In short

The complex plane (Argand diagram) identifies each complex number z = x + iy with the point (x, y). In this plane, the distance between z_1 and z_2 is |z_1 - z_2|, the midpoint of z_1 and z_2 is \frac{z_1 + z_2}{2}, and the point dividing the segment from z_1 to z_2 in the ratio m:n is \frac{mz_2 + nz_1}{m + n}. These are the same formulas you know from coordinate geometry, but written in complex number notation — shorter, cleaner, and ready for rotation and scaling.

Take three points forming a triangle: 1 + i, 4 + i, and 1 + 4i. What is the distance between the first two? What is the midpoint of the longest side? In coordinate geometry, you would pull out the distance formula and the midpoint formula -- separate tools for separate jobs. With complex numbers, both answers are just modulus and addition: the distance is |4 + i - 1 - i| = 3, and the midpoint is \frac{(4 + i) + (1 + 4i)}{2} = \frac{5 + 5i}{2}.

The complex plane turns every point into a number you can add, subtract, multiply, and divide. That means geometry problems become algebra problems -- and the formulas are shorter and cleaner than their Cartesian counterparts.

The complex plane (Argand diagram)

The complex plane has two axes:

Every complex number z = x + iy corresponds to exactly one point (x, y), and every point on the plane corresponds to exactly one complex number. The correspondence is:

z = x + iy \;\longleftrightarrow\; (x, y)

The origin is 0. The number 1 is one unit to the right. The number i is one unit up. The number -1 is one unit to the left. The number -i is one unit down.

The Argand diagram showing several complex numbers plotted as pointsA coordinate plane with real axis horizontal and imaginary axis vertical. Five complex numbers are plotted: 3 plus 2i in the first quadrant, negative 2 plus 3i in the second quadrant, negative 1 minus 2i in the third quadrant, 4 minus i in the fourth quadrant, and 2i on the positive imaginary axis. Dashed lines from each point project onto both axes showing the real and imaginary parts. ReIm 1234 −1−2−3 123 −1−2 3 + 2i −2 + 3i −1 − 2i 4 − i 2i Each complex number is a point. Each point is a complex number.
The Argand diagram with five complex numbers plotted. The number $3 + 2i$ sits at $(3, 2)$ in the first quadrant — the dashed lines show the real part (horizontal projection) and imaginary part (vertical projection). Every quadrant is reachable: second quadrant for negative real, positive imaginary; third for both negative; fourth for positive real, negative imaginary. The purely imaginary number $2i$ sits on the vertical axis.

The modulus |z| = \sqrt{x^2 + y^2} is the distance from z to the origin. The conjugate \bar{z} = x - iy is the reflection of z across the real axis. These two geometric operations — distance from origin and reflection — are the building blocks for everything that follows.

Distance between two points

In coordinate geometry, the distance between (x_1, y_1) and (x_2, y_2) is \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.

In complex number language, the two points are z_1 = x_1 + iy_1 and z_2 = x_2 + iy_2. Their difference is:

z_1 - z_2 = (x_1 - x_2) + i(y_1 - y_2)

The modulus of this difference is:

|z_1 - z_2| = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

This is exactly the distance formula. So:

The distance between two complex numbers z_1 and z_2 is |z_1 - z_2|.

One formula, no coordinates to unpack. The complex number z_1 - z_2 is the vector from z_2 to z_1, and its modulus is the length of that vector.

Distance between two complex numbers z1 and z2 as the modulus of their differenceTwo points z1 equals 4 plus 3i and z2 equals 1 plus 1i on the complex plane. A line segment connects them, labelled with the distance formula: the absolute value of z1 minus z2 equals the absolute value of 3 plus 2i equals the square root of 13. The difference vector 3 plus 2i is also shown as an arrow from z2 to z1. ReIm 1234 123 3 (real part of difference) 2 (imaginary part) z₁ = 4 + 3i z₂ = 1 + i |z₁ − z₂| = √13
The distance between $z_1 = 4 + 3i$ and $z_2 = 1 + i$ is $|z_1 - z_2| = |3 + 2i| = \sqrt{9 + 4} = \sqrt{13}$. The difference $3 + 2i$ is the displacement vector from $z_2$ to $z_1$ — $3$ units right and $2$ units up. The dashed lines show the right triangle whose hypotenuse is the distance.

This reformulation might look like just a notational shortcut. But it has a real advantage: the distance formula |z_1 - z_2| is algebraically manipulable in ways that \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} is not. You can use properties of the modulus — the triangle inequality |z_1 + z_2| \le |z_1| + |z_2|, the multiplicativity |z_1 z_2| = |z_1| |z_2| — to prove geometric results without ever expanding into coordinates.

