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

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Why doesn't $1+z^{2^n}$ have zeros on the unit disc?
 
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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
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...

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