Find the $z\in \Bbb C$ such that $\bar z=z^3$

Question

Here is what I've done. Let $z\in \Bbb C$ be of the form $a+ib$, lets assume in this case that $a,b\not=0$. We have that $\bar z=a-ib$, and $z^3=a^3-3ab^2+i(3a^2b-b^3)$, so to find the correct $z$, it has to satisfy: $$\begin{cases} a= a^3-3ab^2\\ -b= 3a^2b-b^3\\ \end{cases}$$ wich is the same as: $$\begin{cases} a=a( a^2-3b^2)\\ -b=b (3a^2-b^2)\\ \end{cases} \Rightarrow \begin{cases} 1=a^2-3b^2\\ 1=b^2 -3a^2\\ \end{cases}$$ then, $a^2-3b^2=b^2 -3a^2 \Rightarrow 4a^2=4b^2 \Rightarrow a^2=b^2 \Rightarrow \begin{cases} a=b \;\text{ or,}\\ a=-b\\ \end{cases}$

I thought this was correct, and to try it out I went to Wolfram, and it turns out that the examples that came to my mind didn't satisfied this. What did I do wrong?

What about the cases $a=0$ or $b=0$ ? For $b=0$ we have $a=1,-1$, for example. — Dietrich Burde, Feb 19 '14 at 20:56
Just because two equations are equal to a number (here 1) does not mean that the two equations can be set equal? — imranfat, Feb 19 '14 at 20:59
@DietrichBurde I was considering the case when both numbers are $\not=0$ — Ana Galois, Feb 19 '14 at 21:00

A Blumenthal · Accepted Answer · 2014-02-19T21:22:26.073

6

You should use radial coordinates. If $z \neq 0$, then your equation is equivalent to $|z|^2 = z^4$.

For $r$, we have $r^2 = r^4$, which amounts to $r^2 - 1 = (r - 1)(r + 1) = 0$. Since $r > 0$ by hypothesis, we have $r = 1$.

Pick out polar and radial parts: if $z = r e^{i \theta}$, then we must have that $4 \theta$ is a multiple of $2 \pi$, hence $\theta = 0, \pi/2, \pi, 3 \pi/2$.

The amusing thing is that in assuming $a,b$ both nonzero, you excluded all the possible solutions, which are $1, i, -1, -i$, and $0$.

edited Feb 19 '14 at 21:22

answered Feb 19 '14 at 20:58

A Blumenthal

5,048

Oops, forgot the solution $z = 0$. – A Blumenthal Feb 19 '14 at 21:22

score 6 · Answer 2 · edited Apr 13 '17 at 12:20

6

Note that $|\bar{z}|=|z|$. Thus, $|z|=|\bar{z}|=|z|^3\implies|z|\in\{0,1\}$.

If $|z|=0$, then $z=0$.

If $|z|=1$, then $\bar{z}=1/z$. Thus, $1/z=\bar{z}=z^3\implies z^4=1$. Thus, $z\in\{1,-1,i,-i\}$.

Just to be clear, there are $5$ solutions: $z\in\{0,1,-1,i,-i\}$

To answer Ana's comment: this generalizes to $\bar{z}=z^n$. $|z|=|\bar{z}|=|z|^n\implies|z|\in\{0,1\}$.

If $|z|=0$, then $z=0$.

If $|z|=1$, then $\bar{z}=1/z$. Thus, $1/z=\bar{z}=z^n\implies z^{n+1}=1$. That is, $$ z\in\{0\}\cup\{e^{i2k\pi/(n+1)}:k\in\mathbb{Z}\text{ and }0\le k\le n\} $$ These are precisely the solutions to $z^{n+2}-z=0$.

edited Apr 13 '17 at 12:20

Community

1

answered Feb 19 '14 at 21:05

robjohn

345,667

1

So $0$ is indeed an solution, isn't it? – imranfat Feb 19 '14 at 21:11
@imranfat: it is indeed! I have stated that explicitly to be clearer. – robjohn Feb 19 '14 at 21:13
Yes, yes, I see it :) – imranfat Feb 19 '14 at 21:16

score 2 · Answer 3 · answered Feb 19 '14 at 21:01

2

Ana, take a look at this:

Say we have $a+b=1$ and $a-b=1$ It follows very easily that $a=1$ AND $b=0$. However, with your approach, we could say $a+b=a-b$ and here we have infinite options for $a$

answered Feb 19 '14 at 21:01

imranfat

10,029

oh silly me... thanks – Ana Galois Feb 19 '14 at 21:03
So indeed in the original question if you substitute $a^2=b^2$ in $1=a^2-3b^2$, you see that there is no solution. This means that there is no solution with $a\ne0$, $b\ne0$, which is true. – Carsten S Feb 19 '14 at 21:30

Find the $z\in \Bbb C$ such that $\bar z=z^3$

3 Answers3

Linked