AP Precalculus: The Complex Plane

Visualize and analyze complex numbers geometrically

📍 Plotting 📏 Distance 🎯 Midpoint 🔄 Conjugates

📚 Understanding the Complex Plane

The complex plane (also called the Argand diagram) provides a geometric way to visualize complex numbers. Each complex number corresponds to a unique point, enabling us to apply geometric concepts like distance, midpoint, and reflections to complex number operations.

1 The Complex Plane Structure

The complex plane is a 2D coordinate system where complex numbers are represented as points. It combines algebra with geometry.

Horizontal Axis

Real axis (Re)
Represents the real part \(a\)

Vertical Axis

Imaginary axis (Im)
Represents the imaginary part \(b\)

Origin

\((0, 0)\)
Represents \(z = 0\)

Any Point

\((a, b)\)
Represents \(z = a + bi\)

Plotting Rule Complex number \(z = a + bi\) → Point \((a, b)\) on the plane

📌 Examples

\(z = 3 + 4i\) → Plot at \((3, 4)\)

\(z = -2 + i\) → Plot at \((-2, 1)\)

\(z = 5\) → Plot at \((5, 0)\) on real axis

\(z = -3i\) → Plot at \((0, -3)\) on imaginary axis

2 Modulus (Absolute Value)

The modulus of a complex number is its distance from the origin in the complex plane.

Modulus Formula \(|z| = |a + bi| = \sqrt{a^2 + b^2}\)

Geometric Meaning

Distance from origin to \((a, b)\)

Uses the Pythagorean theorem

Properties

\(|z| \geq 0\) and \(|z| = 0\) iff \(z = 0\)

Always non-negative

📌 Example

Find: \(|3 - 4i|\)

Calculate: \(\sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

Meaning: The point \((3, -4)\) is 5 units from the origin

💡 Connection to Conjugates

\(z \cdot \bar{z} = |z|^2\). So \(|z|^2 = a^2 + b^2\).

3 Addition & Subtraction Geometrically

Adding complex numbers is equivalent to vector addition in the complex plane.

Addition

\((a+bi) + (c+di) = (a+c) + (b+d)i\)

Parallelogram rule: combine vectors

Subtraction

\((a+bi) - (c+di) = (a-c) + (b-d)i\)

Add the negative of the second vector

Vector Addition Visualization

To add \(z_1\) and \(z_2\):
1. Draw vector from origin to \(z_1\)
2. Draw vector from origin to \(z_2\)
3. Complete the parallelogram — the diagonal from origin is \(z_1 + z_2\)

📌 Example

Add: \(z_1 = 2 + 3i\) and \(z_2 = 4 + i\)

Sum: \((2 + 4) + (3 + 1)i = 6 + 4i\)

Geometrically: Vector from \((0,0)\) to \((2,3)\) combined with vector to \((4,1)\) gives \((6,4)\)

4 Complex Conjugate Graphically

The complex conjugate \(\bar{z}\) is the reflection of \(z\) across the real axis.

Conjugate If \(z = a + bi\), then \(\bar{z} = a - bi\)

Original Point

\(z = a + bi\) → \((a, b)\)

Conjugate Point

\(\bar{z} = a - bi\) → \((a, -b)\)

Same real part: Both have x-coordinate \(a\)
Opposite imaginary part: y-coordinates are negatives of each other
Same distance from origin: \(|z| = |\bar{z}|\)
Mirror image: Reflected across the real (x) axis

📌 Example

Given: \(z = 3 + 4i\) at point \((3, 4)\)

Conjugate: \(\bar{z} = 3 - 4i\) at point \((3, -4)\)

Both have modulus: \(|z| = |\bar{z}| = 5\)

5 Distance Between Two Complex Numbers

The distance between two complex numbers is the modulus of their difference.

Distance Formula \(d(z_1, z_2) = |z_1 - z_2| = \sqrt{(a-c)^2 + (b-d)^2}\)

where \(z_1 = a + bi\) and \(z_2 = c + di\)

📌 Example

Find distance between: \(z_1 = 1 + 2i\) and \(z_2 = 4 + 6i\)

Difference: \(z_1 - z_2 = (1-4) + (2-6)i = -3 - 4i\)

Distance: \(|{-3 - 4i}| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

💡 Same as 2D Distance

This is exactly the Euclidean distance formula from geometry! Points \((a, b)\) and \((c, d)\) have distance \(\sqrt{(a-c)^2 + (b-d)^2}\).

6 Midpoint Between Two Complex Numbers

The midpoint of two complex numbers is the average of their positions.

Midpoint Formula \(z_m = \frac{z_1 + z_2}{2} = \frac{(a+c)}{2} + \frac{(b+d)}{2}i\)

where \(z_1 = a + bi\) and \(z_2 = c + di\)

📌 Example

Find midpoint between: \(z_1 = 2 + 6i\) and \(z_2 = 8 + 2i\)

Sum: \(z_1 + z_2 = (2+8) + (6+2)i = 10 + 8i\)

Midpoint: \(z_m = \frac{10 + 8i}{2} = 5 + 4i\)

As coordinates: Midpoint of \((2, 6)\) and \((8, 2)\) is \((5, 4)\)

7 Sets of Points (Loci)

Equations involving \(|z|\) or \(|z - z_0|\) describe geometric shapes in the complex plane.

Circle centered at origin

\(|z| = r\)

All points distance \(r\) from origin

Circle centered at \(z_0\)

\(|z - z_0| = r\)

All points distance \(r\) from \(z_0\)

Disk (inside circle)

\(|z - z_0| < r\)

All points within distance \(r\)

Perpendicular Bisector

\(|z - z_1| = |z - z_2|\)

Equidistant from \(z_1\) and \(z_2\)

📌 Example

Describe: \(|z - (2 + 3i)| = 4\)

Answer: Circle with center \((2, 3)\) and radius 4

📋 Quick Reference

Plot \(z = a + bi\)

Point \((a, b)\)

Modulus

\(|z| = \sqrt{a^2 + b^2}\)

Distance

\(|z_1 - z_2|\)

Midpoint

\(\frac{z_1 + z_2}{2}\)

Conjugate

Reflect across real axis

Circle at \(z_0\)

\(|z - z_0| = r\)

📚 Understanding the Complex Plane

1 The Complex Plane Structure

2 Modulus (Absolute Value)

3 Addition & Subtraction Geometrically

4 Complex Conjugate Graphically

Original Point

Conjugate Point

5 Distance Between Two Complex Numbers

6 Midpoint Between Two Complex Numbers

7 Sets of Points (Loci)

📋 Quick Reference

Plot \(z = a + bi\)

Modulus

Distance

Midpoint

Conjugate

Circle at \(z_0\)

Need Help with Complex Numbers?