$$X^5 +1 = 0$$ I need to find all the five fifth roots of unity.
My method :
$$\begin{align}X &=(-1)^{1/5} \\ &= \left(\;\cos(2\pi k+\pi)+\sin(2\pi k+\pi)\;\right)^{\;1/5} \end{align}$$ where $k = 0,1,2,3,4$. Then by using De Moivre's Theorem, I further solved it.
But I'm getting stuck at $k=2$, as I'm again getting $X=-1$ at both $k=2$ and $k=0$.
I am not able to understand as where I'm going wrong. If anyone could provide me with a detailed answer it would be very helpful for me
On
$$X^5 = -1 = e^{(2k+1)\pi i}$$
$$X = e^{\frac{(2k+1)}{5}\pi i}, \quad (k=0,1,...,4)$$
$$X \in \{ e^{\pi i /5}, e^{3\pi i /5}, e^{5\pi i /5}, e^{7\pi i /5}, e^{9\pi i /5} \}$$
As you wrote, apply De Moivre's theorem.
On
Abbreviation: $Cis(x)=\cos x +i\sin x.$ Not very common, but I like it.
"Unity" is $1.$ If $X$ is a $5$th root of unity then $X^5-1=0,$ not $X^5+1=0.$
If $t\in \Bbb R$ then $[Cis(t)=1 \iff t/2\pi\in\Bbb Z].$ Therefore if $t\in\Bbb R$ and $X=Cis(t),$ we have $$[ X^5=1] \iff [ Cis(5t)=1] \iff$$ $$\iff [5t/2\pi\in \Bbb Z] \iff$$ $$\iff [\exists n\in \Bbb Z\,(t=2\pi n/5)\,] \iff$$ $$\iff [X\in S=\{Cis(2\pi n/5): n\in \{0,1,2,3,4\}\,\}\,]$$ because if $n\in\Bbb Z$ and $n\not\in \{0,1,2,3,4\}$ then $Cis (2\pi n/5)\in S$ anyway.
So $S$ is a set of $5$ solutions to $X^5-1=0$. And $\Bbb C$ is a $field$. In a field, a polynomial equation in one variable, of degree $m,$ with $m\in\Bbb Z^+$, cannot have more than $m$ solutions. So $S$ is the set of $all$ the $5$th roots of unity.
You can also observe that if $X\in \Bbb C$ but $X\ne Cis(t)$ for any $t\in\Bbb R$ then $\exists r,t\in \Bbb R\,(|r|\ne 1\land X=r\cdot Cis(t)\,),$ but then $|X^5|=|r|^5\ne 1,$ so $X^5\ne 1.$
As you suggest, by de Moivre the roots are $$z_k =\cos \frac{\pi + 2k\pi}{5} + i\sin \frac{\pi + 2k\pi}{5}$$
etc.; in particular, $z_0\neq z_2$. Perhaps you forgot to divide by $5$?