Solving Complex Equation: $$ \bar{z} = z^n $$

[Rectifier]

The problem
I would like to solve:
$$ \bar{z} = z^n $$ where $n$ is a positive integer.

The attempt
$z = r e^{i \theta} \\ \\ \overline{ r e^{i \theta} } = r^n e^{i \theta n} \\ r e^{-i \theta} = r^n e^{i \theta n}$
$r = r^n \Leftrightarrow true \ \ if \ \ n=1 \ \ or \ \ r=1$
$e^{-i \theta} = e^{i \theta n} \\ -\theta = \theta n + 2 \pi k, \ k = 0,1,2...n \\ -\theta - \theta n = 2 \pi k \\ -\theta(1+n) = 2 \pi k \\ \theta(n+1) = 2 \pi k_1, \ k_1 = 0,-1,-2,...,-n \\ \theta = \frac{2 \pi k_1}{(n+1)}$

I don't understand the answer to this problem:

If $n=1$, then all real numbers are the solution.
If $n>1$, then $z=0$ or $z=e^{i\frac{2 \pi k_1}{(n+1)}}$ for $\ k = 0,1,2...n$

I am not sure how all real numbers could be a solution when n=1 and where they got $z=0$ and the condition $n>1$ from.

 

[fresh_42]

The problem
I would like to solve:
$$ \bar{z} = z^n $$ where $n$ is a positive integer.

The attempt
$z = r e^{i \theta} \\ \\ \overline{ r e^{i \theta} } = r^n e^{i \theta n} \\ r e^{-i \theta} = r^n e^{i \theta n}$
$r = r^n \Leftrightarrow true \ \ if \ \ n=1 \ \ or \ \ r=1$
$e^{-i \theta} = e^{i \theta n} \\ -\theta = \theta n + 2 \pi k, \ k = 0,1,2...n \\ -\theta - \theta n = 2 \pi k \\ -\theta(1+n) = 2 \pi k \\ \theta(n+1) = 2 \pi k_1, \ k_1 = 0,-1,-2,...,-n \\ \theta = \frac{2 \pi k_1}{(n+1)}$

I don't understand the answer to this problem:I am not sure how all real numbers could be a solution when n=1 and where they got $z=0$ and the condition $n>1$ from.

$r=r^n$ is also true for $r=0$ which you have forgotten. This results in $z=0$
For $z \in \mathbb{R}$ and $n=1$ the condition reads $\bar{z}=z$ which is certainly true.
Now to the role of $k$. Shouldn't it be $k \in \mathbb{Z}$ and the positive ones are just a matter of representation?

 

[Rectifier]

Thank you for your comment!

If the r=0 doesn't it matter what n could be? I mean, n could be 1 since $0 = 0^n = 0^1$, right?
I understand why z=0 if r=0 now, but I don't understand why n has to be > 1 and couldn't be = 1 there.

For k, I guess, that its just matter of representation since I get the same answer as the book if I put $2 \pi k$
on a different side when I compared exponents.

 

[fresh_42]

Thank you for your comment!

If the r=0 doesn't it matter what n could be? I mean, n could be 1 since $0 = 0^n = 0^1$, right?

Yes.

I understand why z=0 if r=0 now, but I don't understand why n has to be > 1 and couldn't be = 1 there.

Yes, it can also be one. However, this case ($z=0$) is already covered by $z \in \mathbb{R}$, the first case.

For k, I guess, that its just matter of representation since I get the same answer as the book if I put $2 \pi k$
on a different side when I compared exponents.

The $k$ says in which direction full circles are added, but both directions work: a full circle is a full circle, no matter in which direction. So the book as well as your solution should better be $k \in \mathbb{Z}$ in both cases. On the other hand, it doesn't really matter.

 

[Rectifier]

I understand the part with k but, unfortunately, I still don't get the rest of it, sorry. I suspect that I misunderstand some very basic algebra.

Let me try again and provide a few more steps to my calculation:
$r=r^n \\ r \left(1-r^{n-1} \right) = 0$ I understand this as $r_1 = 0$ if $n \neq 1$. Since if $n = 1$ then we have $r \cdot \left(1-r^{1-1} \right) \Rightarrow r \cdot 0 = 0$ which is solved by all $r$:s.

$1-r^{n-1} = 0$ leads to $1=r^{n-1}$ which leads to $r_2 = 1^{\frac{1}{n-1}}$ therefore $r_2 = 1$.

Since $z=re^{i\theta n}$ then $r_1 = 0$ leads to $z=0 \cdot e^{i\theta n} \Rightarrow z = 0$ when $n \neq 1$. We now that $n \in \mathbb{Z^+}$ "without 1" - $n \in \mathbb{Z^+} \setminus \{1\}$ which is bascially $n>1$. It is bascially the same thing for z where $r_2 = 1$ and we'll get the same result as in my first post. This part I understand, somewhat. Now to the next one.Now for $n=1$. How do I reason here?
I guess I could go back to $r=r^n \\ r \left(1-r^{n-1} \right) = 0$ and set $n =1$ and get $r \cdot 0 = 0$. All r:s solve it but I am expecting to get $z=r$ from $z=re^{i\theta n}$ . In that case we would need $\theta = 0$ since $n = 1$, $z=re^{i 0 \cdot 1}$. But how could one reason here?

 

[fresh_42]

I understand the part with k but, unfortunately, I still don't get the rest of it, sorry. I suspect that I misunderstand some very basic algebra.

Let me try again and provide a few more steps to my calculation:
$r=r^n \\ r \left(1-r^{n-1} \right) = 0$

This is the correct way to look at it, yes.

I understand this as $r_1 = 0$ if $n \neq 1$. Since if $n = 1$ then we have $r \cdot \left(1-r^{1-1} \right) \Rightarrow r \cdot 0 = 0$ which is solved by all $r$:s.

