Answer
Solutions of the provided equation in polar form are:
$\sqrt[3]{2}\left[ \cos 100{}^\circ +i\sin 100{}^\circ \right],\sqrt[3]{2}\left[ \cos 220{}^\circ +i\sin 220{}^\circ \right]\text{ and }\sqrt[3]{2}\left[ \cos 340{}^\circ +i\sin 340{}^\circ \right]$
Solutions of the provided equation in rectangular form are:
$-0.2188+1.2408i,-0.9652-0.8099i\text{ and }1.1839-0.4309i$.
Work Step by Step
We apply Demoivre’s Theorem.
Demoivre’s Theorem 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.
$\begin{align}
& r=z_{k}^{n} \\
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ k}{n} \right) \right] \\
\end{align}$
Here $k$ is the number of distinct ${{n}^{th}}$ solutions and $\theta $ is in degrees.
There are exactly three solutions of the provided equation as the degree of the equation is $3$ .
$\begin{align}
& {{x}^{3}}=\left( 1-\sqrt{3}i \right) \\
& \sqrt[3]{{{x}^{3}}}=\sqrt[3]{1-\sqrt{3}i} \\
& x=\sqrt[3]{1-\sqrt{3}i} \\
& =\sqrt[3]{2\left( \cos 300{}^\circ +i\sin 300{}^\circ \right)}
\end{align}$
Apply Demoivre’s Theorem to find all solutions of the provided equation.
Here, $n=3$ as the number of solutions is three.
${{z}_{k}}=\sqrt[3]{2}\left[ \cos \left( \frac{300{}^\circ +360{}^\circ k}{3} \right)+i\sin \left( \frac{300{}^\circ +360{}^\circ k}{3} \right) \right]$
Here, $k=0,1,2$.
Insert above values of $k$ to find three distinct solutions of the provided equation.
For $k=0$
$\begin{align}
& {{z}_{0}}=\sqrt[3]{2}\left[ \cos \left( \frac{300{}^\circ +360{}^\circ \cdot \left( 0 \right)}{3} \right)+i\sin \left( \frac{300{}^\circ +360{}^\circ \cdot \left( 0 \right)}{3} \right) \right] \\
& =\sqrt[3]{2}\left[ \cos 100{}^\circ +i\sin 100{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos 100{}^\circ $ and $\sin 100{}^\circ $ to find solution in rectangular form.
$\begin{align}
& {{z}_{0}}=\sqrt[3]{2}\left[ -0.1736+0.9848i \right] \\
& \approx -0.2188+1.2408i
\end{align}$
For $k=1$
$\begin{align}
& {{z}_{1}}=\sqrt[3]{2}\left[ \cos \left( \frac{300{}^\circ +360{}^\circ \cdot \left( 1 \right)}{3} \right)+i\sin \left( \frac{300{}^\circ +360{}^\circ \cdot \left( 1 \right)}{3} \right) \right] \\
& =\sqrt[3]{2}\left[ \cos 220{}^\circ +i\sin 220{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos 220{}^\circ $ and $\sin 220{}^\circ $ to find solution in rectangular form.
$\begin{align}
& {{z}_{1}}=\sqrt[3]{2}\left[ -0.7660+\left( -0.6428 \right)i \right] \\
& \approx -0.9652-0.8099i
\end{align}$
For $k=2$
$\begin{align}
& {{z}_{2}}=\sqrt[3]{2}\left[ \cos \left( \frac{300{}^\circ +360{}^\circ \cdot \left( 2 \right)}{3} \right)+i\sin \left( \frac{300{}^\circ +360{}^\circ \cdot \left( 2 \right)}{3} \right) \right] \\
& =\sqrt[3]{2}\left[ \cos 340{}^\circ +i\sin 340{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos 340{}^\circ $ and $\sin 340{}^\circ $ to find solution in rectangular form.
$\begin{align}
& {{z}_{2}}=\sqrt[3]{2}\left[ 0.9397+\left( -0.3420 \right)i \right] \\
& \approx 1.1839-0.4309i
\end{align}$
