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?

*

*Your final step is invalid (discarding solutions!) because it
misapplies De Moivre's theorem, which does not generally hold for
non-integer
powers
like $\frac13.$ So, for example,
\begin{align}1^\frac12&=\left(\cos(2\pi)+i\sin(2\pi)\right)^{\frac12}\\&\neq\cos(\frac12\times2\pi)+i\sin(\frac12\times2\pi)\\&=-1.\end{align}
This is because in the complex world, raising a number to a
non-integer power generally outputs multiple values.

*In your exercise, the goal is to determine $(1-i)^\frac13,$ i.e., the
third roots of unity of $1-i.$ This process (abbreviating $(\cos
   x+i\sin x)$ as $\text{cis }x)$ invokes not De Moivre's theorem but
the multi-valued
definition of
$e^z:$ \begin{align}(1-i)^\frac13&=\left[\sqrt2 \text{ cis}
   \left(-\frac\pi4\right)\right]^\frac13\\&=2^\frac16\text{ cis}
   \left(\frac{-\frac\pi4+2k\pi}3\right)\\&=2^\frac16\text{ cis}
   \left(\frac{8k-1}{12}\pi\right)\\&=2^\frac16\text{ cis}
   \left(\frac{-\pi}{12}\right),\,2^\frac16\text{ cis}
   \left(\frac{7\pi}{12}\right)\text{ or }\,2^\frac16\text{ cis}
   \left(\frac{5\pi}{4}\right).\end{align}
