# Pentagram coordinates

1. Jan 23, 2005

### arcnets

Hi,
I tried to find out how to plot a pentagram in the complex plane. Let the radius of the pentagram be 1, then all the corners of the pentagram satisfy the equation
x^5 = 1 (where x is a complex number).

The answer is, of course, x = exp (2n*pi*i/5).

But I wanted to express that in roots, not in exponentials.
I could not find anything on the Web, so I tried it myself.
It can be done, but the answer is not beautiful.
The prime root is:
x = a + i*a / sqrt(1-2/sqrt(5)),
where a = [ (9-4*sqrt(5)) / (16*sqrt(5)-16) ]^(1/5).
Next, I tried to express this in terms of the "Golden Ratio" Phi. Because I knew that Phi appears in the pentagram a lot.
Since
Phi^2 = 1*Phi + 1
Phi^3 = 2*Phi + 1
Phi^4 = 3*Phi + 2
Phi^5 = 5*Phi + 3
... and so on (that's the Fibonacci series twice),
we get
x = (Phi - 1)/2 + sqrt(Phi + 2)*i/2
and also
x^2 = -Phi/2 + sqrt(-Phi + 3)*i/2.
(x^3 and x^4 are obviously symmetrical wrt. the real axis)
That looks better.

But not perfect. I tried to get rid of the square root in the imaginary part, because I wanted everything to be linear in Phi. But no success.

Any help?

Last edited: Jan 23, 2005
2. Jan 23, 2005

### Hurkyl

Staff Emeritus
I'm pretty sure that one coordinate will require a square root nested inside a square root.

3. Jan 23, 2005

### dextercioby

Well,they do.Using
$$\varphi=:\frac{1+\sqrt{5}}{2}$$
,u see that sqrt of \phi implies sqrt from an sqrt...

Daniel.

4. Jan 23, 2005

### Hurkyl

Staff Emeritus
Right -- I was suggesting that he's stuck with something involving the square root of phi.

IMHO, it's better to explicitly write out what it is, rather than suppressing the square root of 5 into phi.

5. Jan 24, 2005

### arcnets

Thanks Hurkyl.
Of course, in some cases, a root containing Phi can be removed. For instance,
sqrt(Phi + 1) = phi
sqrt(3*Phi + 2) = phi + 1,
etc.

Do you think it's impossible for
sqrt(Phi + 2) = ?
sqrt(-Phi + 3) = ?

6. Jan 24, 2005

### Hurkyl

Staff Emeritus
Suppose it can be written as a * phi + b. Now square both sides and try to solve for a and b...

7. Jan 25, 2005

### arcnets

Yes. It works, but the coefficients have a square root inside a square root.