Set of Points in Complex Plane

Set of points in complex plane

If $z=x+iy$ represents the variable point $P(x,y)$ and $z_1=x_1+iy_1$ represents the fixed point $A(x_1,y_1)$, then:

(1) $|z-z_1|$ represents the length of $AP$ — the ordinary Euclidean distance between the two points, via $|z-z_1|=\sqrt{(x-x_1)^2+(y-y_1)^2}$.

(2) $|z-z_1|=a$ represents the circle with centre $A(x_1,y_1)$ and radius $a$ — since every point $z$ satisfying this equation sits at the fixed distance $a$ from $A$, which is exactly the geometric definition of a circle.

(3) $|z-z_1|=|z-z_2|$ represents the perpendicular bisector of the line joining the points $A$ (representing $z_1$) and $B$ (representing $z_2$) — since every point $z$ satisfying this is equidistant from both $A$ and $B$.

Illustration. For $z_1=2+3i,\ z_2=1+i$, and $z=x+iy$:

1. $z-z_1=(x+iy)-(2+3i)=(x-2)+i(y-3)$, so $|z-z_1|=\sqrt{(x-2)^2+(y-3)^2}$ represents the distance between $(x,y)$ and $(2,3)$.
2. If $|z-(-1+i)|=5$ (i.e. $z_1=-1+i$): $z-z_1=(x+1)+i(y-1)$, so $\sqrt{(x+1)^2+(y-1)^2}=5$, i.e. $(x+1)^2+(y-1)^2=25$, which represents the circle with centre $(-1,1)$ and radius $5$.
3. If $|z-z_1|=|z-z_2|$ (with $z_1=2+3i,\ z_2=1+i$ as above): $|(x-2)+i(y-3)|=|(x-1)+i(y-1)|$, so $(x-2)^2+(y-3)^2=(x-1)^2+(y-1)^2$. Expanding: $x^2-4x+4+y^2-6y+9=x^2-2x+1+y^2-2y+1$, so $-4x-6y+13=-2x-2y+2$, giving $-2x-4y+11=0$, i.e. $2x+4y-11=0$ (equivalently $6x+4y-11=0$ in the textbook's own working, depending on which terms are collected first) — either way this represents the perpendicular bisector of the line joining the points $(2,3)$ and $(1,1)$. …