Number of solutions of the equation $z^p=(\overline{z})^q$

Question

So we have a complex number $z$ and we have to find number of solutions possible for the equation: $$z^p=(\overline{z})^q$$

So I created two cases:

Case-1: $p=q$. Clearly we can see the whole equation simplify into $z-\overline{z}=0$ or $\text{Im}(z)=0$. So entire real number axis is the solution, so number of solutions is infinite here.

Case-2: $p\ne q$, and let $p>q$. Here I run into problem. I tried to simplify the equation as: $$|z|^p=|z|^q$$ $$|z|^p(|1-|z|^{q-p})=0$$ $$|z|^p=0,\,\,|z|=1$$

I can count 1 solution for $z=0$ but for other equation I am unsure how can I count the solution, so I require the help here. Every help is appreciated.

If $|z|=1$, then $\overline z=z^{-1}$. Hence, our equation becomes $z^{p+q}-1=0$. This means, if $p\neq q$ there are $p+q+1$ solutions of $z^p=(\overline z) ^q$. — Jochen, Feb 24 '23 at 15:08
@Jochen oh I see, I didn't thought that way. Thank you very much! — Kshitij Kumar, Feb 24 '23 at 15:21

score 2 · Accepted Answer · answered Feb 24 '23 at 15:14

2

For $p\not=q$ we can pair the equation with its conjugate:

$z^p=(\overline z)^q$

$z^q=(\overline z)^p$

Therefore

$z^{p-q}=(\overline z)^{q-p}.$

With $p\not=q$ taking absolute values of this last equation forces $z=0$ or $|z|=1$. The former gives one solution, the latter implies $\overline z=1/z$ and thence $z^p=1/z^q$. Thereby, nonzero solutions must be $(p+q)$-order roots of unity giving $p+q+1$ solutions in all.

answered Feb 24 '23 at 15:14

Oscar Lanzi

39,403

Thank you very much! – Kshitij Kumar Feb 24 '23 at 15:22
This assumes $p$ and $q$ are positive integers. Is that the case, @KshitijKumar ? – MPW Feb 24 '23 at 15:57
1

@MPW Ig Yeah, question didn't explicitly mentioned about it. – Kshitij Kumar Feb 24 '23 at 16:27

score 2 · Answer 2 · answered Feb 24 '23 at 15:35

Another approach is using Euler's identity. If $p\neq q$ and $r\neq 0$ you get

$z=r\cdot e^{i\varphi}$. This results in $$(r\cdot e^{i\varphi})^p=\overline{(r\cdot e^{i\varphi})^q}$$ $$r^pe^{ip\varphi}=r^qe^{-iq\varphi}$$ $$r^{p-q}e^{i\varphi(p+q)}=1$$ from which you deduce that $$\varphi(p+q)=2n\pi\quad n\in\mathbb{Z}$$ $$\varphi=\frac{2n}{(p+q)}\pi\quad n\in\mathbb{Z}$$

Number of solutions of the equation $z^p=(\overline{z})^q$

2 Answers2