[[Noteworthy Uses]]:: *[[Noteworthy Uses]]* [[Proved By]]:: [[De Moivre's formula]], [[The Fundamental Theorem of Algebra]] Intuition:: *[[Intuition]]* Examples:: [[]] Specializations:: *[[Specializations]]* Generalizations:: *[[Generalizations]]* ---- > [!theorem] Theorem. ([[nth root theorem]]) > If $z$ is a [[complex numbers|complex number]] written in [[polar form of a complex number|polar form]] as $z=r(\cos x + i\sin x)$, then the $n$ $n^{th}$ roots of $z$ are given by $\sqrt[n]{ r } \left( \cos \frac{x+2\pi k}{n} + \sin \frac{x+2\pi k}{n} \right), $ where $k \in \{ 0, \dots, n-1 \}.$ That is, the $n^{th}$ roots of $z$ lie on an origin-centered circle of radius $|z|$ and are spaced out evenly by angles of $\frac{2\pi}{n}$. \ **Remark.** The $\sqrt[n]{ r }$ notation refers to the *unique* nonnegative $n^{th}$ root of $|z |$. > [!proof]- Proof. ([[nth root theorem]]) > We have via [[De Moivre's formula]] (**remark**) : $\begin{align} z^{1/n} = & r ^{\frac{1}{n}}(\cos x + i\sin x)^{\frac{1}{n}} \\ = & r ^{\frac{1}{n}} \left( \cos \frac{x+2\pi k}{n} + i\sin \frac{x + 2\pi k}{n} \right)^{\frac{1}{n}} , \end{align}$ for $k \in \zz$. By [[The Fundamental Theorem of Algebra]], we only have $n$ distinct roots ($n$ distinct solutions to $c ^{n}=z$). Which are they? We see that $k=0$ yields one, as does $k=1$; continuing in this fashion we have $k \in \{ 0, \dots, n-1 \}$ as required. > [!basicexample]- > ###### Cube roots of Unity. What are the solutions to $1=z^{3}$? The [[modulus]] and [[argument]] of $1$ are $1$ and $0$ respectively. Using the [[nth root theorem]] we have $1^{1/3}=(e^{j 0})^{1/3}=(\cos 0 + j \sin 0)^{\frac{1}{3}} = \left( \cos \frac{0+2 \pi k}{3} + i \sin \frac{0 + 2 \pi k}{3} \right),$ where $k$ varies from $0$ to $2$. In particular, we have three solutions: > - $z_{0} = \cos 0 + i \sin 0 = 1$, > - $z_{1} = \cos \frac{2 \pi}{3} + i\sin \frac{2 \pi}{3}=-\frac{1}{2}+\frac{\sqrt{ 3 }}{2} j$, and >- $z_{2} = \cos \frac{4 \pi }{3} + j\sin \frac{4 \pi}{3}=-\frac{1}{2} - \frac{\sqrt{ 3 }}{2}j$. ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```