complex numbers $z^n = 1$ complex plane I would like to know if I answered the question below correctly or if I need to change something.

Question: Take $n ∈ \mathbb N$. Find all solutions for $z^n=1$. Where can we find the solutions in the complex plane? Draw them in the complex plane for some values of $n$.**

$z^n = 1$ and $ n ∈ \mathbb N$. Assumption: $ 1 = e^{i0} = e^{i2π} = e^{i4π} = e^{i2(2π)} $.
$z^n = 1$ and $ n ∈ \mathbb N$.
= $ \cos (0) + i \sin (0)  $, (as $r = 1$)
= $ \cos (0+2kπ) + i \sin (0+2kπ)$
= $ \cos (2kπ) + i \sin (2kπ) $,    this gives us $ e^{i(2kπ)} $ with $k ∈ \mathbb Z$.
$z^n = 1$, so $z=\sqrt[n]{1 }$  and $z = 1^{1/n}$.
Using deMoivre's theorem, we get $ z = [e^{i(2kπ)}]^{1/n} $ =
$ z = e^{i2π(k/n)} ,  k = 0,1,2,...,n-1 $ with $ n ∈ \mathbb N$ .
Using deMoivre's theorem again, we get $ z = [\cos(2kπ) + i \sin (2kπ)]^{1/n} = \cos (2kπ/n) + i \sin (2kπ/n) $.

Now, we draw the solutions for some values of $n$ in the complex plane.
$ n = 1  \to z = 1 $ gives $ \cos (2kπ) + i \sin (2kπ) $, so $ z = 1 $.
$ n = 2  \to z = 1^{1/2} $ gives $ \cos (2kπ/2) + i \sin (2kπ/2) = \cos (kπ/2) + i \sin (kπ/2) $ , so $ z = 1, -1 $.
$ n = 3   \to z = 1^{1/3} $ gives $ \cos (2kπ/3) + i \sin (2kπ/3) $ , so $ z = -1/2 + 1/2√3 i, -1/2 - 1/2√3 i, 1 $ .
$ n = 4  \to z = 1^{1/4} $ gives $ \cos (2kπ/4) + i \sin (2kπ/4) = \cos (kπ/2) + i \sin (kπ/2) $, so $z = i, -1, -i, 1$.
Did I do it right? Should I change something in my notation?

An easier way to solve would be:
$$
\begin{align*}
z^{n} &= 1 \Rightarrow z^{n} = |z^{n}|\\
z &= |z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}}
(|z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\
|z|^{n} \cdot (\mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\
|z|^{n} \cdot \mathrm{e}^{n \cdot \arg(z) \cdot \mathrm{i}} &= 1\\
\mathrm{e}^{n \cdot \arg(z) \cdot \mathrm{i}} &= 1 \quad\mid\quad (\text{ 
})^{\frac{1}{2}}\\
\mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= 1^{\frac{1}{2}}\\
\mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= \pm \sqrt{1}\\
\mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= \pm 1\\
&\Rightarrow \mathrm{e}^{\arg(z) \cdot \mathrm{i}} = \Re(\mathrm{e}^{\arg(z) \cdot \mathrm{i}})\\
&=\Rightarrow \arg{z} = 2 \cdot \frac{k_{k \in \mathbb{Z}} + n}{n} \cdot \pi
\end{align*}
$$
That means that there are n solutions with $$ k = 1, ~2, ~..., ~n - 1, $$
wich means that there are mure solutios for z then $$ z = \left\{1, ~i, ~-1, ~-i\right\}, $$ aka you should change this.
e.g.
$$
\begin{align*}
z &= \frac{\mathrm{1}}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}}\\
z^{4} &= (\frac{\mathrm{1}}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}})^{4}\\
z^{4} &= (1^{\frac{1}{4}})^{4}\\
z^{4} &= 1
\end{align*}
$$
