Polar Form, Exponential Form, and Roots of Unity
Pillar: SHAPE --- "Polar form reveals the geometry hiding inside complex multiplication."
Why a New Form?
In Lesson 1, we represented complex numbers as \(z = a + bi\) --- the rectangular form. Addition was easy: just add components. But multiplication produced a formula \((ac - bd) + (ad + bc)i\) that felt mechanical and opaque.
There's a deeper structure hiding in that formula. Multiplication doesn't just scramble coordinates --- it rotates and scales. Polar form makes this explicit. Once you see it, operations like raising to a power, extracting roots, and understanding the FFT become almost trivial.
Polar Coordinates for Complex Numbers
Every point in the complex plane can be described by two numbers:
- \(r\) = distance from the origin = \(|z| = \sqrt{a^2 + b^2}\)
- \(\theta\) = angle measured counterclockwise from the positive real axis
The connection to rectangular form:
So any complex number can be written:
This is the polar form. The shorthand \(\text{cis}\,\theta\) means \(\cos\theta + i\sin\theta\), so:
Converting Rectangular to Polar
Given \(z = a + bi\):
Use \(\text{atan2}\) (not \(\arctan\)) because it correctly handles all four quadrants.
| \(z\) | \(r\) | \(\theta\) |
|---|---|---|
| \(1 + i\) | \(\sqrt{2}\) | \(\pi/4\) (45 degrees) |
| \(-1 + i\) | \(\sqrt{2}\) | \(3\pi/4\) (135 degrees) |
| \(3 + 4i\) | \(5\) | \(\arctan(4/3) \approx 0.927\) |
| \(-2\) | \(2\) | \(\pi\) (180 degrees) |
| \(5i\) | \(5\) | \(\pi/2\) (90 degrees) |
Converting Polar to Rectangular
Given \(r\) and \(\theta\):
Example: \(z = 4\,\text{cis}(\pi/3) = 4\cos(\pi/3) + 4i\sin(\pi/3) = 4 \cdot \frac{1}{2} + 4i \cdot \frac{\sqrt{3}}{2} = 2 + 2\sqrt{3}\,i\)
Multiplication in Polar Form
Here is where polar form pays off. If \(z_1 = r_1\,\text{cis}\,\theta_1\) and \(z_2 = r_2\,\text{cis}\,\theta_2\), then:
Multiply the magnitudes, add the angles.
This is why multiplying by \(i = 1 \cdot \text{cis}(\pi/2)\) rotates by \(90^\circ\) --- you're adding \(\pi/2\) to the angle.
Division works the same way:
Divide the magnitudes, subtract the angles.
Example: Multiply \(z_1 = 2\,\text{cis}(\pi/6)\) by \(z_2 = 3\,\text{cis}(\pi/4)\):
No FOIL. No combining real and imaginary parts. One step.
De Moivre's Theorem
Since multiplication adds angles, repeated multiplication (powers) multiplies the angle:
This is De Moivre's Theorem. For a number on the unit circle (\(r = 1\)):
Example: Compute \((1 + i)^8\).
First, convert to polar: \(1 + i = \sqrt{2}\,\text{cis}(\pi/4)\).
Apply De Moivre: \((\sqrt{2})^8\,\text{cis}(8 \cdot \pi/4) = 16\,\text{cis}(2\pi) = 16 \cdot 1 = 16\).
Check: \((1+i)^2 = 2i\), \((2i)^2 = -4\), \((-4)^2 = 16\). Confirmed.
Euler's Formula
The most celebrated equation in mathematics connects the exponential function to trigonometry:
We state this without proof (the proof uses Taylor series). Combined with the magnitude:
This is the exponential form of a complex number. It's the most compact and the most useful for algebra.
| Form | Expression | Best for |
|---|---|---|
| Rectangular | \(a + bi\) | Addition, subtraction |
| Polar | \(r\,\text{cis}\,\theta\) | Geometric intuition |
| Exponential | \(re^{i\theta}\) | Multiplication, powers, roots |
In exponential form, multiplication becomes:
Just the laws of exponents. And De Moivre's Theorem becomes:
Setting \(r = 1\) and \(\theta = \pi\): \(e^{i\pi} = \cos\pi + i\sin\pi = -1\), giving Euler's identity: \(e^{i\pi} + 1 = 0\).
Finding \(n\)-th Roots of a Complex Number
If \(w = Re^{i\phi}\), the \(n\) solutions to \(z^n = w\) are:
There are exactly \(n\) roots. They have the same magnitude \(R^{1/n}\) and are evenly spaced around a circle, separated by angles of \(2\pi/n\).
Example: Find the cube roots of \(8\).
\(8 = 8e^{i \cdot 0}\), so \(R = 8\), \(\phi = 0\), \(n = 3\).
| \(k\) | Angle | Root |
|---|---|---|
| 0 | \(0\) | \(2\) |
| 1 | \(2\pi/3\) | \(2(\cos 120^\circ + i\sin 120^\circ) = -1 + \sqrt{3}\,i\) |
| 2 | \(4\pi/3\) | \(2(\cos 240^\circ + i\sin 240^\circ) = -1 - \sqrt{3}\,i\) |
Three points equally spaced on a circle of radius 2. Only one is real (\(z = 2\)). The other two are complex conjugates.
