I am trying to find the principal value of, $$\left(\left(\frac{1+\sqrt{3}i}{2}\right)^{-3}\right)^{1-i}.$$
My attempt:
I use the fact that $PVz^a=\exp(a\text{Log}(z))$ for $z\in\mathbb{C}$\ $0$ and $a\in\mathbb{C}$. So, \begin{align} PV\left(\left(\frac{1+\sqrt{3}i}{2}\right)^{-3}\right)^{1-i}&=\exp\left((3i-3)\text{Log}\left(\frac{1+\sqrt{3}i}{2}\right)\right) \\ &=\exp\left((3i-3)\left(\ln\left|\frac{1+\sqrt{3}i}{2}\right|+i\text{Arg}\left(\frac{1+\sqrt{3}i}{2}\right)\right)\right) \\ &=\exp\left((3i-3)\left(i\frac{\pi}{3}\right)\right) \\ &=-\exp(-\pi) \end{align} Is this correct? I am unsure if my first step (to determine $a$) is valid.
edit
following through the same steps, but using $z=-1$ and $a=1-i$, I get the answer $-\exp(\pi)$. Why does, $$PV\left(\left(\frac{1+\sqrt{3}i}{2}\right)^{-3}\right)^{1-i}\neq PV\left(\frac{1+\sqrt{3}i}{2}\right)^{3i-3}??$$
i'm denoting cube roots of unity i.e, $(1)^{\frac{1}{3}}$by $ 1 \ or \ \omega^3,\omega, \omega^2 $
therefore,
$ 1+x+x^2 $ has roots $\omega , \omega^2$
$\implies$ $1-x+x^2$ has roots $-\omega,\omega^2$ and product of it's roots i.e, $-\omega^3 =1\implies \omega^{-3}=-1 $
therefore your equation can be written as
$\left[(-\omega)^{-3}\right]^{1-i}=\left[-(\omega)^{-3}\right]^{1-i}=\left[-1 \times -1 \right]^{1-i}=(1)^{1-i}=e^{2n \pi \ i(1-i)}=e^{2n\pi (1+i)}=e^{2n\pi}$ it is purely real with principle argument zero and magnitude $e^{2n\pi}$
where ,
$n=0,1,2.......$