Problem 1. Solve in the set of complex numbers the equation $|z-| z+1||=|z+| z-1||$.
Solution. Let $z=a+bi, a, b \in \mathbb{R}$. The equation can be written as $(a-|z+1|)^{2}+b^{2}=$ $(a+|z-1|)^{2}+b^{2}$, from which we get $a-|z+1|= \pm(a+|z-1|)$.

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We obtain $2 a=|z+1|-|z-1|$, equivalent to $2 a=\sqrt{(1+a)^{2}+b^{2}}-$ $\sqrt{(1-a)^{2}+b^{2}}$.

1 point

It is observed that $a=0$ satisfies, and any $z=bi, b \in \mathbb{R}$ is a solution.

1 point

For $a \neq 0$, multiplying by $\sqrt{(1+a)^{2}+b^{2}}+\sqrt{(1-a)^{2}+b^{2}}$ we deduce that $\sqrt{(1+a)^{2}+b^{2}}+\sqrt{(1-a)^{2}+b^{2}}=2$, from which $\sqrt{(1+a)^{2}+b^{2}}=1+a$ and $\sqrt{(1-a)^{2}+b^{2}}=1-a$.

1 point

For $|a|>1$ there are no solutions. For $a \in[-1,1]$ we get $b=0$, so any $z=a, a \in[-1,1]$ is a solution.

1 point