Midpoint formula

The midpoint of the segment from z_1 to z_2 in coordinate geometry is \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).

In complex notation:

\text{Midpoint} = \frac{z_1 + z_2}{2}

This is the average of the two complex numbers. It works because addition of complex numbers is component-wise: \frac{(x_1 + iy_1) + (x_2 + iy_2)}{2} = \frac{x_1 + x_2}{2} + i\frac{y_1 + y_2}{2}, which is exactly the coordinate midpoint.

For three points, the centroid of the triangle with vertices z_1, z_2, z_3 is:

G = \frac{z_1 + z_2 + z_3}{3}

Again, just the average. The centroid divides each median in the ratio 2:1 — this classical result follows directly from the algebra of complex numbers.

Midpoint and centroid of a triangle using complex numbersA triangle with vertices z1 equals 1 plus 4i, z2 equals 5 plus 2i, and z3 equals 3 minus 2i on the complex plane. The midpoint of z1 and z2 is marked as M equals 3 plus 3i. The centroid G equals 3 plus four-thirds i is marked inside the triangle. A median from z3 to M passes through G. ReIm 12345 12 −1 M = 3 + 3i G = 3 + 4i/3 z₁ = 1 + 4i z₂ = 5 + 2i z₃ = 3 − 2i
A triangle with vertices $z_1 = 1 + 4i$, $z_2 = 5 + 2i$, $z_3 = 3 - 2i$. The midpoint of $z_1$ and $z_2$ is $M = \frac{z_1 + z_2}{2} = 3 + 3i$. The centroid is $G = \frac{z_1 + z_2 + z_3}{3} = \frac{9 + 4i}{3} = 3 + \frac{4}{3}i$. The dashed median from $z_3$ to $M$ passes through $G$, dividing it in ratio $2:1$.

Section formula

The midpoint is the special case of a more general question: what point divides the segment from z_1 to z_2 in a given ratio?

Internal division. The point that divides the segment from z_1 to z_2 in the ratio m:n (internally) is:

z = \frac{mz_2 + nz_1}{m + n}

External division. The point that divides the segment externally in the ratio m:n is:

z = \frac{mz_2 - nz_1}{m - n}

These are identical to the coordinate geometry section formulas, but with complex numbers replacing coordinate pairs. The derivation is the same: the point z lies on the line through z_1 and z_2, at the appropriate fraction of the way.

Section formula (complex form)

The point dividing the segment from z_1 to z_2 internally in the ratio m:n is

z = \frac{mz_2 + nz_1}{m + n}.

The point dividing it externally in the ratio m:n is

z = \frac{mz_2 - nz_1}{m - n}.

The midpoint is the special case m = n = 1.

The section formula illustrated for a 2 to 1 internal divisionA line segment from z1 equals 1 plus i on the left to z2 equals 4 plus 4i on the right. A point P divides this segment in the ratio 2 to 1 from z1 to z2. P is located at 3 plus 3i, two-thirds of the way from z1 to z2. The distances from z1 to P and from P to z2 are labelled in the ratio 2 to 1. ReIm 12345 12 z₁ = 1 + i z₂ = 4 + 4i P = 3 + 3i m = 2 n = 1 P divides z₁z₂ internally in ratio 2:1.
The point $P$ divides the segment from $z_1 = 1 + i$ to $z_2 = 4 + 4i$ in the ratio $2:1$. Using the section formula: $P = \frac{2(4+4i) + 1(1+i)}{2+1} = \frac{9 + 9i}{3} = 3 + 3i$. The point is two-thirds of the way from $z_1$ to $z_2$.

Why complex notation is better

At this point, you might ask: these are the same formulas as coordinate geometry, just written differently. Why bother?

The answer becomes clear when you combine distance and angle. In coordinate geometry, rotating a point by angle \theta about the origin requires a 2 \times 2 rotation matrix:

\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}

In complex notation, the same rotation is:

z' = z \cdot e^{i\theta}

One multiplication replaces a matrix equation. Scaling by a factor r and rotating by \theta simultaneously is z' = rz \cdot e^{i\theta}. Reflecting across the real axis is z' = \bar{z}. These operations compose by ordinary arithmetic — no matrices, no trigonometric expansions.