Roots of Unity
The \(n\)-th roots of unity are the solutions to \(z^n = 1\). Since \(1 = e^{i \cdot 0}\):
The primitive \(n\)-th root of unity is \(\omega = e^{2\pi i/n}\). All roots are powers of \(\omega\):
Geometric Picture
The \(n\)-th roots of unity are the vertices of a regular \(n\)-gon inscribed in the unit circle, with one vertex at \(1\).
| \(n\) | Shape | Roots |
|---|---|---|
| 2 | Line segment | \(1, -1\) |
| 3 | Equilateral triangle | \(1, e^{2\pi i/3}, e^{4\pi i/3}\) |
| 4 | Square | \(1, i, -1, -i\) |
| 6 | Regular hexagon | \(1, e^{i\pi/3}, e^{2i\pi/3}, -1, e^{4i\pi/3}, e^{5i\pi/3}\) |
The Sum Property
The \(n\)-th roots of unity sum to zero:
Proof: This is a geometric series with ratio \(\omega \neq 1\):
Geometrically: the vectors from the origin to the vertices of a regular polygon cancel out. Pull any regular polygon by attaching strings to each vertex --- the net force is zero.
The Cancellation Lemma
For the FFT, the following property is critical. If \(n\) is even and \(\omega_n = e^{2\pi i/n}\):
And \((\omega_n^k)^2 = \omega_{n/2}^k\) --- squaring the \(n\)-th roots gives you the \((n/2)\)-th roots. This halving property is what lets the FFT recursively split a polynomial evaluation into two half-sized problems.
Why This Matters: The FFT Connection
The Fast Fourier Transform (Ch11) evaluates a polynomial \(P(x)\) at all \(n\)-th roots of unity simultaneously:
Why these specific points? Because of the cancellation lemma. The roots of unity split into two interleaved sets when squared, enabling a divide-and-conquer algorithm that runs in \(O(n \log n)\) instead of \(O(n^2)\).
You don't need to understand the FFT now. But everything in this lesson --- polar form, exponential form, roots of unity, the sum property, the cancellation lemma --- is the mathematical foundation for it.
C++ Implementation
int n = 8;
complex<double> omega = polar(1.0, 2.0 * M_PI / n);
complex<double> sum_roots(0, 0);
for (int k = 0; k < n; k++) {
complex<double> root = pow(omega, k);
cout << "omega^" << k << " = " << root << "\n";
sum_roots += root;
}
cout << "Sum = " << sum_roots << "\n"; // approximately 0
Key functions: polar(r, theta) creates \(re^{i\theta}\), abs(z) gives \(r\), arg(z) gives \(\theta\).
Practice Problems
Problem 1. Convert \(z = -1 + \sqrt{3}\,i\) to polar form \(r\,\text{cis}\,\theta\).
Answer
\(r = \sqrt{1 + 3} = 2\). The point \((-1, \sqrt{3})\) is in the second quadrant: \(\theta = \pi - \arctan(\sqrt{3}/1) = \pi - \pi/3 = 2\pi/3\). So \(z = 2\,\text{cis}(2\pi/3)\).
Problem 2. Use De Moivre's Theorem to compute \((1 + i)^{12}\).
Answer
\(1 + i = \sqrt{2}\,\text{cis}(\pi/4)\). By De Moivre: \((\sqrt{2})^{12}\,\text{cis}(12\pi/4) = 64\,\text{cis}(3\pi) = 64\,\text{cis}(\pi) = -64\).
Problem 3. Find all fourth roots of \(-16\).
Answer
\(-16 = 16e^{i\pi}\), so \(R = 16\), \(\phi = \pi\), \(n = 4\). \(z_k = 2\,e^{i(\pi + 2\pi k)/4}\) for \(k = 0, 1, 2, 3\). \(z_0 = 2\,e^{i\pi/4} = \sqrt{2} + \sqrt{2}\,i\), \(z_1 = 2\,e^{i3\pi/4} = -\sqrt{2} + \sqrt{2}\,i\), \(z_2 = 2\,e^{i5\pi/4} = -\sqrt{2} - \sqrt{2}\,i\), \(z_3 = 2\,e^{i7\pi/4} = \sqrt{2} - \sqrt{2}\,i\).
Problem 4. Prove that \(\omega^k + \omega^{n-k} = 2\cos(2\pi k/n)\) where \(\omega = e^{2\pi i/n}\).
Answer
\(\omega^k = e^{2\pi i k/n} = \cos(2\pi k/n) + i\sin(2\pi k/n)\). \(\omega^{n-k} = e^{2\pi i(n-k)/n} = e^{2\pi i - 2\pi ik/n} = e^{-2\pi ik/n} = \cos(2\pi k/n) - i\sin(2\pi k/n)\). Adding: \(\omega^k + \omega^{n-k} = 2\cos(2\pi k/n)\). The imaginary parts cancel.
Problem 5. Compute \(\sum_{k=0}^{5} \omega^{3k}\) where \(\omega = e^{2\pi i/6}\).
Answer
\(\omega^3 = e^{2\pi i \cdot 3/6} = e^{i\pi} = -1\). So \(\sum_{k=0}^{5}(-1)^k = 1 - 1 + 1 - 1 + 1 - 1 = 0\).