- Thread starter
- #1

- Thread starter dwsmith
- Start date

- Thread starter
- #1

- Jan 26, 2012

- 890

All its zeros are on the unit circle, aren't they?Why doesn't $1+z^{2^n}$ have zeros on the unit disc?

CB

- Thread starter
- #3

I don't think so. If we solve for z, we have $z = (-1)^{1/2^n}$All its zeros are on the unit circle, aren't they?

CB

- Jan 26, 2012

- 890

\(z^{2n}=-1=e^{(2k+1)\pi i}, k \in \mathbb{Z}\)I don't think so. If we solve for z, we have $z = (-1)^{1/2^n}$

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