Zeros of z^a-1

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  • #1
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What are the values of [itex]z[/itex] for which

[tex]z^\alpha-1=0[/tex]

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?

Thanks,
Harald.
 

Answers and Replies

  • #2
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By definition
[tex]z^{a}=\exp (a\ln z)[/tex].​
Denote
[tex]u:=\ln z[/tex].​
So you have to solve
[tex]\exp(a u)=1[/tex].​
Hence
[tex]au=i2k\pi[/tex],​
where [tex]k[/tex] is arbitrary integer. It gives
[tex]u=\frac{i2k\pi}{a}[/tex].​
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}+
\frac{i2k\pi\text{Re}\,a}{\left|a\right|^2}[/tex].​
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]​
and
[tex]\varphi=\frac{2k\pi\text{Re}\,a}{\left|a\right|^2}+
2\ell\pi[/tex]​
where [tex]\ell[/tex] is an arbitrary integer.
 

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