# Plotting the roots of unity on the complex plane

1. Apr 25, 2012

### Cottontails

1. The problem statement, all variables and given/known data
Find the 6th complex roots of √3 + i.

2. Relevant equations
z^6=2(cos(π/6)+isin(π/6))
r^6=2, r=2^1/6
6θ=π/6+2kπ, θ=π/36+kπ/3

3. The attempt at a solution
When k=0, z = 2^1/6(cos(π/36)+isin(π/36)),
When k=1, z = 2^1/6(cos(13π/36)+isin(13π/36)),
When k=2, z = 2^1/6(cos(25π/36)+isin(25π/36)),
When k=3, z = 2^1/6(cos(37π/36)+isin(37π/36)),
When k=4, z = 2^1/6(cos(49π/36)+isin(49π/36)),
When k=5, z = 2^1/6(cos(61π/36)+isin(61π/36)).
I just want help with plotting these roots on the complex plane. So, I am just wondering, are all roots of unity on the complex plane the same, regardless of the equation? By this, I mean, are the position of the roots the same, regardless of what the equation is?

2. Apr 25, 2012

### LCKurtz

These aren't the roots of unity. They are the roots of $\sqrt 3 +i$. I'm not sure what you mean by the position being the same. What they are is distributed evenly around the circle. In this example, each one is rotated $\frac \pi 3$ from the next, so by the time you do that 6 times, you are all the way around.