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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?

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closed as off-topic by Nosrati, José Carlos Santos, John B, Leucippus, Cesareo Nov 7 '18 at 0:54

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  • $\begingroup$ Write down $\bar z$ on the lhs. $\endgroup$ – Wuestenfux Nov 6 '18 at 14:14
  • $\begingroup$ Not sure what do you mean lhs? $\endgroup$ – bm1125 Nov 6 '18 at 14:18
  • $\begingroup$ Lhs is usually Left Hand Side, that is, what is at the left of the $=$ $\endgroup$ – ajotatxe Nov 6 '18 at 14:22
  • $\begingroup$ If $z=r(\cos(\theta)+i\cdot\sin(\theta))$, then $\bar{z}=r(\cos(\theta)-i\cdot\sin(\theta))$ $\endgroup$ – cansomeonehelpmeout Nov 6 '18 at 14:22
  • $\begingroup$ Oh thanks!! So if $\ r^n = r $ then $\ r = 1 $ ? and if $\ i \sin \theta = - i \sin n \theta $ then does it mean $\ \theta = 0 $ ?? $\endgroup$ – bm1125 Nov 6 '18 at 14:26
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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 $b$.

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HINT

We have that

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

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  • $\begingroup$ if $\ z^{n+1} = |z|^2 = 1 $ then $\ r = 1 $ and then what? $\endgroup$ – bm1125 Nov 6 '18 at 15:09
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\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}

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