Answer
Solutions of the provided equation in polar form are:
$\sqrt[3]{2}\left[ \cos 20{}^\circ +i\sin 20{}^\circ \right],\sqrt[3]{2}\left[ \cos 140{}^\circ +i\sin 140{}^\circ \right]\text{ and }\sqrt[3]{2}\left[ \cos 260{}^\circ +i\sin 260{}^\circ \right]$
Solutions of the provided equation in rectangular form are:
$1.1839+0.4309i,-0.9652+0.8099i\text{ and }-0.2188+1.2408i$.
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 $3$ .
$\begin{align}
& {{x}^{3}}=\left( 1+\sqrt{3}i \right) \\
& \sqrt[3]{{{x}^{3}}}=\sqrt[3]{1+\sqrt{3}i} \\
& x=\sqrt[3]{1+\sqrt{3}i} \\
& =\sqrt[3]{2\left( \cos 60{}^\circ +i\sin 60{}^\circ \right)}
\end{align}$
Apply Demoivre’s Theorem to find all solutions of the provided equation.
Here, $n=3$ as the number of solutions is three.
${{z}_{k}}=\sqrt[3]{2}\left[ \cos \left( \frac{60{}^\circ +360{}^\circ k}{3} \right)+i\sin \left( \frac{60{}^\circ +360{}^\circ k}{3} \right) \right]$
Here, $k=0,1,2$
Insert above values of $k$ to find three distinct solutions of the provided equation.
For $k=0$
$\begin{align}
& {{z}_{0}}=\sqrt[3]{2}\left[ \cos \left( \frac{60{}^\circ +360{}^\circ \cdot \left( 0 \right)}{3} \right)+i\sin \left( \frac{60{}^\circ +360{}^\circ \cdot \left( 0 \right)}{3} \right) \right] \\
& =\sqrt[3]{2}\left[ \cos 20{}^\circ +i\sin 20{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos 20{}^\circ $ and $\sin 20{}^\circ $ to find the solution in rectangular form.
$\begin{align}
& {{z}_{0}}=\sqrt[3]{2}\left[ 0.9397+0.3420i \right] \\
& \approx 1.1839+0.4309i
\end{align}$
For $k=1$
$\begin{align}
& {{z}_{1}}=\sqrt[3]{2}\left[ \cos \left( \frac{60{}^\circ +360{}^\circ \cdot \left( 1 \right)}{3} \right)+i\sin \left( \frac{60{}^\circ +360{}^\circ \cdot \left( 1 \right)}{3} \right) \right] \\
& =\sqrt[3]{2}\left[ \cos 140{}^\circ +i\sin 140{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos 140{}^\circ $ and $\sin 140{}^\circ $ to find the solution in rectangular form.
$\begin{align}
& {{z}_{1}}=\sqrt[3]{2}\left[ -0.7660+0.6428i \right] \\
& \approx -0.9652+0.8099i
\end{align}$
For $k=2$
$\begin{align}
& {{z}_{2}}=\sqrt[3]{2}\left[ \cos \left( \frac{60{}^\circ +360{}^\circ \cdot \left( 2 \right)}{3} \right)+i\sin \left( \frac{60{}^\circ +360{}^\circ \cdot \left( 2 \right)}{3} \right) \right] \\
& =\sqrt[3]{2}\left[ \cos 260{}^\circ +i\sin 260{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos 260{}^\circ $ and $\sin 260{}^\circ $ to find the solution in rectangular form.
$\begin{align}
& {{z}_{2}}=\sqrt[3]{2}\left[ -0.1736+\left( -0.9848 \right)i \right] \\
& \approx -0.2188-1.2408i
\end{align}$
