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Zeros of z^a-1

  1. Jan 22, 2008 #1
    What are the values of [itex]z[/itex] for which


    when [itex]z[/itex] and [itex]\alpha[/itex] are complex? Trivially,

    [tex]r\cdot e^{k 2\pi i/\beta}[/tex]

    is a solution for every integer value [itex]k[/itex] if [itex]r=1[/itex] and [itex]\beta=\alpha[/itex].

    Can every solution be written in this form? Again this is trivial for real [itex]\alpha[/itex], but how about [itex]\alpha[/itex] with a non-zero imaginary part? What are the values of [itex]r[/itex] and [itex]\beta[/itex] then?

  2. jcsd
  3. Jan 26, 2008 #2
    By definition
    [tex]z^{a}=\exp (a\ln z)[/tex].​
    [tex]u:=\ln z[/tex].​
    So you have to solve
    [tex]\exp(a u)=1[/tex].​
    where [tex]k[/tex] is arbitrary integer. It gives
    Now we solve
    [tex]\ln z=\frac{i2k\pi}{a}=\frac{i2k\pi\bar{a}}{\left|a\right|^2}=
    \frac{2k\pi \text{Im}\,a}{\left|a\right|^2}+
    If we find [tex]z[/tex] in the form [tex]z=r\exp(i\varphi)[/tex], then we obtain
    [tex]\ln\,r=\frac{2k\pi \text{Im}\,a}{\left|a\right|^2}[/tex]​
    where [tex]\ell[/tex] is an arbitrary integer.
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