## Precalculus (6th Edition) Blitzer

The three cube roots are $\,\,\frac{-3}{2}+i\frac{3\sqrt{3}}{2}\,\,\text{ and }\,\,\frac{-3}{2}-i\frac{3\sqrt{3}}{2}$.
We will rewrite the provided number as a complex number: $z=27+0i$ Now we will convert this into polar form. For a complex number $x+iy$, the polar form is $r\left( \cos \theta +i\sin \theta \right)$. Here, \begin{align} & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ & \theta ={{\tan }^{-1}}\left( \frac{y}{x} \right) \\ \end{align} For the provided complex number z: \begin{align} & r=\sqrt{{{27}^{2}}+{{0}^{2}}} \\ & =\sqrt{729} \\ & =27 \end{align} And, \begin{align} & \theta ={{\tan }^{-1}}\left( \frac{0}{27} \right) \\ & ={{\tan }^{-1}}\left( 0 \right) \\ & =0 \end{align} This gives the polar form as: $r\left( \cos \theta +i\sin \theta \right)=27\left( \cos 0+i\sin 0 \right)$ Now use deMoivre’s theorem for find the roots, ${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right]$ …… (1) Now we will put the values of n,r, k and $\theta$ in equation (1) to find out the roots. For cube roots $n=3$; k is $0,1,$ and $2$ for three roots. Now for $k=0$, \begin{align} & {{z}_{0}}=\sqrt[3]{27}\left[ \cos \left( \frac{0+2\pi \cdot 0}{3} \right)+i\sin \left( \frac{0+2\pi \cdot 0}{3} \right) \right] \\ & ={{\left( 27 \right)}^{\frac{1}{3}}}\left[ \cos \left( 0 \right)+i\sin \left( 0 \right) \right] \\ & =3\left[ 1 \right] \\ & =3 \end{align} So the first root of the equation is 3. For $k=1$, \begin{align} & {{z}_{1}}=\sqrt[3]{27}\left[ \cos \left( \frac{0+2\pi \cdot 1}{3} \right)+i\sin \left( \frac{0+2\pi \cdot 1}{3} \right) \right] \\ & ={{\left( 27 \right)}^{\frac{1}{3}}}\left[ \cos \left( \frac{2\pi }{3} \right)+i\sin \left( \frac{2\pi }{3} \right) \right] \\ & =3\left[ \frac{-1}{2}+i\frac{\sqrt{3}}{2} \right] \\ & =\frac{-3}{2}+i\frac{3\sqrt{3}}{2} \end{align} The second root of the equation is $\frac{-3}{2}+i\frac{3\sqrt{3}}{2}$. For $k=2$, \begin{align} & {{z}_{1}}=\sqrt[3]{27}\left[ \cos \left( \frac{0+2\pi \cdot 2}{3} \right)+i\sin \left( \frac{0+2\pi \cdot 2}{3} \right) \right] \\ & ={{\left( 27 \right)}^{\frac{1}{3}}}\left[ \cos \left( \frac{4\pi }{3} \right)+i\sin \left( \frac{4\pi }{3} \right) \right] \\ & =3\left[ \frac{-1}{2}-i\frac{\sqrt{3}}{2} \right] \\ & =\frac{-3}{2}-i\frac{3\sqrt{3}}{2} \end{align} The third root of the equation is $\frac{-3}{2}-i\frac{3\sqrt{3}}{2}$.