solution of $\bar{z}=\xi z$

Question

Does $\bar{z}=\xi z$ has solution in $z$ (complex number) for all values of $\xi \in \mathbb{C}$?

I was trying to use normal method but its coming quite complicated.

Answer 1

First, for any $\;\xi\in\Bbb C\;,\;\;z=0\;$ is a solution to $\;\overline z=\xi z\;$. Now, if we want non-zero solutions to this equation, we get that

$$\overline z=\xi z\implies |z|=|\xi|\,|z|\stackrel{\text{since}\;z\neq 0}\iff |\xi|=1\;$$

Thus, the answer to question in the first line of your post is yes, since $\;z=0\;$ does the job , and if we consider only $\;z\neq 0\iff z\in\Bbb C^*\;$, the answer then is no, as only the complex elements $\;\xi\in S^1=\{z\in\Bbb C\;:\;\;|z|=1\}\;$ have a chance to solve the given equation .

Added on request: If $\;\xi=\cos\theta+i\sin\theta\in S^1\;$, putting $\;z=r(\cos t+i\sin t)\;$ we get

$$\overline z=r(\cos t-i\sin t)=r\left(\cos\theta\cos t-\sin\theta\sin t+(\cos\theta\sin t+\sin\theta\cos t)i\right)\implies$$

$$\cos t-i\sin t=\cos(t+\theta)+i\sin(t+\theta)$$

This gives us two equations:

$$\begin{align*}&\cos t=\cos(t+\theta)\iff t=t+\theta\,,\,\,\text{or}\;\;t=-t-\theta\\{}\\ &\sin t=-\sin(t+\theta)\iff t=-t-\theta\;\;or\;\;t=\pi+t+\theta\end{align*}$$

So we can always find a non-zero solution.

Answer 2

As mentioned in another answer, $z=0$ is always a solution and we must have $|\xi|=1$ in order for a nonzero solution $z$ to exist.

To find a solution, consider this: if $|\alpha|=1$ then $\bar{\alpha} = \alpha^{-1}$. Thus $$\bar z = \xi z \quad \Leftrightarrow \quad \xi^{-\frac{1}{2}} \bar z = \xi^{\frac{1}{2}} z \quad \Leftrightarrow \quad \overline{\xi^{\frac{1}{2}}z} = \xi^{\frac{1}{2}}z$$ So what can you say about $\xi^{\frac{1}{2}}z$ in relation to $z$ being a solution to $\bar z = \xi z$?