## Precalculus (6th Edition) Blitzer

Two square roots of $9$ using DeMoivre’stheorem are $3$ and $-3$.
The roots of a complex number represented in polar form can be found by using DeMoivre’s theorem. DeMoivre’stheorem for finding complex roots states that every complex number has two distinct complex square roots, three distinct complex cube roots, four distinct complex fourth roots, and so on. Each root has the same modulus. \begin{align} & r=z_{k}^{n} \\ & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \end{align} Here, $k$ is the number of distinct $n\text{th}$ roots and $\theta$ is in radians. There are exactly two square roots of $9$ as per DeMoivre’s theorem. First write the provided number in rectangular form as follows: $9=9+0i$ Convert the above expression into polar form as follows: \begin{align} & 9=r\left( \cos \theta +i\sin \theta \right) \\ & =9\left( \cos 0+i\sin 0 \right) \end{align} Apply DeMoivre’s theorem to find the two roots of $9$. ${{z}_{k}}=\sqrt[2]{1}\left[ \cos \left( \frac{0+2\pi k}{2} \right)+i\sin \left( \frac{0+2\pi k}{2} \right) \right]$ Here, $k=0,1$. Insert the above values of $k$ to find two distinct square roots of the provided number. For $k=0$, \begin{align} & {{z}_{0}}=\sqrt[2]{9}\left[ \cos \left( \frac{0+2\pi \left( 0 \right)}{2} \right)+i\sin \left( \frac{0+2\pi \left( 0 \right)}{2} \right) \right] \\ & =3\left[ \cos 0+i\sin 0 \right] \\ & =3 \end{align} For $k=1$, \begin{align} & {{z}_{1}}=\sqrt[2]{9}\left[ \cos \left( \frac{0+2\pi \left( 1 \right)}{2} \right)+i\sin \left( \frac{0+2\pi \left( 1 \right)}{2} \right) \right] \\ & =3\left[ \cos \left( \pi \right)+i\sin \left( \pi \right) \right] \end{align} Substitute the values of $\cos \left( \pi \right)$ and $\sin \left( \pi \right)$. Then, \begin{align} & {{z}_{1}}=3\left[ -1.0+i\left( 0 \right) \right] \\ & =-3 \end{align} The two square roots of the complex number $9$ are $3$ and $-3$.