<s>[INST] Find $z\in\mathbb{C}$ such that $|z|=2i(\bar{z}+1)$ I was trying to solve the following problem:

Find $z\in\mathbb{C}$ such that $$|z|=2i(\bar{z}+1).$$

I tried to solve the problem by letting $z=x+iy$ and found out that the solutions are $z=-1+i\frac{1}{\sqrt{3}},-1-i\frac{1}{\sqrt{3}}$. But as tried to justify the solutions $z=-1-i\frac{1}{\sqrt{3}}$ does not satisfy the equation. 
I couldn't find any error in my calculation. Any help would be highly appreciated. Thanks in advance.
 [/INST] The equation you should get is
$$\sqrt{x^2+y^2} = 2i(x-iy + 1)\\
\sqrt{x^2 + y^2} = y + 2(x+1)i$$
which means that 


*

*$2y=\sqrt{x^2+y^2}$

*$2(x+1)=0$


From the second equation, you get $x=-1$, and the first equation becomes
$$2y=\sqrt{y^2+1}$$
Now, you SQUARE the equation and get $$4y^2=y^2+1$$
and get $$y^2=\frac{1}{3}$$

Now, the equation $y^2=\frac13$ has two solutions, but since you got that equation by squaring some other equation, you should know that some of the solutions of the second equation may not solve the first equation. In your case, $y=\frac{1}{\sqrt{3}}$ solves the equation, because $$2\cdot \frac1{\sqrt{3}}=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+1}$$
but $y=-\frac{1}{\sqrt{3}}$ does not solve the equation, because $$2\cdot\left(- \frac1{\sqrt{3}}\right)\neq 2\cdot\frac1{\sqrt{3}}=\sqrt{\left(-\frac{1}{\sqrt{3}}\right)^2+1}$$
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