1. **Problem statement:** Given points $A=1$, $A'=-1$, $B=i$, $B'=-i$ in the complex plane, and a point $M$ with complex number $z$ different from $0, A, A', B, B'$, define points $M_1(z_1)$ and $M_2(z_2)$ such that triangles $BMM_1$ and $AMM_2$ are right-angled isosceles with angles $(\overrightarrow{M_1B}, \overrightarrow{M_1M}) = (\overrightarrow{M_2M}, \overrightarrow{M_2A}) = \frac{\pi}{2}$. Find all points $M$ such that triangle $OM_1M_2$ is equilateral. 2. **Key formulas and rules:** - A right-angled isosceles triangle with angle $\frac{\pi}{2}$ means the legs are equal and the angle between them is $90^\circ$. - Rotation by $90^\circ$ in the complex plane corresponds to multiplication by $i$ (counterclockwise) or $-i$ (clockwise). - For a point $M$ with complex number $z$, the point $M_1$ satisfies $\overrightarrow{M_1B} \perp \overrightarrow{M_1M}$ and $|M_1B|=|M_1M|$. - Similarly for $M_2$ with respect to $A$ and $M$. - The triangle $OM_1M_2$ is equilateral means $|OM_1|=|OM_2|=|M_1M_2|$ and angles between sides are $60^\circ$. 3. **Find $M_1$ given $M$:** - Let $z$ be the complex number of $M$. - Since $B=i$, the vector $\overrightarrow{M_1B} = z_1 - i$ and $\overrightarrow{M_1M} = z_1 - z$. - The right angle condition and isosceles condition imply: $$z_1 - z = i (z_1 - i)$$ (rotation by $90^\circ$ counterclockwise of $\overrightarrow{M_1B}$ gives $\overrightarrow{M_1M}$) - Solve for $z_1$: $$z_1 - z = i z_1 - i^2 = i z_1 + 1$$ $$z_1 - i z_1 = z + 1$$ $$z_1(1 - i) = z + 1$$ $$z_1 = \frac{z + 1}{1 - i}$$ 4. **Find $M_2$ given $M$:** - Similarly, $A=1$, so: $$z_2 - z = i (z_2 - 1)$$ - Solve for $z_2$: $$z_2 - z = i z_2 - i$$ $$z_2 - i z_2 = z - i$$ $$z_2(1 - i) = z - i$$ $$z_2 = \frac{z - i}{1 - i}$$ 5. **Simplify denominators:** - Note $1 - i = \sqrt{2} e^{-i \pi/4}$, so dividing by $1 - i$ rotates by $\pi/4$ and scales by $\frac{1}{\sqrt{2}}$. 6. **Condition for $OM_1M_2$ equilateral:** - Points are $O=0$, $M_1=z_1$, $M_2=z_2$. - Equilateral triangle means: $$|z_1| = |z_2| = |z_1 - z_2|$$ and the angle between vectors $z_1$ and $z_2$ is $\pm \frac{\pi}{3}$. 7. **Calculate $|z_1|$ and $|z_2|$:** $$|z_1| = \left| \frac{z + 1}{1 - i} \right| = \frac{|z + 1|}{|1 - i|} = \frac{|z + 1|}{\sqrt{2}}$$ $$|z_2| = \frac{|z - i|}{\sqrt{2}}$$ 8. **Calculate $|z_1 - z_2|$:** $$z_1 - z_2 = \frac{z + 1}{1 - i} - \frac{z - i}{1 - i} = \frac{(z + 1) - (z - i)}{1 - i} = \frac{1 + i}{1 - i}$$ - Note $\frac{1 + i}{1 - i} = i$ (since multiplying numerator and denominator by conjugate $1 + i$ gives $\frac{(1+i)^2}{1^2 + 1^2} = \frac{1 + 2i + i^2}{2} = \frac{1 + 2i -1}{2} = \frac{2i}{2} = i$). - So $|z_1 - z_2| = |i| = 1$. 9. **Equilateral conditions become:** $$|z_1| = |z_2| = 1$$ $$\Rightarrow \frac{|z + 1|}{\sqrt{2}} = \frac{|z - i|}{\sqrt{2}} = 1$$ $$\Rightarrow |z + 1| = |z - i| = \sqrt{2}$$ 10. **Interpretation:** - $|z + 1| = \sqrt{2}$ means $z$ lies on a circle centered at $-1$ with radius $\sqrt{2}$. - $|z - i| = \sqrt{2}$ means $z$ lies on a circle centered at $i$ with radius $\sqrt{2}$. 11. **Find intersection of circles:** - Circle 1: $(x + 1)^2 + y^2 = 2$ - Circle 2: $x^2 + (y - 1)^2 = 2$ 12. **Solve system:** - Expand: $$x^2 + 2x + 1 + y^2 = 2$$ $$x^2 + y^2 - 2y + 1 = 2$$ - Simplify: $$x^2 + y^2 + 2x = 1$$ $$x^2 + y^2 - 2y = 1$$ - Subtract second from first: $$(x^2 + y^2 + 2x) - (x^2 + y^2 - 2y) = 1 - 1$$ $$2x + 2y = 0$$ $$x + y = 0$$ 13. **Substitute $y = -x$ into one circle:** $$(x + 1)^2 + (-x)^2 = 2$$ $$x^2 + 2x + 1 + x^2 = 2$$ $$2x^2 + 2x + 1 = 2$$ $$2x^2 + 2x - 1 = 0$$ 14. **Solve quadratic:** $$x^2 + x - \frac{1}{2} = 0$$ $$x = \frac{-1 \pm \sqrt{1 + 2}}{2} = \frac{-1 \pm \sqrt{3}}{2}$$ 15. **Find $y$ values:** $$y = -x = -\frac{-1 \pm \sqrt{3}}{2} = \frac{1 \mp \sqrt{3}}{2}$$ 16. **Final points $z = x + iy$:** $$z_1 = \frac{-1 + \sqrt{3}}{2} + i \frac{1 - \sqrt{3}}{2}$$ $$z_2 = \frac{-1 - \sqrt{3}}{2} + i \frac{1 + \sqrt{3}}{2}$$ **Answer:** The points $M$ such that triangle $OM_1M_2$ is equilateral are exactly the two points $$z = \frac{-1 \pm \sqrt{3}}{2} + i \frac{1 \mp \sqrt{3}}{2}.$$