Let $z = i – 1\in\mathbb{C}$. Let $n\in\mathbb{N}$. Show that $z^n \in\mathbb{N}$ when $n$ is divisible by $8$.
I note that $z$ can be written as
$$z = \sqrt{2}(\cos(\arctan(-1)) + i \sin(\arctan(-1))) = \sqrt{2}\left(\cos\left(- \frac{\pi}{4}\right) + i \sin\left(- \frac{\pi}{4}\right)\right)$$
Now, using De Moivre's thereom, which states that $(\cos(\theta) + i \sin(\theta))^{n}=\cos(n\theta) + i \sin(n\theta)$,
$$z^{n} = \left(\sqrt{2}\left(\cos\left(- \frac{\pi}{4}\right) + i \sin\left(- \frac{\pi}{4}\right)\right)\right)^{n}
= \sqrt{2}^{n}\left(\cos\left(-\frac{n\pi}{4}\right) + i \sin\left(- \frac{n\pi}{4}\right)\right)$$
For $n$ increasing by $8$, $n = 8k$, where $k$ is natural number. Thus
$$z^{n} = \sqrt{2}^{8k}\left(\cos\left(-\frac{8k\pi}{4}\right) + i \sin\left(- \frac{8k\pi}{4}\right)\right).$$
Since $\cos\left(-\frac{8k\pi}{4}\right) = \cos(-2k\pi) = \cos(0) = 1$ and $\sin\left(- \frac{8k\pi}{4}\right)= \sin(-2kπ) = \sin(0) = 0$, $z^{n} = \sqrt{2}^{8k}$ is indeed a natural number for $n$ increasing by $8$.
Is my proof valid? Can I make any improvements anywhere? I have not constructed many proofs yet, and will gladly welcome any feedback and critique!
Best Answer
Your proof looks valid, though you could simply say $(i-1)^2=-2i$, so $(i-1)^4=-4$, so $(i-1)^8=16\in\mathbb N$, so $(i-1)^{8k}=16^k\in\mathbb N$.