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[SOLVED] zeros on D(0,1)

dwsmith

Well-known member
Feb 1, 2012
1,673
Why doesn't $1+z^{2^n}$ have zeros on the unit disc?
 

CaptainBlack

Well-known member
Jan 26, 2012
890

dwsmith

Well-known member
Feb 1, 2012
1,673

CaptainBlack

Well-known member
Jan 26, 2012
890
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