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?


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, José Carlos Santos, John B, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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

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

Therefore, the given equation becomes,


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 $b$.



We have that

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

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

\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}.


\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}


Not the answer you're looking for? Browse other questions tagged or ask your own question.