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Hello all I am trying to find out if given that $m,n$ are integers such that $m \gt$ $n \gt 0$ , how I can find all the solutions $\in \mathbb{C}$ to

$$z^{m}=\bar z^{n}$$

What I have tried.

I wrote let $z=a+bi$ then $\bar z= a-bi$

We also have that $$|z|=|\bar z|=\sqrt{a^2+b^2}=r$$

I thought maybe I could use the form like

$$z^{m}=r^{m}(cos(m\theta)+isin(m\theta))=(re^{i\theta})^{m}$$

$$\bar z^{n}=r^{n}(cos(n\theta)+isin(n\theta))=(re^{i\theta})^{n}$$

Now am not sure,

I was thinking maybe like these are equal when $$r^{m}=r^{n}$$ and when

$$cos(m\theta)+isin(m\theta)=cos(n\theta)+isin(n\theta)$$

and then equating the real parts for that or something. Is this any way on right track? any hints or solution?

Thank you.

PS. Is it even possible that I will be able to get $r^{m}=r^{n}$ when $m \gt n$, or is there no solution?

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    $\begingroup$ Your second approach (utilizing de Moivre's law) is more likely to lead to success. $\endgroup$ Sep 26 '15 at 20:32
  • $\begingroup$ Your formula for $\bar{z}^n$ is wrong: $\bar{z}=re^{-i\theta}$, so $\bar{z}^n=(re^{-i\theta})^n$. $\endgroup$ Sep 27 '15 at 2:06
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Taking absolute values of both sides, we find $r^m=r^n$. Since $m>n$, the only way this can happen is if $r=0$ or $r=1$. If $r=0$ then $z=0$, which is a solution for any $m$ and $n$. If $r=1$, then $z=e^{i\theta}$ and $\bar{z}=e^{-i\theta}$, so we have $e^{i m\theta}=e^{-in\theta}$. Now use the following fact: if $s,t\in\mathbb{R}$, then $e^{is}=e^{it}$ iff $s-t$ is an integer multiple of $2\pi$.

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  • $\begingroup$ thanks , this makes sense . $\endgroup$
    – Quality
    Sep 27 '15 at 2:55
  • $\begingroup$ Could you possibly elaborate on how you know that last fact and such? $\endgroup$
    – Quality
    Sep 27 '15 at 18:46
  • $\begingroup$ Well, $e^{is}=e^{it}$ iff $e^{i(s-t)}=1$ iff $\cos(s-t)=1$ and $\sin(s-t)=0$. (Actually, I have seen books whose definition of $\pi$ is the number which makes this fact true.) $\endgroup$ Sep 27 '15 at 18:49
  • $\begingroup$ Could I maybe do this also by expanding it into the form cos(x)+isin(y) and then making use of even and off properties to get a system cos(x)=cos(y) and sin(x)=-sin(y) ? $\endgroup$
    – Quality
    Sep 27 '15 at 18:53
  • $\begingroup$ Yeah, you can do it directly like that too. $\endgroup$ Sep 27 '15 at 18:55
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Using knowledge of polar coordinates in $\mathbf{R}^{2}$, you are correct in intuiting that $r^{m}, r^{n}$ need to be equal. Putting aside the solution $z=0$ to your equation, we focus on $r>0$. For such $r$, $r^{m}=r^{n}$ implies $m=n$. But this contradicts your assumption that $m>n$.

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  • $\begingroup$ Hmm so what would that suggest then? That only zero works? $\endgroup$
    – Quality
    Sep 26 '15 at 22:16

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