This is the real reason geometry with complex numbers exists as a separate topic. The basic formulas (distance, midpoint, section) are just coordinate geometry in disguise. The deeper results (rotation, collinearity conditions, circle equations) use the multiplicative structure of complex numbers and have no clean coordinate-geometry equivalent.

Interactive: distance and midpoint explorer

Drag the red point on the plane below. The readouts show the position of the point and its distance from the fixed point 2 + i, as well as the midpoint of the segment joining them. Watch how the midpoint tracks halfway between the two points as you move.

Interactive distance and midpoint explorer on the complex planeA complex plane from negative 5 to 5 on both axes. A red draggable point represents z. A fixed point at 2 plus i is marked. Readouts show the coordinates of z, the distance from z to 2 plus i, and the midpoint of the two points, all updating in real time. ReIm 1234 −1−2−3−4 2 + i (fixed) drag the red point
Drag the red point anywhere on the plane. The readouts show the distance $|z - (2+i)|$ and the midpoint coordinates $\left(\frac{\operatorname{Re}(z)+2}{2},\, \frac{\operatorname{Im}(z)+1}{2}\right)$. Move the red point in a circle around the fixed point — the distance stays constant (the locus $|z - (2+i)| = r$ is a circle), while the midpoint traces a smaller circle.

Worked examples

Example 1: Find the distance between $z_1 = 3 + 4i$ and $z_2 = -1 + i$, and the point dividing the segment in ratio $3:1$

A computation using the distance and section formulas in complex notation.

Step 1. Compute the difference.

z_1 - z_2 = (3 + 4i) - (-1 + i) = 4 + 3i

Why: the difference z_1 - z_2 is the displacement vector from z_2 to z_1. Its components are 4 (rightward) and 3 (upward).

Step 2. Find the distance.

|z_1 - z_2| = |4 + 3i| = \sqrt{16 + 9} = \sqrt{25} = 5

Why: this is the familiar 3-4-5 right triangle. The distance between the two points is exactly 5.

Step 3. Find the point dividing the segment internally in ratio 3:1.

P = \frac{3z_2 + 1 \cdot z_1}{3 + 1} = \frac{3(-1 + i) + (3 + 4i)}{4} = \frac{-3 + 3i + 3 + 4i}{4} = \frac{0 + 7i}{4} = \frac{7}{4}i

Why: the section formula with m = 3, n = 1 gives the point three-quarters of the way from z_1 to z_2. The result \frac{7}{4}i is purely imaginary — it sits on the imaginary axis.

Step 4. Verify: check that |P - z_1| : |P - z_2| = 3 : 1.

P - z_1 = \frac{7}{4}i - 3 - 4i = -3 - \frac{9}{4}i, \qquad |P - z_1| = \sqrt{9 + \frac{81}{16}} = \sqrt{\frac{225}{16}} = \frac{15}{4}
P - z_2 = \frac{7}{4}i + 1 - i = 1 + \frac{3}{4}i, \qquad |P - z_2| = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}

Ratio: \frac{15/4}{5/4} = 3. So |P - z_1| : |P - z_2| = 3 : 1. \checkmark

Why: the ratio of distances confirms the section formula is correct. The point P is three times as far from z_1 as from z_2.

Result: Distance = 5. The dividing point is P = \frac{7}{4}i.

Distance and section formula worked example on the complex planeA complex plane showing z1 equals 3 plus 4i in the first quadrant and z2 equals negative 1 plus i in the second quadrant. A line segment connects them with length 5. The point P equals 7i over 4 on the imaginary axis divides the segment in ratio 3 to 1. The right triangle with legs 4 and 3 is shown with dashed lines. ReIm 1234 −1 1234 4 3 z₁ = 3+4i z₂ = −1+i P = 7i/4 distance = 5
The points $z_1 = 3 + 4i$ and $z_2 = -1 + i$ are connected by a segment of length $5$ (the hypotenuse of a $3$-$4$-$5$ triangle). The point $P = \frac{7}{4}i$ divides the segment in ratio $3:1$ and sits exactly on the imaginary axis.

The fact that P lands on the imaginary axis is not a coincidence you could easily spot from the coordinates alone. In complex notation, P = \frac{7}{4}i immediately tells you the real part is zero — the point is purely imaginary.

Example 2: Show that the three points $z_1 = 2$, $z_2 = -1 + i\sqrt{3}$, $z_3 = -1 - i\sqrt{3}$ form an equilateral triangle

A geometric proof using only the distance formula in complex notation.

Why these points? They are the three cube roots of 8: 2, 2\omega, 2\omega^2 where \omega = e^{2\pi i/3}. These sit on a circle of radius 2 at angles 0°, 120°, 240°.

