I use an online calculator in order to calculate $x^5-1=0$
I get the results
I know that this is the correct answer because my roots have to be on the complex plane but I do not understand how can I get the results!
$$ x^5 = 1 = e^{2k\pi i} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k \in \mathbb N$$ $$ x = e^{\frac {2k\pi i}5}$$ Since it has 5 roots, for $k = 0,1,2,3,4$ $$ x = 1, e^{\frac {2\pi i}5}, e^{\frac {4\pi i}5}, e^{\frac {6\pi i}5}, e^{\frac {8\pi i}5}$$
Now you can use $e^{ix} = \cos (x) + i \sin (x)$
Instead of using $x = a + bi$ type of complex numbers for the soultion, you should consider using $x = r\operatorname{cis}\theta$ to solve this. For example you can do : $$x^5 - 1 = 0$$ $$x^5 = 1$$ $$r^5(\operatorname{cis}(5\theta)) = 1\operatorname{cis}0$$ $$r^5 = 1$$ $$r=1$$ $$5\operatorname\theta = 0 + 360k$$ $$\operatorname\theta = 72k$$ $$x = 1\operatorname{cis}(72k), k = 1,2,3,4,5$$
This solution is easier and more basic than Bhuvan Agrawal solution, but his one is good too.
$$x^5-1=(x-1)(x^4+x^3+x^2+x+1)$$
If $x-1=0,x=1$
Else $x^4+x^3+x^2+x+1=0$
Now follow Equation with high exponents
