Loci and regions in the Argand diagramEdexcel International A Level Further Maths: Revision notes

$|z-a|$ is the distance between the points representing $z$ and $a$. So $|z-a|=b$ is a circle with centre $a$ and radius $b$. Take care with signs: $|z-3+4i|=|z-(3-4i)|$ has centre $3-4i$. In Cartesian form, with $z=x+iy$ and $a=p+iq$: $(x-p)^2+(y-q)^2=b^2$. The greatest and least values of $|z|$ on the circle are the distance from the origin to the centre, plus or minus the radius. For $|z-3+4i|=5$: $5+5=10$ and $5-5=0$.

If $|z-a|=|z-b|$ then $z$ is equally distant from $a$ and $b$: the perpendicular bisector of the line joining them. For $|z-2|=|z+4i|$, $(x-2)^2+y^2=x^2+(y+4)^2$ gives $x+2y+3=0$. If $|z-a|=k|z-b|$ with $k\ne1$, the locus is a circle. Square both sides and set $z=x+iy$. For $|z+3|=\frac12|z-6|$: $4\left[(x+3)^2+y^2\right]=(x-6)^2+y^2$, so $(x+6)^2+y^2=36$: centre $-6$, radius $6$.

$\arg(z-a)=\beta$ is a half-line starting at the point $a$ (which is not included), making an angle $\beta$ with the positive real direction. For $\arg(z-2+i)=\frac\pi3$ it starts at $2-i$ and has equation $y+1=\sqrt3(x-2)$ with $x>2$. Convert with $\tan\beta=\frac{y-q}{x-p}$, and choose the half-line using the sign of $x-p$ or $y-q$ for the quadrant of $\beta$.

$\arg\frac{z-a}{z-b}=\arg(z-a)-\arg(z-b)$ is the angle between the lines from $z$ to $b$ and to $a$. The locus is an arc of a circle through $a$ and $b$. If $\beta=\frac\pi2$ it is a semicircle with diameter $ab$. Algebraically, multiply by the conjugate of the denominator and use the real and imaginary parts. For $\arg\frac{z+3}{z-6}=\frac\pi2$: real part $=0$ gives $\left(x-\frac32\right)^2+y^2=\frac{81}{4}$, and imaginary part $-9y>0$ gives $y<0$: the lower semicircle. Test one point to decide which side the arc lies on.

Replace $=$ by an inequality to describe a region. $|z-a|\le b$ is the closed disc inside and on the circle; $|z-a|>b$ is the outside. $|z-a|\le|z-b|$ is the half-plane of points closer to $a$ than to $b$: for $|z-2|\le|z+4i|$, $x+2y+3\ge0$. A solid boundary includes the line or circle; a dashed boundary (strict inequality) does not. Check a test point such as $z=a$ to see which side satisfies the inequality.