In short

A locus in the Argand plane is the set of all complex numbers z that satisfy a given condition. The most important loci are: |z - z_1| = |z - z_2| (perpendicular bisector), \arg(z - z_1) = \theta (ray from z_1), |z - z_1| + |z - z_2| = k (ellipse), and |z - z_1| - |z - z_2| = k (hyperbola branch). A rotation about the origin is z' = ze^{i\theta}, and a rotation about a general point z_0 is z' - z_0 = (z - z_0)e^{i\theta}. Every one of these conditions has a clean geometric meaning — the complex equation is just a compact way of writing it.

Take the condition |z - 2| = 3. You know from Geometry with Complex Numbers — Circles that this is a circle with centre 2 and radius 3. But what about |z - 1| = |z + 1|? Or \arg(z - i) = \pi/4? Or |z - 1| + |z + 1| = 4?

Each of these is a constraint on z. The set of all z that satisfies the constraint is a locus — a curve (or sometimes a region) on the Argand plane. The word comes from Latin: locus means "place." The locus is the place where z is allowed to live.

The power of complex numbers is that many loci that take several lines to describe in Cartesian coordinates collapse into a single equation involving |z|, \arg(z), or e^{i\theta}. This article catalogues the standard loci and shows how to decode each one.

Locus 1: |z - z_1| = |z - z_2| — the perpendicular bisector

The condition says: the distance from z to z_1 equals the distance from z to z_2. The set of all points equidistant from two fixed points is the perpendicular bisector of the segment joining them.

Perpendicular bisector as the locus of |z minus z1| equals |z minus z2|A complex plane with two fixed points z1 at (negative 2, 0) and z2 at (2, 0). The perpendicular bisector is a vertical line through the origin, drawn in red. A general point z on the line has equal dashed distances to z1 and z2. The midpoint of z1 z2 is marked at the origin. ReIm0−22z₁z₂z|z−z₁||z−z₂||z − z₁| = |z − z₂| : perpendicular bisector
The locus $|z - z_1| = |z - z_2|$ is the perpendicular bisector of the segment $z_1 z_2$. Every point on the red vertical line is equidistant from $z_1 = -2$ and $z_2 = 2$. The midpoint of $z_1 z_2$ lies on the bisector, and the bisector is perpendicular to the segment.

To verify algebraically: square both sides of |z - z_1| = |z - z_2|.

|z - z_1|^2 = |z - z_2|^2
(z - z_1)(\bar{z} - \bar{z}_1) = (z - z_2)(\bar{z} - \bar{z}_2)

Expanding both sides and cancelling the z\bar{z} term:

-\bar{z}_1 z - z_1\bar{z} + |z_1|^2 = -\bar{z}_2 z - z_2\bar{z} + |z_2|^2

Rearranging:

(\bar{z}_2 - \bar{z}_1)z + (z_2 - z_1)\bar{z} = |z_2|^2 - |z_1|^2

This is a linear equation in z and \bar{z} — the equation of a line, as expected. The line passes through the midpoint \frac{z_1 + z_2}{2} and is perpendicular to the direction z_2 - z_1.

For the specific case z_1 = -2, z_2 = 2: the equation becomes (2 - (-2))z + (2 - (-2))\bar{z} = 4 - 4, so 4z + 4\bar{z} = 0, giving z + \bar{z} = 0, which means 2x = 0, so x = 0 — the imaginary axis.

Locus 2: \arg(z - z_1) = \theta — a ray

The condition \arg(z - z_1) = \theta says: the argument (angle) of the vector from z_1 to z is fixed at \theta. Geometrically, this means z lies on a ray starting from z_1, making angle \theta with the positive real direction.

The locus is a half-line (ray), not a full line. The point z_1 itself is excluded (the argument is undefined at the origin of the vector). Points on the opposite ray (which would have argument \theta + \pi or \theta - \pi) do not satisfy the condition.

Ray as the locus of arg(z minus z1) equals thetaA complex plane with a fixed point z1 at (1, 1). A red ray emanates from z1 at angle pi/4 (45 degrees) to the positive real direction. The ray extends upward to the right. The angle theta is marked with an arc at z1. The opposite direction is shown as a thin dashed line, excluded from the locus. ReIm012312θ = π/4real directionz₁ = 1+izopposite ray excludedarg(z − z₁) = π/4 : ray from z₁
The locus $\arg(z - z_1) = \pi/4$ is a ray from $z_1 = 1 + i$ at angle $45°$ to the positive real direction. The point $z_1$ is excluded (hollow in concept, though drawn solid here for visibility). The opposite ray, at angle $\pi/4 + \pi = 5\pi/4$, is not part of the locus.

