The Moivre's theorem can be used to find roots of complex numbers. Given a complex number:
[tex]z=r(\cos(\theta)+i\sin(\theta))[/tex]Then, the theorem says, to find the n-th root:
[tex]\sqrt[n]{z}=\sqrt[n]{r}(\cos(\frac{\theta+2\pi k}{n})+i\sin(\frac{\theta+2\pi k}{n}))[/tex]Where n and k are natural numbers. The argument of the sine and cosine functions is the angle of the roots, the n-th root of the module is the module of the n-th root of the complex number.
Thus, to find the fourth root, n =4:
[tex]\sqrt[4]{z}=\sqrt[4]{r}(\cos(\frac{\theta}{4})+i\sin(\frac{\theta}{4}))[/tex]Since in step 1 the polar form of the number is written, we can write the formula for the 4-th roots:
[tex]\sqrt[4]{z}=\sqrt[4]{2}(\cos(\frac{\frac{\pi}{3}+2\pi k}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi k}{4}))=\sqrt[4]{2}(\cos(\frac{\pi}{12}+\frac{\pi k}{2})+i\sin(\frac{\pi}{12}+\frac{\pi k}{2})[/tex]Thus:
[tex]Module=\sqrt[4]{2}[/tex]And the angles are:
[tex]\begin{gathered} k=0\Rightarrow\frac{\pi}{12}+\frac{\pi\cdot0}{2}=\frac{\pi}{12}\frac{}{} \\ . \\ k=1\Rightarrow\frac{\pi}{12}+\frac{\pi\cdot1}{2}=\frac{7\pi}{12} \\ . \\ k=2\Rightarrow\frac{\pi}{12}+\frac{\pi\cdot2}{2}=\frac{13\pi}{12} \\ . \\ k=3\Rightarrow\frac{\pi}{12}+\frac{\pi\cdot3}{2}=\frac{19\pi}{12} \end{gathered}[/tex]Those are the angles of the fourth roots.
Next, for part 3, we can put all this together to write the roots in polar form:
[tex]\begin{gathered} z_1=\sqrt[4]{2}(\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})) \\ . \\ z_2=\sqrt[4]{2}(\cos(\frac{7\pi}{12})+i\sin(\frac{7\pi}{12})) \\ . \\ z_3=\sqrt[4]{2}(\cos(\frac{13\pi}{12})+i\sin(\frac{13\pi}{12})) \\ . \\ z_4=\sqrt[4]{2}(\cos(\frac{19\pi}{12})+i\sin(\frac{19\pi}{12})) \end{gathered}[/tex]