$z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . Show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. As in the title: $z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . I need to show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. I know how to do it "non creative" way. So, I transform both complex numbers into trigonometric form and divide:
$$\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}} = \frac{2[\cos(\frac{7}{4}\pi)+\sin(\frac{7}{4}\pi)i]}{2[\cos(\frac{5}{6}\pi)+\sin(\frac{5}{6}\pi)i]}= \cos \left(\frac{11}{12}\pi \right)+\sin \left( \frac{11}{12}\pi \right)i$$
Now, I see that this is $165^{\circ}$, and that module $= 1$. The problem is, that the only way I see to find the number of elements of the X set is to calculate degrees:
$$165^{\circ}, 330^{\circ}, 135^{\circ}, 300^{\circ}, 105^{\circ}, 270^{\circ}, 75^{\circ}, 240^{\circ}, 45^{\circ}, 210^{\circ}, 15^{\circ}, 180^{\circ}, 345^{\circ}, 150^{\circ}, 315,^{\circ} 120^{\circ}, 285^{\circ}, 90^{\circ}...$$
So I feel like there has to be some more effective way to solve that problem and here comes my question. Do you know any?
 A: Note that $$z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}} =\frac{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i}{\frac{1}{2}−\frac{\sqrt{3}}{2}i}=\frac{e^{-\frac{\pi}{4}i}}{e^{-\frac{\pi}{3}i}}=e^{\frac{\pi}{12}i}$$
Thus $$z^n=e^{\frac{n\pi }{12}i}$$
The proof follows from the fact that $z^{24}=1$
A: Hint
Calculate separatelly
$$\frac{\sqrt{2}(1-i)}{2}$$
and
$$\frac{1−i\sqrt{3}}2$$
A: A way is to find a polynomial satisfied by $z$.  (For low degrees, this is likely to be a minimal polynomial.  To show this requires computing elements of $X$ successively and searching for a linear relation among the powers of $z$ produced so far.  The first such relation is the minimal polynomial.  All we can actually promise is that we get a polynomial multiple of a minimal polynomial.) \begin{align*}
\frac{z}{1+\mathrm{i}\sqrt{3}} 
    &= \frac{\sqrt{2}(1+\mathrm{i})}{1+3}  \\
4z &= \sqrt{2}(1+\mathrm{i})(1+\mathrm{i}\sqrt{3})  \\
16z^2 &= 2(1+\mathrm{i})^2(1+\mathrm{i}\sqrt{3})^2  \\
8z^2 &= 2\mathrm{i}(1+\mathrm{i}\sqrt{3})^2  \\
4z^2 &= \mathrm{i}(-2 + 2\mathrm{i}\sqrt{3})  \\
2z^2 &= \mathrm{i}(-1 + \mathrm{i}\sqrt{3})  \\
4z^4 &= \mathrm{i}^2(-1 + \mathrm{i}\sqrt{3})^2    \\
-4z^4 &= (-2 - 2\mathrm{i}\sqrt{3})  \\
-2z^4 &= (-1 - \mathrm{i}\sqrt{3})  \tag{$\star$}  \\
4z^8 &= (-1 - \mathrm{i}\sqrt{3})^2  \\
4z^8 &= (-2 + 2\mathrm{i}\sqrt{3})  \\
2z^8 &= (-1 + \mathrm{i}\sqrt{3})  \\
2z^8 - 2z^4 &= -2  &  &  \text{using ($\star$)}  \\
z^8 - z^4 + 1 &= 0  \\
z^8 &= z^4 - 1  \text{.}
\end{align*}
(Using other tools, we can show that this is the minimal polynomial of $z$, but this is not critical.)
So, $\require{enclose}$
$$  X = \{1,z, z^2, \dots, z^7, z^4-1, z^5-z, \dots, z^8 - z^4 = -1, -z, -z^2, \dots, -z^7, -z^4+1, -z^5+z, \dots, \enclose{horizontalstrike}{-z^8+z^4 = 1}\}  \text{,}  $$
a set of only $24$ elements.
We haven't explicitly shown that the norm of $z$ is $1$, it is implicit in the fact that the powers of $z$ are cyclic, equivalently, $X$ is finite.
A: If $q = a/b$ is any fraction in lowest terms, then by de Moivre's identity, $$\left( \cos(2\pi q) + i \sin(2 \pi q) \right)^b = \cos(2\pi a) + i \sin(2 \pi a) = 1,$$ and conversely, if we raise $\cos(2\pi q) + i \sin(2 \pi q)$ to any power $0 < n < b$, we won't get an integer multiple of $2\pi$ inside the cosine and sine, and so the power will not be equal to 1. This means the order must be $b$.
In your case, $2\pi q = 11\pi / 12$, so $q = 11/24$ and the order is 24.
