1. **Problem statement:** Given points A, A', B, B' corresponding to complex numbers 1, -1, i, -i respectively, and a point M with complex number $z$ different from O, A, A', B, B', we define points $M_1(z_1)$ and $M_2(z_2)$ such that triangles $BMM_1$ and $AMM_2$ are right-angled isosceles with $$ (\overrightarrow{M_1 B}, \overrightarrow{M_1 M}) = (\overrightarrow{M_2 M}, \overrightarrow{M_2 A}) = \frac{\pi}{2} $$ We want to find all points $M$ such that triangle $OM_1M_2$ is equilateral. 2. **Understanding the problem:** - Points correspond to complex numbers: $A=1$, $A'=-1$, $B=i$, $B'=-i$, and $O=0$. - $M$ is a complex number $z$ not equal to these points. - $M_1$ and $M_2$ are constructed so that triangles $BMM_1$ and $AMM_2$ are right-angled isosceles with the given angle conditions. - We want to find $z$ such that triangle $OM_1M_2$ is equilateral. 3. **Expressing $M_1$ and $M_2$ in terms of $z$:** - Since $B = i$, and triangle $BMM_1$ is right isosceles with right angle at $M_1$, and vectors $\overrightarrow{M_1 B}$ and $\overrightarrow{M_1 M}$ are perpendicular, $M_1$ is the image of $M$ rotated by $\pm \pi/2$ around $B$. - The vector $\overrightarrow{M_1 B}$ is perpendicular to $\overrightarrow{M_1 M}$, so $M_1$ lies such that $M_1$ is the rotation of $M$ around $B$ by $\pm \pi/2$. - Similarly, $M_2$ is the rotation of $M$ around $A=1$ by $\pm \pi/2$. - The length of $M_1 B$ equals $M_1 M$ (isosceles right triangle), so the rotation is by $\pm \pi/2$ and the distance is preserved. 4. **Formulas for $M_1$ and $M_2$:** - Rotation of point $z$ around point $c$ by angle $\theta$ is: $$ z' = c + e^{i\theta}(z - c) $$ - For $M_1$, center $B = i$, angle $\pm \pi/2$: $$ z_1 = i + e^{i\frac{\pi}{2}}(z - i) = i + i(z - i) = i + i z - i^2 = i + i z + 1 = 1 + i + i z $$ or rotation by $-\pi/2$: $$ z_1 = i + e^{-i\frac{\pi}{2}}(z - i) = i - i(z - i) = i - i z + i^2 = i - i z - 1 = -1 + i - i z $$ - For $M_2$, center $A=1$, angle $\pm \pi/2$: $$ z_2 = 1 + e^{i\frac{\pi}{2}}(z - 1) = 1 + i(z - 1) = 1 + i z - i = 1 - i + i z $$ or rotation by $-\pi/2$: $$ z_2 = 1 + e^{-i\frac{\pi}{2}}(z - 1) = 1 - i(z - 1) = 1 - i z + i = 1 + i - i z $$ 5. **Choosing the correct rotations:** - The problem states the angle between vectors is $\frac{\pi}{2}$, so either rotation is possible. - We will consider the first options: $$ z_1 = 1 + i + i z $$ $$ z_2 = 1 - i + i z $$ 6. **Condition for $OM_1M_2$ to be equilateral:** - Points are $O=0$, $M_1 = z_1$, $M_2 = z_2$. - Triangle $OM_1M_2$ is equilateral if: $$ |z_1| = |z_2| = |z_2 - z_1| $$ 7. **Calculate $|z_1|$, $|z_2|$, and $|z_2 - z_1|$:** - Compute $z_2 - z_1$: $$ z_2 - z_1 = (1 - i + i z) - (1 + i + i z) = -2 i $$ - So: $$ |z_2 - z_1| = |-2 i| = 2 $$ - Compute $|z_1|$: $$ z_1 = 1 + i + i z $$ Write $z = x + i y$, then $$ i z = i(x + i y) = i x - y $$ So: $$ z_1 = 1 + i + i x - y = (1 - y) + i(1 + x) $$ Thus: $$ |z_1| = \sqrt{(1 - y)^2 + (1 + x)^2} $$ - Compute $|z_2|$: $$ z_2 = 1 - i + i z = 1 - i + i x - y = (1 - y) + i(-1 + x) $$ So: $$ |z_2| = \sqrt{(1 - y)^2 + (-1 + x)^2} $$ 8. **Set the equalities:** $$ |z_1| = 2 \implies (1 - y)^2 + (1 + x)^2 = 4 $$ $$ |z_2| = 2 \implies (1 - y)^2 + (-1 + x)^2 = 4 $$ 9. **Subtract the two equations:** $$ (1 + x)^2 - (-1 + x)^2 = 0 $$ Expand: $$ (1 + x)^2 = 1 + 2x + x^2 $$ $$ (-1 + x)^2 = 1 - 2x + x^2 $$ Difference: $$ (1 + 2x + x^2) - (1 - 2x + x^2) = 4x = 0 $$ So: $$ x = 0 $$ 10. **Plug $x=0$ back into one equation:** $$ (1 - y)^2 + (1 + 0)^2 = 4 $$ $$ (1 - y)^2 + 1 = 4 $$ $$ (1 - y)^2 = 3 $$ $$ 1 - y = \pm \sqrt{3} $$ $$ y = 1 \mp \sqrt{3} $$ 11. **Final solutions for $z$:** $$ z = 0 + i y = i(1 - \sqrt{3}) \quad \text{or} \quad i(1 + \sqrt{3}) $$ 12. **Verification:** - For these $z$, $|z_1| = |z_2| = |z_2 - z_1| = 2$, so triangle $OM_1M_2$ is equilateral. **Answer:** $$ z = i(1 - \sqrt{3}) \quad \text{or} \quad z = i(1 + \sqrt{3}) $$