<s>[INST] Equation with complex numbers Solve the following equation in $ \mathbb{C} $:
$ |z - |z + 1|| = |z + |z - 1|| $
I started it but I don't know how to finish it. Here is what I did so far:
$ |z - |z + 1||^2 = (z - |z + 1|)(\bar{z} - \overline{|z + 1|}) = |z|^2 - z \cdot \overline{|z + 1|} - \bar{z} \cdot |z + 1| + |z + 1|^2 $
$ |z + |z - 1||^2 = (z + |z - 1|)(\bar{z} + \overline{|z - 1|}) = |z|^2 + z \cdot \overline{|z - 1|} + \bar{z} \cdot |z - 1| + |z - 1|^2 $
Now, $ |z - |z + 1|| = |z + |z - 1|| \Rightarrow |z - |z + 1||^2 = |z + |z - 1||^2 $
So we have: $ |z|^2 - z \cdot \overline{|z + 1|} - \bar{z} \cdot |z + 1| + |z + 1|^2 = |z|^2 + z \cdot \overline{|z - 1|} + \bar{z} \cdot |z - 1| + |z - 1|^2 $
$ z(\overline{|z + 1| + |z - 1|}) + \bar{z}(|z + 1| + |z - 1|) = (|z + 1| + |z - 1|)(|z + 1| - |z - 1|) $
We divide all by |z + 1| + |z - 1| and we get:
$ z + \frac{||z + 1| + |z - 1||^2}{(|z + 1| + |z - 1|)^2} + \bar{z} = |z + 1| - |z - 1|$
But $ |z + 1| + |z - 1| $ is a positive number so $ ||z + 1| + |z - 1|| = |z + 1| + |z - 1| $
So $ z + \bar{z} = |z + 1| - |z - 1| $.
And now I don't know what to do with this equation.
 [/INST] $z+\overline{z}=|z+1|-|z-1|$
You might square twice
$$(z+\overline{z})^2=2z\overline{z}+2-2|z+1||z-1|\\
4(z+1)(\overline{z}+1)(z-1)(\overline{z}-1)=(2-z^2-\overline{z}^2)^2\\
4(z^2-1)(\overline{z}^2-1)=4-4z^2-4\overline{z}^2+z^4+2z^2\overline{z}^2+\overline{z}^4\\
0=(z^2-\overline{z}^2)^2$$
so $z^2$ is real, so $z=x$ or $z=iy$.
If $z=iy$, then $z+\overline{z}=0=|iy+1|-|iy-1|$, so the whole imaginary axis is included.
If $z=x$, then just the interval $[-1,1]$ is included.
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