Expanding $(1-i)^\frac{1}{3}$ using De Moivre's formula I want to rewrite $(1-i)^ \frac 1 3$ using de Moivre's formula.
I defined $z := 1 - i$, then $r_z = \sqrt{2}$ and
$1 = \sqrt2\cos\theta$ and $-1 = \sqrt2 \sin\theta \Rightarrow \theta = -\frac \pi 4$
So:
$$1 - i = \sqrt{2}(\cos(\frac \pi 4) + i\sin(\frac \pi 4))$$
$$(1-i)^\frac 1 3 = 2^ \frac 1 6 (\cos(\frac \pi {12} + i\sin(\frac \pi {12}))$$
Am I correct with this derivation?

You have only take the principal root, indeed we have (there is also a typo for the angle in the $\sin$ term):
$$1 - i = \sqrt{2}\left(\cos\left(\frac \pi 4+2n\pi\right) + i\sin\left(-\frac \pi 4+2n\pi\right)\right)$$
and therefore the three possible roots are
$$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}\right) + i\sin\left(-\frac \pi {12}\right)\right)$$
$$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}+\frac 2 3 n\right) + i\sin\left(-\frac \pi {12}+\frac 2 3 n\right)\right)$$
$$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}+\frac 4 3 n\right) + i\sin\left(-\frac \pi {12}+\frac 4 3 n\right)\right)$$
with $n=0,1,2$.
