x^2+y^2=4$; imaginary part $>0$ gives $y>0$: the upper semicircle.

$\alpha\leq\arg(z-z_1)\leq\beta$ is the wedge between two half-lines from $z_1$. $p\leq\mathrm{Re}(z)\leq q$ is the strip between the vertical lines $x=p$ and $x=q$. Combining conditions means finding the intersection of the regions. Example: $\frac{\pi}{6}\leq\arg(z-1)\leq\frac{\pi}{3}$ with $\mathrm{Re}(z)\leq3$ is a triangle with vertices $(1,0)$, $\left(3,\frac{2}{\sqrt3}\right)$, $\left(3,2\sqrt3\right)$, of area $\frac{4\sqrt3}{3}$.

$|z-a|\leq|z-b|$ is the half-plane on $a$'s side of the perpendicular bisector. For $|z-2|\leq|z-2i|$: $(x-2)^2+y^2\leq x^2+(y-2)^2\Rightarrow y\leq x$. Together with $|z|\leq4$ this is half of a disc of radius $4$ (the line passes through the centre), with area $8\pi$. For $z\neq0$ in this region $-\frac{3\pi}{4}\leq\arg z\leq\frac{\pi}{4}$, and the greatest $\mathrm{Im}(z)$ is $2\sqrt2$, at the corner of the half-disc where $y=x$ meets the circle.

For a circle with centre $c$ and radius $r$ the least and greatest $|z-w|$ are $|c-w|-r$ and $|c-w|+r$ (if $w$ is outside), attained on the line through $w$ and $c$. Always check that the point found really lies on the locus: on an arc such as the upper semicircle, a nearest point might be an excluded end point. Example: on the upper semicircle $|z|=2$, the least $|z-5i|$ is $5-2=3$ at $z=2i$, which has $y>0$.