Circles and Bisectors on the Argand Diagram

A modulus is a distance

For z = x + yi, $|z|=\sqrt{x^2+y^2}$|z| = √(x² + y²), the length of the arrow from the origin to z. So |z| = 2 asks for every point exactly 2 from the origin: the circle of radius 2 centered at 0.

Squaring gives its Cartesian equation, $x^2+y^2=4$x² + y² = 4. The points 2i, −2 and $\sqrt{2}+\sqrt{2}i$√2 + √2 i are on it: each is 2 from the origin.

The circle |z| = 2: every point 2 from the origin, including 2i, −2 and $\sqrt{2}+\sqrt{2}i$√2 + √2 i.

Measured from a

z − a is the arrow from a to z: it is what is added to a to reach z. So |z − a| is the distance from a to z, and |z − a| = r is the circle of radius r centered at a.

For |z − 3 − 4i| = 2, write it as |z − (3 + 4i)| = 2: the circle of radius 2 about 3 + 4i. With z = x + yi its equation is ${\left(x-3\right)}^2+{\left(y-4\right)}^2=4$(x − 3)² + (y − 4)² = 4. Test a point: z = 5 + 4i makes z − (3 + 4i) = 2, whose modulus is 2, so 5 + 4i is on the circle; so is 3 + 6i.

The circle |z − 3 − 4i| = 2, centered at 3 + 4i. The gold arrow is z − (3 + 4i) for z = 5 + 4i: it runs from the center to z, and its length is 2.

Read the center off

Write the inside as z minus one number. That number is the center, and the number on the other side is the radius.

|z − 5i| = 3 is the circle of radius 3 about 5i. |z − 2 + 3i| = 4 is |z − (2 − 3i)| = 4, the circle of radius 4 about 2 − 3i; the point 6 − 3i is on it, since 6 − 3i − (2 − 3i) = 4. |z + 1| = 3 is |z − (−1)| = 3, the circle of radius 3 about −1.

A number multiplying z comes out first: |2z − 4| = 6 is 2|z − 2| = 6, so |z − 2| = 3, the circle of radius 3 about 2.

Equal distances: the perpendicular bisector

|z − a| = |z − b| asks for every point as far from a as from b. Those points make the perpendicular bisector of the segment from a to b: the line through its midpoint at right angles to it.

Take a = −2 and b = 4 + 2i. With z = x + yi, square both sides: ${\left(x+2\right)}^2+y^2={\left(x-4\right)}^2+{\left(y-2\right)}^2$(x + 2)² + y² = (x − 4)² + (y − 2)². The $x^2$x² and $y^2$y² cancel, leaving 4x + 4 = −8x − 4y + 20, so 12x + 4y = 16, which is y = −3x + 4.

The midpoint of a and b is 1 + i, and 1 = −3 × 1 + 4, so the line passes through it. The segment from a to b has gradient $\frac{2}{6}=\frac{1}{3}$2/6 = 1/3, and $\frac{1}{3}\times \left(-3\right)=-1$1/3 × (−3) = −1, so the line crosses it at right angles. Test a point: z = 2 − 2i is on the line, and $|z-a|=|4-2i|=\sqrt{20}$|z − a| = |4 − 2i| = √20 while $|z-b|=|-2-4i|=\sqrt{20}$|z − b| = |−2 − 4i| = √20.

a = −2 and b = 4 + 2i lie on the plain line. The gold line, y = −3x + 4, crosses it at right angles at the midpoint 1 + i. Every point on the gold line is as far from a as from b: 2 − 2i is $\sqrt{20}$√20 from each.

Drag z. The two bands show its distances to $z_1$z₁ and to $z_2$z₂, and the dashed line is the set where they are equal. Move z onto it and the two readings agree; move $z_2$z₂ and the line moves with it. The other button asks for $|z-z_1|=2$|z − z₁| = 2 instead, and the set becomes a circle about $z_1$z₁.

A circle or a line

Fix one distance, |z − a| = r, and the points lie on a circle. Set two distances equal, |z − a| = |z − b|, and they lie on a straight line. The equation |z − 1| = |z + 1| is the line through 0 at right angles to the real axis: the imaginary axis. The point 3i is $\sqrt{10}$√10 from 1 and $\sqrt{10}$√10 from −1.

Only equal distances give a line. |z| = 2|z − 3| asks for points twice as far from 0 as from 3. Squaring, $x^2+y^2=4\left({\left(x-3\right)}^2+y^2\right)$x² + y² = 4((x − 3)² + y²), which simplifies to ${\left(x-4\right)}^2+y^2=4$(x − 4)² + y² = 4: the circle of radius 2 about 4. The point 2 is 2 from 0 and 1 from 3; the point 6 is 6 from 0 and 3 from 3.

The usual mistakes

Reading the center with the wrong sign. |z − 2 + 3i| = 4 is centered at 2 − 3i, not −2 + 3i or 2 + 3i: rewrite it as |z − (2 − 3i)| first.

Swapping the center and the radius. |z − 5i| = 3 has center 5i and radius 3.

Drawing the line through a and b. The equal-distance points lie on the perpendicular bisector, at right angles to the segment ab; the real axis passes through 1 and −1, and every point on it except 0 is nearer one of them.

Calling |z − a| = r a line. One fixed distance gives a circle, including |z| = 1, centered at the origin.

A drone between two masts

The application below joins the two loci: the drone flies along the bisector of the masts at 1 + i and 5 + 5i, and must stay inside $|z-\left(3+4i\right)|\le 5$|z − (3 + 4i)| ≤ 5, a disc whose edge is the circle of radius 5 about 3 + 4i.