I need to find fifth root of unity in the form $x+iy$. I'm new to this topic and would appreciate a detailed "dummies guide to..." explanation!
I understand the formula, whereby for this question I would write: $1^{1/5} = r^{1/5}e^{2ki\pi/5}$. However, I don't know what to do next.
Any help is appreciated.
On
Let's do it the hard way. We want to solve the equation $$x^5-1=(x-1)(x^4+x^3+x^2+x+1)=0.$$ Then we are interested in solving $x^4+x^3+x^2+x+1=0$. Note the symmetry: if $r$ is a root, $1/r$ is also a root. Why? Thereafter, we have two ways out. First, we divide everything by $x^2$ to get $$x^2+x+1+\frac 1x+\frac{1}{x^2}=\left(x+\frac1x\right) +\left(x+\frac1x\right)^2-1=u^2+u-1=0.$$ What did we do? We noted $\left(x+x^{-1}\right)^2-2=x^2+x^{-2}$ and substituted $u=x + \frac{1}{x}$. The solutions for $u^2+u=1$ are $-\varphi$ and $\varphi-1$, $\varphi$ being the golden ratio. We now have two equations: $$x+\frac1x=-\varphi\implies x^2+\varphi x+1=0\implies x=\frac{\pm\sqrt{\varphi-3}-\varphi}{2}$$ $$x+\frac{1}{x}=\varphi-1\implies x^2+(1-\varphi)x+1=0\implies x=\frac{\pm\sqrt{-\varphi-2}+\varphi-1}{2}.$$
You can now manipulate it to get a $a+bi$ form. Alternatively, you can multiply $(x-r)(x-r^{-1})$, divide $x^4+x^3+x^2+x+1$ by it and discover which $r$ makes remainder zero. This would be somewhat long, so I won't explictly do it, but here is the idea.
You pretty much have the right idea. We have:
$1^{1/5} = (e^{2\pi ki})^{1/5} = e^{2\pi k i/5}=\cos(2 \pi k/5) + i\sin(2 \pi k/5)$
And that's in $a+bi$ form. Letting $k = 0,\dots,4$ gives you all $5$ fifth roots of unity.