$1-r^{n-1} = 0$ leads to $1=r^{n-1}$ which leads to $r_2 = 1^{\frac{1}{n-1}}$ therefore $r_2 = 1$.

Yes. You could also just say: $r=0 \,\, \text{ OR } \,\, 1-r^{n-1}=0$.

This way you have the logic or in between, which means, and is allowed. They can both be zero, which answers your previous question in post #3. The distinction in the book starts with $n$. $n$ is either $1$ or greater than $1$. This separates all further consideration into these two parts. As the result is a combination of $n,r,\theta$, the combinations will be distinguished by $n$, not by the other two numbers. There can and will still be $(1,0,\theta)$ and $(n>1,0,\theta)$, which are different as triples, although resulting in the same number for $z$. Similar happens to $\theta$ where you have many values resulting in the same number $e^{i \theta}$. However, they are still different as triples, because we separated the cases $n=1$ and $n>1$. I hope this clarifies your statement:

Since $z=re^{i\theta n}$ then $r_1 = 0$ leads to $z=0 \cdot e^{i\theta n} \Rightarrow z = 0$ when $n \neq 1$. We now that $n \in \mathbb{Z^+}$ "without 1" - $n \in \mathbb{Z^+} \setminus \{1\}$ which is bascially $n>1$. It is bascially the same thing for z where $r_2 = 1$ and we'll get the same result as in my first post. This part I understand, somewhat. Now to the next one.

further.

Now for $n=1$. How do I reason here?
I guess I could go back to $r=r^n \\ r \left(1-r^{n-1} \right) = 0$ and set $n =1$ and get $r \cdot 0 = 0$. All $r$ solve it but I am expecting to get $z=r$ from $z=re^{i\theta n}$ . In that case we would need $\theta = 0$ since $n = 1$, $z=re^{i 0 \cdot 1}$. But how could one reason here?

You have forgotten the step before $r \left(1-r^{n-1} \right) = 0$.
The reason why we got this equation is, that we looked at $\overline{re^{i \theta}} = r^ne^{i n \theta}$ and compared only their length:
$$
\overline{re^{i \theta}} = r^ne^{i n \theta} \Rightarrow |\overline{re^{i \theta}}| = |r^ne^{i n \theta}| \Rightarrow r = r^n \Rightarrow r(1-r^{n-1}=0 \Rightarrow r=0 \,\, \text{ OR }\,\, n=1
$$
That means we deduced a necessary condition which must hold, if $\bar{z}=z^n$. It means the absolute value (length) of our complex numbers $\bar{z},z^n$ has to be equal. It does not mean, that automatically all such numbers are a solution. So $r=0 \,\, \text{ OR } \,\, n=1$ has always to hold. Now we can distinguish the cases $n=1$ and $n>1$. One of them has to be true, and only one.

CASE 1: $n=1$
This means $\bar{z}=z^1=z$ which is equivalent to $z \in \mathbb{R}$ which automatically means $\theta = 0$ per convention.
Now we formally will have to check, whether all reals are actually a solution. O.k. this is easy and true. Thus all $(n,r,\theta) = (1,r,0)$ are a solution, which still means $z$ is real.

CASE 2: $n>1$
This means $r=0 \Leftrightarrow z=0 \,\, \text{ OR } \,\, r=1$ are the only solutions. The case $r=0$ is clear. Now from here we have to consider more cases:
Given $n>1, r=1$, which are necessary condition on $\theta\,?$ You have found that under these circumstances, $\theta = \dfrac{2 \pi k}{n+1}$ for all $k \in \mathbb{Z}$. This is still a necessary condition, because we only have considered conclusions from $\bar{z}=z^n$. Now we formally will have to check, whether all those complex numbers $z=1 \cdot e^{i \theta}$ are actually a solution. This is the case, so all possibilities are checked.

In sum we have
$$
\{z=re^{i\theta}\triangleq (n,r,\theta)\,\vert \,\bar{z}=z^n\} = \{(n,r,\theta)\,\vert \,n=1 \wedge r\in \mathbb{R}\wedge \theta \stackrel{(*)}{=} 0\} \cup \{(n,r,\theta)\,\vert \, n>1 \wedge r=0 \wedge \theta \stackrel{(*)}{=} 0 ) \} \cup \{(n,r,\theta)\,\vert \, n>1 \wedge r=1 \wedge \theta =\frac{2 \pi k}{n+1} (k \in \mathbb{Z}) \}
$$
and the unions are not all disjoint and $(*)$ per convention of how the reals are considered as complex numbers.

By the way, the formal step from $1 \cdot e^{-i \frac{2 \pi k}{n+1}}= 1 \cdot e^{i \frac{2 \pi k n}{n+1}}$ is to multiply the equation by $e^{+ i \frac{2 \pi k}{n+1}}$ and investigate when the new exponent will be zero to match the $1=e^0$ on the left hand side.

 

[Rectifier]

Thank you for your answer! I understand it better now!

What does the triangle above $=$ stand for here:

$$
\{z=re^{i\theta}\triangleq (n,r,\theta)\,\vert \,\bar{z}=z^n\}
$$.

 

[fresh_42]

Thank you for your answer! I understand it better now!

What does the triangle above $=$ stand for here:

I used it for "corresponds to" or "is represented by". It is not an official symbol and here we have neither a form of equality nor a bijection. It just should say: "Let's have a solution $z=re^{i\theta}=\bar{z}^n\,\triangleq\,$Let's have a solution $(n,r,\theta) \ldots$". I simply wanted to abbreviate the long, formal road between the set of $z's$ and the set of triplets $(n,r,\theta)$.