# for what values of $\ \overline z = z^n$ [closed]

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, CesareoNov 7 '18 at 0:54

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

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

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 $$b$$.

HINT

We have that

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

• if $\ z^{n+1} = |z|^2 = 1$ then $\ r = 1$ and then what? – 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}.

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}