for what values of $\ \overline z = z^n $

Question

Suppose I have a matrix $\ D $ with the determinant $\ \det D = \overline z - z^n $ and I want to know when this expression is $$\ \overline z - z^n = 0 \\ \overline z = z^n \\ ?? = r^n(\cos n\theta + i \sin n\theta) $$

not sure how to procceed from here?

Answer 1

Start by letting $z=a+bi$, where $a$ and $b$ are real numbers.

Therefore, the given equation becomes,

$(a+bi)^n=a-bi$

Expanding the left hand side using binomial theorem.

Now, equate the real and imaginary parts across the two sides.

Solve the system of equations for $a$ and $b$.

Answer 2

HINT

We have that

$$\bar z = z^n \implies z=0 \quad \lor \quad z^{n+1}=|z|^2=1$$

Answer 3

\begin{align} \overline{z}&=z^n \tag{1}\label{1} . \end{align}

Assuming $n\in\mathbb{N},\ n>0$ we have a solution $z=0$.

Another trivial case:

\begin{align} n&=1 ,\\ \overline{z}&=z \tag{2}\label{2} . \end{align} In this case any $z\in\mathbb{R}$ is a solution.

Let $z\ne0$, $n>1$. Then for $z=|z|\exp(\theta\cdot i)$ we have

\begin{align} \overline{|z|\exp(\theta\cdot i)}&= (|z|\exp(\theta\cdot i))^n ,\\ |{z}|\exp(-\theta\cdot i)&=|z|^n\exp(n\theta\cdot i) \tag{3}\label{3} ,\\ \exp(-\theta\cdot i)&=|z|^{n-1}\exp(n\theta\cdot i) \tag{4}\label{4} , \end{align}.

hence

\begin{align} |z|&=1 ,\\ n\theta &=-\theta +2\pi k ,\quad k\in\mathbb{Z} ,\\ \theta &=\frac{2\pi k}{n+1} . \end{align}

Thus, for $n>1$ we have non-trivial solutions of the form

\begin{align} z&= \cos\left(\frac{2\pi k}{n+1}\right) + i\cdot\sin\left(\frac{2\pi k}{n+1}\right) ,\quad k\in\mathbb{Z} . \end{align}