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

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Discussion Overview

The discussion centers on the zeros of the function $1+z^{2^n}$ within the unit disc, exploring whether any zeros exist inside the disc or if they are confined to the unit circle. The scope includes mathematical reasoning and exploration of complex roots.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant questions why $1+z^{2^n}$ does not have zeros on the unit disc.
  • Another participant asserts that all zeros must lie on the unit circle.
  • A different participant challenges this by proposing a solution for $z$, suggesting that $z = (-1)^{1/2^n}$ leads to $z^{2n} = -1$.
  • This participant further elaborates that the solutions can be expressed as $z = e^{\frac{(2k+1)\pi}{2n} i}$ for integer values of $k$, indicating that while there are $2n$ distinct solutions, they all lie on the unit circle.

Areas of Agreement / Disagreement

Participants express differing views on the location of the zeros, with some asserting they are on the unit circle while others question this assertion. The discussion remains unresolved regarding the presence of zeros within the unit disc.

Contextual Notes

There are unresolved assumptions regarding the nature of the roots and their distribution, particularly concerning the implications of the solutions derived from the equation.

Dustinsfl
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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
 

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