Precalculus / Unit 7: Polar Coordinates and Complex Numbers
7.6 De Moivre's Theorem and Roots
Notes
De Moivre's Theorem: $[r\,\text{cis}\,\theta]^n = r^n\,\text{cis}\,(n\theta)$.
- Example: $\displaystyle (1 + i)^8 = (\sqrt{2})^8\,\text{cis}\,2\pi = 16$.
- The $n$ distinct nth roots of $r\,\text{cis}\,\theta$ are $\displaystyle r^{1/n}\,\text{cis}\,\frac{\theta + 2\pi k}{n}$ for $k = 0, 1, \dots, n - 1$.
- The nth roots are equally spaced around a circle.
- Example: the cube roots of 8 are $2$, $\displaystyle -1 + \sqrt{3}i$ and $\displaystyle -1 - \sqrt{3}i$.
Practice
Compute $(1 + i)^8$.
$\displaystyle 1 + i = \sqrt{2}\,\text{cis}\,\frac{\pi}{4}$, so the power is $2^4\,\text{cis}\,2\pi = 16$.
Compute $\displaystyle (\sqrt{3} + i)^3$.
$\displaystyle \sqrt{3} + i = 2\,\text{cis}\,\frac{\pi}{6}$, so the cube is $\displaystyle 8\,\text{cis}\,\frac{\pi}{2} = 8i$.
Which is a cube root of 8?
It has modulus 2 and argument $\displaystyle \frac{2\pi}{3}$, so cubing gives $8\,\text{cis}\,2\pi = 8$.
Ready for more? Take the Unit 7 test.