find the roots of the following: $z^3=2-2i$ in $Re^{i\theta}$ form Question
find the roots of the following: $z^3=2-2i$ in $Re^{i\theta}$ form
my steps:
So i first find the magnitude of the root to get $|2-2i|^{\frac{1}{3}}$
and the argument of the root is $\frac{\frac{-\pi}{4}}{3}$
thus the first root must be $|2-2i|e^{i*\frac{\frac{-\pi}{4}}{3}}$
But then i was confused in finding the other roots as well
 A: Cube roots are equally distributed around a circle centered at the pole and having radius equal to the cube root of the modulus (absolute value).
In the case of $z^3=2-2\,i$ the modulus is, as you stated, $\vert 2-2\,i\vert=2\sqrt{2}$. So the modulus of the roots is $\sqrt[3]{2\sqrt{2}}=\sqrt{2}$.The argument for $2-2\,i$ is $-\frac{\pi}{4}$ so the argument for one of the cube roots is one-third of that, namely as you state, $-\frac{\pi}{12}$.
All the three cube roots of $2-2\,i$ are equally spaced about the circle containing them, so they are separated by angles of $\frac{2\pi}{3}=\frac{8\pi}{12}$. So the three arguments of the roots are $-\frac{\pi}{12},\,\frac{7\pi}{12}$ and $\frac{15\pi}{12}=\frac{5\pi}{4}$.
Thus, the three cube roots of $2-2\,i$ are
$$ \sqrt{2}\exp\left(-\frac{\pi}{12}\right),\quad \sqrt{2}\exp\left(\frac{7\pi}{12}\right),\quad \sqrt{2}\exp\left(\frac{5\pi}{4}\right) $$
A: Let $z^3 = (R(\cos\theta + i \sin\theta))^3 = R^3(\cos3\theta + i \sin3\theta)$ using De Moivre's Theorem.
Recalling that $\cos\theta + i \sin\theta = e^{i\theta}$ we can solve for three values of $\theta$ by considering
$R^3(\cos3\theta + i \sin3\theta)=2-2i$
and solving for $r$ and $\theta$ in
(1) $R^3\sin3\theta = -2$
(2) $R^3\cos3\theta = 2$
before subbing back into $Re^{i\theta}$ as desired.
Hint $\cos^2\alpha + \sin^2\alpha = 1$ and $\tan\alpha=\displaystyle\frac{\sin\alpha}{\cos\alpha}$