Step 1. Compute all three side lengths.

z_1 - z_2 = 2 - (-1 + i\sqrt{3}) = 3 - i\sqrt{3}, \qquad |z_1 - z_2| = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}
z_2 - z_3 = (-1 + i\sqrt{3}) - (-1 - i\sqrt{3}) = 2i\sqrt{3}, \qquad |z_2 - z_3| = 2\sqrt{3}
z_3 - z_1 = (-1 - i\sqrt{3}) - 2 = -3 - i\sqrt{3}, \qquad |z_3 - z_1| = \sqrt{9 + 3} = 2\sqrt{3}

Why: all three sides have length 2\sqrt{3}. The three distances are equal because the cube roots of 8 are equally spaced on a circle of radius 2 — the angles between consecutive roots are all 120°, forcing an equilateral triangle.

Step 3. Confirm the centroid lies at the origin.

\frac{z_1 + z_2 + z_3}{3} = \frac{2 + (-1 + i\sqrt{3}) + (-1 - i\sqrt{3})}{3} = \frac{0}{3} = 0

Why: the sum of the cube roots of 8 is 0 (the sum of nth roots of any number is 0 for n \ge 2, since the roots of unity sum to zero). The centroid at the origin confirms the triangle is centred there.

Step 4. Verify: the circumradius should equal 2 (the modulus of each vertex).

|z_1| = 2, \quad |z_2| = \sqrt{1 + 3} = 2, \quad |z_3| = 2

All vertices are at distance 2 from the centroid (origin), so the circumradius is 2. For an equilateral triangle with side s = 2\sqrt{3}, the circumradius is \frac{s}{\sqrt{3}} = \frac{2\sqrt{3}}{\sqrt{3}} = 2. \checkmark

Why: the circumradius formula for an equilateral triangle is R = s/\sqrt{3}, and it matches the modulus of the vertices.

Result: The three cube roots of 8 — namely 2, -1+i\sqrt{3}, -1-i\sqrt{3} — form an equilateral triangle with side length 2\sqrt{3}, centroid at the origin, and circumradius 2.

The equilateral triangle formed by the cube roots of 8A complex plane with a circle of radius 2 centred at the origin. Three points on this circle form an equilateral triangle: 2 on the positive real axis, negative 1 plus i root 3 in the second quadrant, and negative 1 minus i root 3 in the third quadrant. Each side is labelled with length 2 root 3. The centroid is marked at the origin. ReIm r = 2 2 −1 + i√3 −1 − i√3 2√3 2√3 2√3 G = 0
The cube roots of $8$ form an equilateral triangle inscribed in a circle of radius $2$. All three sides have length $2\sqrt{3}$. The centroid coincides with the origin, which is also the circumcentre — a hallmark of regular polygons centred at the origin. The complex distance formula $|z_1 - z_2|$ gave us all three side lengths with minimal computation.

This example demonstrates the strategy: to prove a triangle is equilateral, compute the three pairwise distances |z_1 - z_2|, |z_2 - z_3|, |z_3 - z_1| and check they are equal. No angle computations, no slopes, no perpendicular bisectors — just three modulus calculations.

Common confusions

Going deeper

If you can compute distances, midpoints, and section formula results using complex numbers, and you understand why the complex plane upgrades coordinate geometry with multiplication, you have the core. The following ideas show where this leads.

Collinearity and concyclicity

Three points z_1, z_2, z_3 are collinear (lie on a single line) if and only if the ratio \frac{z_3 - z_1}{z_2 - z_1} is a real number. This is because a real ratio means the displacement from z_1 to z_3 is a real multiple of the displacement from z_1 to z_2 — same direction, just scaled.

Four points z_1, z_2, z_3, z_4 are concyclic (lie on a single circle) if and only if the cross-ratio \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)} is real. These conditions have no comparably clean form in coordinate geometry.

Rotation about an arbitrary point

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

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

Subtract c to move the centre of rotation to the origin, rotate by multiplying by e^{i\theta}, then add c back. This three-step pattern — translate, transform, translate back — is the standard technique for handling rotations about non-origin centres.

The triangle inequality on the complex plane

The inequality |z_1 + z_2| \le |z_1| + |z_2| has a direct geometric interpretation: the length of one side of a triangle is at most the sum of the other two sides. Equality holds when z_1 and z_2 point in the same direction (i.e., z_2/z_1 is a positive real number). This connects the algebraic property of the modulus to one of the oldest results in geometry.

Where this leads next