3.9 - Loci in an Argand diagram | Cambridge A-Level Mathematics

An equation in a complex number can describe an entire set of points. You will read modulus conditions as distances, equal moduli as equal distances, and an argument condition as a direction from a fixed point. Combining those ideas will let you describe regions and keep their boundaries precise.

Distances from fixed points

Which points satisfy $|z-(2+i)|=2$ ? If $z=x+iy$ , then $z-(2+i)=(x-2)+i(y-1)$ . Its modulus is the distance from $(2,1)$ to $(x,y)$ , so the equation fixes that distance at $2$ .

Locus

A locus is the set of points satisfying a given condition. In an Argand diagram, $|z-a|$ is the distance from the point representing the fixed complex number $a$ to the point representing $z$ .

This is the same endpoint-minus-start displacement used in complex subtraction. It gives a direct translation between distance language and equations.

Circle and disc

For a fixed complex number $a$ and real $k>0$ :

$|z-a|=k$ is the circle with centre $a$ and radius $k$ .
$|z-a|<k$ is its interior, with the circumference excluded.
$|z-a|\le k$ is the closed disc, including the circumference.
$|z-a|>k$ or $|z-a|\ge k$ describes the exterior, excluding or including the circumference respectively.

These locus rules are not listed in MF19.

Draw an included boundary as a solid line and an excluded boundary as a dashed line. Shading identifies the selected region, rather than changing the meaning of the boundary.

Worked example 1: a fixed distance

Question. Describe the locus $|z-2-i|=2$ , and give its Cartesian equation.

Plan. Rewrite the expression as a displacement from one fixed point.

|z-2-i|=|z-(2+i)|=2 \quad\Longrightarrow\quad (x-2)^2+(y-1)^2=4.

Answer. The circumference of a circle, centre $(2,1)$ , radius $2$ .

Check. $z=4+i$ and $z=2+3i$ each give modulus $2$ . The centre itself gives zero, so it is not on this locus.

Change only equality to a strict inequality. The same circle becomes a boundary rather than the entire set.

Worked example 2: less than the fixed distance

Question. Describe the locus $|z-2-i|<2$ , including its boundary convention.

Plan. Use the same centre and radius, selecting points whose distance is smaller.

\sqrt{(x-2)^2+(y-1)^2}<2 \quad\Longleftrightarrow\quad (x-2)^2+(y-1)^2<4.

Squaring preserves the inequality because both sides are non-negative.

Answer. The interior of the circle with centre $(2,1)$ , radius $2$ , excluding the circumference.

Check. The centre has distance $0<2$ , so it is included. The point $4+i$ has distance exactly $2$ , so it is excluded.

The panels distinguish a circle from its interior.

[DIAGRAM: asset_name: A fixed distance and a distance bound; asset_slug: complex-circle-disc; caption: Equality selects the circumference, while a strict upper bound selects the interior and excludes that circumference; description: Two equal-scale Argand panels have centre 2 comma 1 and radius 2. Left shows a solid circle for modulus z minus 2 minus i equals 2. Right shades the interior pale blue and draws a dashed circumference for modulus z minus 2 minus i less than 2. The centre is marked and included only in the right locus.]
Equality selects the circumference, while a strict upper bound selects the interior and excludes that circumference

Check the signs inside the bracket: $|z+1-2i|=|z-(-1+2i)|$ has centre $(-1,2)$ . Also check the radius condition before describing a circle. If $k=0$ , equality gives just $z=a$ ; a negative $k$ cannot equal a modulus. The strict condition $|z-a|<0$ has no points.

Watch out

$|z-a|<k$ describes a whole region, not just a dashed circle. The dashed circle is the excluded boundary; the interior must also be identified.

Now describe an exterior rather than an interior.

Equal distances

If $|z-a|=|z-b|$ , the point $z$ is equally far from two fixed points. For distinct $a,b$ , the set is the perpendicular bisector of the segment joining them.

Why a straight line? At the segment's midpoint the distances agree. Every point on the perpendicular through that midpoint forms two right triangles with equal perpendicular height and equal half-base, so Pythagoras gives equal distances. Conversely, equating the squared distances cancels the $x^2$ and $y^2$ terms, leaving exactly that line.

Equal-distance locus

For $a\ne b$ ,

|z-a|=|z-b|

is the complete perpendicular bisector of the segment joining $a$ and $b$ . If $a=b$ , both distances are identical for every $z$ , so the locus is the whole plane. This result is not listed in MF19.

Worked example 3: fixed points at the same height

Question. Describe $|z+2-i|=|z-4-i|$ , and find its Cartesian equation.

Plan. Identify the two fixed points and use the midpoint of their horizontal joining segment.

The points are $A(-2,1)$ and $B(4,1)$ . Their midpoint is $(1,1)$ ; the joining segment is horizontal.

Answer. The complete vertical line $x=1$ , the perpendicular bisector of $AB$ .

Check. At a general point $(1,y)$ , the squared distances are both $3^2+(y-1)^2$ . There is no restriction on $y$ .

Move only the second point upwards. The bisector is now oblique, so equating squared distances is convenient.

Worked example 4: an oblique joining segment

Question. Find the locus $|z+2-i|=|z-4-3i|$ as a Cartesian line equation.

Plan. Put $z=x+iy$ , square the two non-negative distances, and simplify.

\begin{aligned} (x+2)^2+(y-1)^2&=(x-4)^2+(y-3)^2\\ 4x-2y+5&=-8x-6y+25\\ 12x+4y&=20\\ 3x+y&=5. \end{aligned}

Answer. The complete line $3x+y=5$ .

Check. The fixed points are $(-2,1)$ and $(4,3)$ , with midpoint $(1,2)$ , which lies on the answer. Their joining gradient is $1/3$ ; the answer has gradient $-3$ , so it is perpendicular.

The perpendicular bisector extends in both directions; its points need not lie between the fixed points.

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3.9 - Loci in an Argand diagram

Distances from fixed points

Equal distances

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