MHB Finding the Zeros of $1+z^{2^n}$ on the Unit Disc

  • Thread starter Thread starter Dustinsfl
  • Start date Start date
  • Tags Tags
    Disc Unit
Dustinsfl
Messages
2,217
Reaction score
5
Why doesn't $1+z^{2^n}$ have zeros on the unit disc?
 
Physics news on Phys.org
dwsmith said:
Why doesn't $1+z^{2^n}$ have zeros on the unit disc?

All its zeros are on the unit circle, aren't they?

CB
 
CaptainBlack said:
All its zeros are on the unit circle, aren't they?

CB

I don't think so. If we solve for z, we have $z = (-1)^{1/2^n}$
 
dwsmith said:
I don't think so. If we solve for z, we have $z = (-1)^{1/2^n}$

\(z^{2n}=-1=e^{(2k+1)\pi i}, k \in \mathbb{Z}\)

so:

\(z=e^{\frac{(2k+1)\pi}{2n}\;i}, k \in \mathbb{Z}\)

of which \(2n\) are distinct, but all lie on the unit circle.

CB
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K