Further complex numbers (Further Pure 2)

Explanation

• Three loci recur. First, $\left|z-a\right|=r$ is the circle of centre $a$ and radius $r$, and $\left|z-a\right|=\left|z-b\right|$ is the perpendicular bisector of the segment joining $a$ and $b$.
• Second, $\left|z-a\right|=k\left|z-b\right|$ with $k>0$ and $k\ne1$ is a circle, found by writing $z=x+\mathrm{i}y$, squaring both moduli and completing the square; it is called a circle of Apollonius.
• Third, $\arg\left(z-a\right)=\theta$ is a half-line from $a$, with $a$ itself excluded, while $\arg\left(\dfrac{z-a}{z-b}\right)=\theta$ is an arc of a circle through $a$ and $b$, with the endpoints excluded, because the angle subtended by the chord $ab$ is constant on each arc.
• Replacing an equality by an inequality gives a region: $\left|z-a\right|<r$ is the inside of the circle, and $\left|z-a\right|<\left|z-b\right|$ is the half-plane containing $a$.
• Sketch the boundary first, then test one convenient point to decide which side to shade.

Explanation

• A transformation $w=f(z)$ maps each point of the $z$-plane to a point of the $w$-plane, and the standard task is to find the image of a given line, circle or region.
• The reliable method is to invert the transformation, obtaining $z$ in terms of $w$, and substitute into the equation of the original locus.
• For $w=z+a$ the map is a translation, for $w=kz$ with $k$ real and positive an enlargement, and for $w=\mathrm{e}^{\mathrm{i}\alpha}z$ a rotation about the origin.
• The important non-trivial map is $w=\dfrac{1}{z}$, and more generally $w=\dfrac{az+b}{cz+d}$: these send circles and lines to circles and lines, but a line through the origin can become a line while a line not through the origin becomes a circle through the origin, and vice versa.
• When the substitution leaves a modulus equation, write $w=u+\mathrm{i}v$ and complete the square; when it leaves an equation of the form $\left|w-p\right|=\left|w-q\right|$, the image is a straight line and there is no need to expand.