How to Solve the Complex Equation zn-1=\bar{z}?

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SUMMARY

The equation \( z^{n-1} = \bar{z} \) for \( z \in \mathbb{C}^* \) can be solved by expressing \( z \) in polar form as \( z = re^{it} \). This leads to the equality \( r^{n-1} e^{(n-1)it} = re^{-it} \), which simplifies to \( r = 1 \) and \( nt \equiv 0 \pmod{2\pi} \). The solutions to this equation are thus characterized by \( t = \frac{2k\pi}{n} \) for integers \( k \), providing a complete set of solutions.

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freddyfish
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Hi

I have come across this equation:

zn-1=[itex]\bar{z}[/itex], z[itex]\in[/itex]ℂ* (ℂ*:=ℂ\0)

There are numerous obvious equalities that can be used, but I don't seem to reach a satisfying final answer.

Any help would be appriciated.

Thanks in advance :)
 
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freddyfish said:
Hi

I have come across this equation:

zn-1=[itex]\bar{z}[/itex], z[itex]\in[/itex]ℂ* (ℂ*:=ℂ\0)

There are numerous obvious equalities that can be used, but I don't seem to reach a satisfying final answer.

Any help would be appriciated.

Thanks in advance :)



Write [itex]\,z=re^{it}\,\,,\,\,0\leq t<2\pi\,[/itex] , so that

$$z^{n-1}=\bar z\Longleftrightarrow r^{n-1}e^{(n-1)it}=re^{-it}\Longrightarrow r=1\,\,,\,(n-1)t=-t\pmod {2\pi}\Longleftrightarrow nt=0\pmod{2\pi}$$

DonAntonio
 
That's what I got too, using a somewhat different method, but I felt that my answer didn't include all the solutions to this equation for some reason. But I suppose my gut feeling betrayed me on this one.

Thank you for your time and effort. :)
 

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