A related condition: \arg(z - z_1) - \arg(z - z_2) = \theta, or equivalently \arg\left(\frac{z - z_1}{z - z_2}\right) = \theta. This is the constant angle condition. The locus is an arc of a circle passing through z_1 and z_2, because the inscribed angle theorem says that all points on an arc subtend the same angle at the chord's endpoints. When \theta = \pi/2, the locus is a semicircle with z_1 z_2 as diameter.

Locus 3: |z - z_1| + |z - z_2| = k — an ellipse

The sum of the distances from z to two fixed points z_1 and z_2 is constant. This is the definition of an ellipse with foci z_1 and z_2.

The constant k must satisfy k > |z_1 - z_2| — the sum of distances must exceed the distance between the foci, otherwise no point can satisfy the condition (the triangle inequality would be violated).

When k = |z_1 - z_2|, the "ellipse" degenerates to the line segment from z_1 to z_2.

Ellipse as the locus of |z minus z1| plus |z minus z2| equals kA complex plane with two focal points z1 at (negative 2, 0) and z2 at (2, 0). An ellipse is drawn around both foci. A point z on the ellipse has dashed lines to both foci. The sum of these two distances equals k, labelled as 6. The semi-major axis a equals 3 and semi-minor axis b equals root 5 are marked. ReIm0−22−33z₁ = −2z₂ = 2z|z−z₁||z−z₂||z−z₁| + |z−z₂| = 6a = 3b = √5
The locus $|z - z_1| + |z - z_2| = 6$ with $z_1 = -2$ and $z_2 = 2$. The foci are $4$ apart, and $k = 6 > 4$. The semi-major axis is $a = k/2 = 3$, and the semi-minor axis is $b = \sqrt{a^2 - c^2} = \sqrt{9 - 4} = \sqrt{5}$, where $c = 2$ is the focal distance.

Similarly, \bigl||z - z_1| - |z - z_2|\bigr| = k (with 0 < k < |z_1 - z_2|) gives a hyperbola with foci z_1 and z_2. Without the absolute value, |z - z_1| - |z - z_2| = k gives one branch of the hyperbola (the branch closer to z_2).

Locus 4: \left|\dfrac{z - z_1}{z - z_2}\right| = k — the circle of Apollonius

When k \neq 1, this is a circle. The locus is the set of all points whose distances to z_1 and z_2 are in a fixed ratio k. The circle is called the circle of Apollonius.

When k = 1, you get |z - z_1| = |z - z_2|, which is the perpendicular bisector — a line, the degenerate case of a circle with infinite radius.

To find the centre and radius, square both sides:

|z - z_1|^2 = k^2|z - z_2|^2

Expand:

(z - z_1)(\bar{z} - \bar{z}_1) = k^2(z - z_2)(\bar{z} - \bar{z}_2)
z\bar{z} - \bar{z}_1 z - z_1\bar{z} + |z_1|^2 = k^2(z\bar{z} - \bar{z}_2 z - z_2\bar{z} + |z_2|^2)

When k \neq 1, the z\bar{z} terms do not cancel, and rearranging gives a circle in general form.

Take a concrete case: z_1 = 0, z_2 = 4, k = 2. The condition is |z| = 2|z - 4|.

x^2 + y^2 = 4\bigl((x-4)^2 + y^2\bigr)
x^2 + y^2 = 4x^2 - 32x + 64 + 4y^2
0 = 3x^2 - 32x + 64 + 3y^2
x^2 + y^2 - \frac{32}{3}x + \frac{64}{3} = 0
\left(x - \frac{16}{3}\right)^2 + y^2 = \frac{256}{9} - \frac{64}{3} = \frac{256 - 192}{9} = \frac{64}{9}

A circle with centre \frac{16}{3} and radius \frac{8}{3}.

Rotation: z' = ze^{i\theta}

Multiplying a complex number by e^{i\theta} = \cos\theta + i\sin\theta rotates it by angle \theta about the origin. The modulus stays the same: |z'| = |z||e^{i\theta}| = |z|. Only the argument changes: \arg(z') = \arg(z) + \theta.

This is one of the deepest facts about complex multiplication. Multiplying by a unit complex number e^{i\theta} is rotation. Multiplying by a general complex number w is rotation by \arg(w) followed by scaling by |w|.

Rotation of a complex number by angle thetaA complex plane showing a point z at angle alpha from the real axis. An arc from z to z prime sweeps through angle theta. The point z prime is at angle alpha plus theta from the real axis, at the same distance from the origin. Both z and z prime are on a dashed circle of radius |z|. ReIm0|z|zz' = ze^(iθ)θαα+θ
Rotation by $\theta$ about the origin. The point $z$ (at angle $\alpha$, distance $|z|$ from the origin) maps to $z' = ze^{i\theta}$ (at angle $\alpha + \theta$, same distance). The red arc shows the rotation. The dashed circle of radius $|z|$ passes through both $z$ and $z'$.

