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Me and my friend try to solve

$${{z}^{3}}=\overline{z}$$ where $z \in \mathbb{C}$.

My way to solve it was:

$\operatorname{cin}(\theta )=\cos(\theta)+\sin(\theta)i$

\begin{align} & z=r \operatorname{cin}(\theta ),\overline{z}=r\operatorname{cin}(-\theta +2\pi k) \\ & {{z}^{3}}={{r}^{3}}\operatorname{cin}(3\theta ) \\ & {{z}^{3}}=\overline{z} \\ & {{r}^{3}}\operatorname{cin}(3\theta )=r\operatorname{cin}(-\theta +2\pi k) \\ & {{r}^{3}}=r\Leftrightarrow {{r}^{3}}-r=0\Leftrightarrow r({{r}^{2}}-1)=0\Leftrightarrow r=\pm 1,r=0\Leftrightarrow r=1 \\ & 3\theta =-\theta +2\pi k\Leftrightarrow 4\theta =2\pi k\Leftrightarrow \theta =\frac{\pi k}{2} \\ \end{align}
So my solution for k=0,1,2 :

$\begin{align} & {{z}_{1}}=\cos (0)+\sin (0)i=1+0i \\ & {{z}_{2}}=\cos (\frac{\pi }{2})+\sin (\frac{\pi }{2})i=0+1i \\ & {{z}_{3}}=\cos (\pi )+\sin (\pi )i=-1+0i \\ \end{align}$

My friend on the other hand solve it like this:

$\begin{align} & {{z}^{3}}=\bar{z} \\ & {{z}^{3}}\cdot z=\bar{z}\cdot z \\ & {{z}^{4}}=|z{{|}^{2}} \\ \end{align}$

He got the same solutions expect he got one more solution than me ${{z}_{4}}=\cos (\pi )+\sin (\pi )i=0-i \\$

What is the right solution and why?

quid
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Guest
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    What is $\bar z$ in $\mathbb Z$ ?? –  May 25 '15 at 15:40
  • Why do you stop at $k=2$? What happens if you fill in $k=3$? – Uncountable May 25 '15 at 15:40
  • @Uncountable when you solve equation with ${{z}^{3}}$ don't you need to have only 3 solutions? – Guest May 25 '15 at 15:52
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    I think you are confusing with polynomials (for a polynomial over $\mathbb{C}$ the maximum number of zeros (roots) is the degree of the polynomial). $f(z)=z^3-\overline{z}$ is not a polynomial over $\mathbb{C}$ (since it cannot be expressed as $a_0+a_1z+\cdots+a_nz^n$ with $a_0,\cdots,a_n\in\mathbb{C}$), so it may have more than $3$ zeros, i.e. $z^3-\overline{z}=0$ can have more than $3$ solutions (and in fact it does). – Uncountable May 25 '15 at 15:56
  • Note that $0$ is a solution! Why/how did you rule that out as a possibility? – Cameron Buie May 25 '15 at 16:17

2 Answers2

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Note that $|z^3|=|z|^3$ and $|\bar{z}|=|z|$ so that for equality we require

$$|z|^3=|z| \underset{z\neq 0}{\Rightarrow} |z|^2=1\underset{|z|\geq 0}{\Rightarrow} |z|=1.$$

Note $0^3=\bar{0}$ so is a solution.

So we have $|z|=1$ so $z=e^{i\theta}$ and we want

$$z^3=\bar{z}\Rightarrow e^{3i\theta}=e^{-i\theta}\Rightarrow 3\theta\equiv -\theta\,\,\text{mod } 2\pi.$$

This gives you, for $k\in \mathbb{Z}$. $$\begin{align} 3\theta&=-\theta+2k\pi \\ \Rightarrow 4\theta&=2k\pi \\ \Rightarrow \theta&=k\frac{\pi}{2}. \end{align}$$

It remains to this gives you the non-zero solutions $\{\pm 1,\pm i\}$.

JP McCarthy
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Your friend is right. Although there is nothing wrong with your method. There is no reason to stop at $k=2$. For $k=3$,

$$\cos\frac{3\pi}2+i\sin\frac{3\pi}2=i$$

When $k=4$, we get $1$ again and so there are no more solutions.

Tim Raczkowski
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  • The problem is that i thought that when you solve ${{z}^{n}}$ you can only get n solutions , and here i get 4 solutions but i solve for solve ${{z}^{3}}$ – Guest May 25 '15 at 15:54
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    That would be true for a polynomial equation. Because $\overline z$ is involved, the equation is not polynomial. – Tim Raczkowski May 25 '15 at 15:57