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


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

Now am not sure,

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


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?

  • 1
    $\begingroup$ 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

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 $2\pi$.

  • $\begingroup$ 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

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 $m>n$.

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

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