Rotation about a general point

To rotate z by angle \theta about a point z_0 (not the origin):

  1. Translate so that z_0 goes to the origin: w = z - z_0.
  2. Rotate: w' = we^{i\theta}.
  3. Translate back: z' = w' + z_0.

Combining: z' - z_0 = (z - z_0)e^{i\theta}, or equivalently:

z' = z_0 + (z - z_0)e^{i\theta}

This is the general rotation formula on the complex plane. It is far more compact than the Cartesian rotation formula, which involves separate expressions for the new x' and y'.

An interactive exploration

Drag the red point below. It is constrained to move on the locus |z - 2| = |z + 2| — the perpendicular bisector of the segment from -2 to 2. The readouts show the distances from the point to both fixed points, confirming they are always equal.

Interactive perpendicular bisector locusA complex plane with two fixed points at negative 2 and 2 on the real axis. A vertical line through the origin is the perpendicular bisector. A red draggable point moves along this line. Readouts show the distances from the point to negative 2 and to 2, which are always equal. ReIm−220z₁=−2z₂=2drag the red point along the bisector
The locus $|z - (-2)| = |z - 2|$ is the imaginary axis — the perpendicular bisector of the segment from $-2$ to $2$. Drag the red point and watch the two squared distances stay equal. The distances are always the same because every point on the bisector is equidistant from the two foci.

Standard loci summary

Here is a reference table of the loci you are most likely to encounter.

Complex equation Locus Key condition
|z - z_0| = r Circle, centre z_0, radius r r > 0
|z - z_1| = |z - z_2| Perpendicular bisector of z_1 z_2 z_1 \neq z_2
\arg(z - z_1) = \theta Ray from z_1 at angle \theta z_1 excluded
\arg\!\left(\frac{z - z_1}{z - z_2}\right) = \theta Arc of a circle through z_1, z_2 0 < \theta < \pi: major arc; \theta = \pi/2: semicircle
|z - z_1| + |z - z_2| = k Ellipse with foci z_1, z_2 k > |z_1 - z_2|
\bigl||z - z_1| - |z - z_2|\bigr| = k Hyperbola with foci z_1, z_2 0 < k < |z_1 - z_2|
|z - z_1|/|z - z_2| = k Circle of Apollonius (k\neq 1); perp. bisector (k=1) k > 0
z' = z_0 + (z - z_0)e^{i\theta} Rotation of z by \theta about z_0

Worked examples

Example 1: Find the locus of $z$ satisfying $|z - 1 - i| = |z - 3 + i|$

Step 1. Identify the fixed points.

z_1 = 1 + i and z_2 = 3 - i.

Why: write |z - z_1| = |z - z_2| and read off z_1 and z_2 directly from the equation.

Step 2. Recognize the locus type.

Equal distances to two fixed points — this is the perpendicular bisector of the segment from z_1 to z_2.

Why: the defining property of the perpendicular bisector is equidistance from the two endpoints.

Step 3. Find the equation by squaring both sides.

|z - 1 - i|^2 = |z - 3 + i|^2

Let z = x + yi.

(x - 1)^2 + (y - 1)^2 = (x - 3)^2 + (y + 1)^2
x^2 - 2x + 1 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 + 2y + 1
-2x - 2y + 2 = -6x + 2y + 10
4x - 4y = 8
x - y = 2

Why: expanding and cancelling the squared terms leaves a linear equation — confirming this is a line, not a circle.

Step 4. Verify the midpoint lies on the line.

Midpoint = \frac{z_1 + z_2}{2} = \frac{(1+i) + (3-i)}{2} = \frac{4}{2} = 2. This corresponds to (2, 0). Check: 2 - 0 = 2. Confirmed.

Result: The locus is the line x - y = 2, which is the perpendicular bisector of the segment from 1 + i to 3 - i.

Perpendicular bisector of the segment from 1 plus i to 3 minus iA complex plane with two points: z1 at (1, 1) and z2 at (3, negative 1). The line x minus y equals 2 is drawn in red, passing through the midpoint (2, 0). A dashed segment connects z1 to z2, and the perpendicular bisector crosses it at right angles at the midpoint. ReIm012341−1z₁ = 1+iz₂ = 3−imidpoint = 2x − y = 2
The locus of $|z - (1+i)| = |z - (3-i)|$ is the line $x - y = 2$ (red), the perpendicular bisector of the segment from $z_1$ to $z_2$ (dashed). The bisector passes through the midpoint $(2, 0)$ and meets the segment at right angles.

