Solve in $\mathbb{C}$ : $\bar{z}(z-1)=z^{2}(\bar{z}-1)$

Question

Solve in $\mathbb{C}$ :

$$\bar{z}(z-1)=z^{2}(\bar{z}-1)~~~~(1)$$

My try : ( please check where my wrong ) :

Take conjugate of equation then :

$$z(\bar{z}-1)=\bar{z^{2}}(z-1)=\bar{z}(z-1).\bar{z}$$

Then from $(1)$

$$z(\bar{z}-1)=\bar{z^{2}}(z-1)=z^{2}(\bar{z}-1).\bar{z}$$

So :

$$z(z-1)(\bar{z}-1)=0$$

Solution :

$$S=\{ 0,1,|z|=1 \}$$

But : $z=\frac{1}{2}+i\frac{\sqrt{3}}{2}$

$|z|=1$ but isn't a solution of $(1)$ why

Also $z=-1$

Please see my solution and I already to see you solution

Answer 1

$z=1$ and $z=0$ are immediately seen to be solutions.

Assuming $z\neq0,1$, take absolute values on both sides. We then get $$ |z^2|\cdot |\bar z-1|=|\bar z|\cdot|z-1| $$ We know that $|z|=|\bar z|$ and $|z-1|=|\bar z-1|$, which gives us $$ |z^2|\cdot |z-1|=|z|\cdot |z-1|\\ |z|=1 $$ Since $|z|=1$, we have $\bar z=\frac1z$, meaning the original equation becomes $$ z^2\left(\frac1z-1\right)=\frac1z(z-1)\\ z(1-z)=\frac1z(z-1)\\ -z(z-1)=\frac1z(z-1)\\ -z=\frac1z $$ This is only true for $z=\pm i$. So those are our solutions: $0,1,i,-i$.

Answer 2

Hint

Conjugating your equation you get $$z(\bar{z}-1)=\bar{z}^{2}(z-1)$$

Then $$\bar{z}(z-1)=z^{2}(\bar{z}-1)=z \left(z(\bar{z}-1)\right)=z\left(\bar{z}^{2}(z-1)\right)= z \bar{z} \left(\bar{z}(z-1)\right) $$

This gives $$\bar{z}=0 \mbox{ or } \\ z-1 = 0 \mbox{ or } \\ z\bar{z}=1 \Rightarrow \bar{z}=\frac{1}{z}$$

From here it should be easy.

Answer 3

From the given,

$$\bar{z}(z-1)=z^{2}(\bar{z}-1)\tag 1$$

we have $$|\bar{z}|^2|z-1|^2=|z^{2}|^2|\bar{z}-1|^2$$

Recognize $|\bar z|^2=|z|^2 = z\bar z$ and $|z-1|^2=|\bar{z}-1|^2$ to factorize,

$$|z|^2(z\bar z-1)|z-1|^2=0$$

Three cases to examine:

1) $|z|^2=0$ yields $z=0$.

2) $|z-1|^2=0$ yields $z=1$.

3) $z\bar z-1=0$. Substitute into (1) to get $(z-1)(1+z^2) = 0$, which yields $z=1$ and $z=\pm i$.

Thus, the solutions from all three cases are $z=0, 1, \pm i$.