De Moivre's Theorem & Roots of Unity Cheat Sheet

De Moivre's theorem connects complex numbers, trigonometry, and powers in a simple formula. This cheat sheet helps students rewrite complex numbers in polar form, raise them to powers, and find roots efficiently. It is especially useful for precalculus, advanced algebra, and early calculus topics involving complex numbers.

The main goal is to make multiplication, powers, and roots easier by using angles and magnitudes instead of rectangular coordinates.

The core idea is that a complex number can be written as $z = r(\cos \theta + i\sin \theta)$ , where $r$ is its modulus and $\theta$ is its argument. De Moivre's theorem says $[r(\cos \theta + i\sin \theta)]^n = r^n(\cos n\theta + i\sin n\theta)$ for integer $n$ . The $n$ th roots of a complex number are found by taking the $n$ th root of the modulus and adding angle increments of $\frac{2\pi}{n}$ .

Roots of unity are the special solutions to $z^n = 1$ , evenly spaced around the unit circle.

Key Facts

The polar form of a complex number is $z = r(\cos \theta + i\sin \theta)$ , where $r = |z| = \sqrt{a^2 + b^2}$ for $z = a + bi$ .
The argument $\theta$ of $z = a + bi$ satisfies $\tan \theta = \frac{b}{a}$ , but the quadrant must be chosen from the signs of $a$ and $b$ .
De Moivre's theorem states that $[r(\cos \theta + i\sin \theta)]^n = r^n(\cos n\theta + i\sin n\theta)$ for any integer $n$ .
Using cis notation, $\operatorname{cis}\theta = \cos \theta + i\sin \theta$ , so De Moivre's theorem becomes $(r\operatorname{cis}\theta)^n = r^n\operatorname{cis}(n\theta)$ .
The $n$ th roots of $z = r\operatorname{cis}\theta$ are $w_k = \sqrt[n]{r}\operatorname{cis}\left(\frac{\theta + 2\pi k}{n}\right)$ for $k = 0, 1, 2, \ldots, n - 1$ .
The $n$ th roots of unity are $z_k = \operatorname{cis}\left(\frac{2\pi k}{n}\right)$ for $k = 0, 1, 2, \ldots, n - 1$ .
All $n$ th roots of unity lie on the unit circle because each has modulus $1$ .
The roots of $z^n = 1$ are evenly spaced by an angle of $\frac{2\pi}{n}$ around the complex plane.

Vocabulary

Complex number: A number of the form $a + bi$ , where $a$ is the real part, $b$ is the imaginary part, and $i^2 = -1$ .
Modulus: The distance of a complex number $z = a + bi$ from the origin, given by $|z| = \sqrt{a^2 + b^2}$ .
Argument: The angle $\theta$ that a complex number makes with the positive real axis in the complex plane.
Polar form: A way to write a complex number as $z = r(\cos \theta + i\sin \theta)$ using its modulus and argument.
De Moivre's theorem: The rule $[r(\cos \theta + i\sin \theta)]^n = r^n(\cos n\theta + i\sin n\theta)$ for powers of complex numbers in polar form.
Root of unity: A complex number that satisfies $z^n = 1$ for a positive integer $n$ .

Common Mistakes to Avoid

Using the wrong quadrant for $\theta$ is wrong because $\tan \theta = \frac{b}{a}$ can give the same reference angle for different complex numbers.
Multiplying the modulus by $n$ in De Moivre's theorem is wrong because the modulus must be raised to the power, so $r$ becomes $r^n$ , not $nr$ .
Forgetting the $2\pi k$ term when finding roots is wrong because it gives only one root instead of all $n$ distinct roots.
Using degrees and radians together is wrong because formulas like $\frac{2\pi}{n}$ assume radian measure unless angles are consistently converted.
Listing more or fewer than $n$ roots for $z^n = a$ is wrong because a nonzero complex number has exactly $n$ distinct $n$ th roots.

Practice Questions

1 Write $z = 1 + i\sqrt{3}$ in polar form $r(\cos \theta + i\sin \theta)$ .
2 Use De Moivre's theorem to find $(2\operatorname{cis}\frac{\pi}{6})^4$ in polar form.
3 Find all cube roots of unity, which are the solutions to $z^3 = 1$ .
4 Explain why the $n$ th roots of unity form a regular polygon on the unit circle in the complex plane.

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De Moivre's Theorem & Roots of Unity Cheat Sheet

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