The slope of z_1 z_2 is \frac{-1 - 1}{3 - 1} = -1, and the slope of x - y = 2 (i.e., y = x - 2) is 1. Since (-1)(1) = -1, the two directions are indeed perpendicular.

Example 2: Find the locus of $z$ satisfying $\left|\dfrac{z - 2}{z + 2}\right| = 2$, and identify the centre and radius

Step 1. Square both sides to remove the modulus.

|z - 2|^2 = 4|z + 2|^2

Why: \left|\frac{z-2}{z+2}\right| = 2 means |z-2| = 2|z+2|, and squaring eliminates the square roots in the modulus.

Step 2. Expand both sides with z = x + yi.

|z-2|^2 = (x-2)^2 + y^2 = x^2 - 4x + 4 + y^2
4|z+2|^2 = 4\bigl((x+2)^2 + y^2\bigr) = 4x^2 + 16x + 16 + 4y^2

Why: |z-a|^2 = (x - \operatorname{Re}(a))^2 + (y - \operatorname{Im}(a))^2 for any real a.

Step 3. Set them equal and simplify.

x^2 - 4x + 4 + y^2 = 4x^2 + 16x + 16 + 4y^2
0 = 3x^2 + 20x + 12 + 3y^2
x^2 + y^2 + \frac{20}{3}x + 4 = 0
\left(x + \frac{10}{3}\right)^2 + y^2 = \frac{100}{9} - 4 = \frac{64}{9}

Why: completing the square in x isolates the circle equation. The y^2 term is already in the right form.

Step 4. Read off the centre and radius.

Centre = -\frac{10}{3} + 0i = -\frac{10}{3}.

Radius = \frac{8}{3}.

Result: The locus is a circle (the circle of Apollonius) with centre -\dfrac{10}{3} and radius \dfrac{8}{3}.

Circle of Apollonius for |z minus 2| over |z plus 2| equals 2A complex plane with two fixed points at (negative 2, 0) and (2, 0). A circle (the Apollonius circle) is drawn with centre at (negative 10/3, 0) and radius 8/3. The circle lies entirely to the left of the origin. A sample point on the circle has dashed lines to both fixed points, with the line to (2, 0) twice as long as the line to (negative 2, 0). ReIm0−1−2−3−41z₁=−2z₂=2−10/3r = 8/3zd₁d₂ = 2d₁
The circle of Apollonius for $|z - 2|/|z + 2| = 2$. The centre is at $-10/3$ on the real axis, and the radius is $8/3$. For any point $z$ on this circle, its distance to $z_2 = 2$ is exactly twice its distance to $z_1 = -2$. The circle lies entirely to the left of the origin because the ratio $k = 2 > 1$ pushes the locus away from $z_2$.

Verify: the rightmost point of the circle is at -10/3 + 8/3 = -2/3. Its distance to -2 is |-2/3 + 2| = 4/3, and its distance to 2 is |2 + 2/3| = 8/3. Ratio: (8/3)/(4/3) = 2. Confirmed.

Common confusions

Going deeper

If you know the five standard loci (perpendicular bisector, ray, ellipse, hyperbola, and Apollonius circle), the rotation formula z' = z_0 + (z - z_0)e^{i\theta}, and the constant-angle arc, you have the complete loci toolkit for JEE-level complex numbers. The material below connects these ideas to broader mathematics.

The inscribed angle theorem via complex numbers

The locus \arg\left(\frac{z - z_1}{z - z_2}\right) = \theta being an arc is a restatement of the inscribed angle theorem: all points on an arc of a circle subtend the same angle at the chord. The complex number formulation makes the proof a one-liner:

If \arg\left(\frac{z - z_1}{z - z_2}\right) = \theta, then the ratio (z - z_1)/(z - z_2) has fixed argument \theta, meaning it lies on a ray from the origin in the complex w-plane. The inverse map z = (z_1 - wz_2)/(1 - w) sends this ray to an arc of a circle in the z-plane. The inscribed angle theorem drops out as an algebraic consequence of the Mobius transformation.

Loci and transformations

Every locus equation can be viewed as the preimage of a simple set under a complex function. For example:

This transformation perspective unifies all loci: find the right map, and the locus becomes obvious.

The spiral similarity

A map of the form z' = \alpha z + \beta (with \alpha \neq 0) is called a spiral similarity — it combines rotation by \arg(\alpha), scaling by |\alpha|, and translation by \beta. When |\alpha| = 1, it is a pure rotation plus translation (a rigid motion). These transformations preserve angles and map circles/lines to circles/lines. Understanding them is key to competition-level geometry problems that use complex coordinates.

Where this leads next

The loci in this article are the geometric consequences of modulus and argument conditions. The next steps explore the algebraic and analytical sides.