$z_1,z_2,z_3$ are the points of an equilateral triangle $\ \iff z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1$ $\implies (-1+i)^2+(2+4i)^2+z_3^2 = (-1+i)(2+4i)+(2+4i)z_3+z_3(-1+i)$ $\implies ( - 2i) + (-12 + 16i) + z_3^2 = (-6 - 2i) + z_3(1+5i)$ $\implies z_3^2 - (1+5i)z_3 - (6 + 16i) = 0$ $\implies z_3 = \frac{(1+5i) \pm \sqrt{74i}}{2}$ Now, we must simplify $\sqrt{74i}$: let $\sqrt{74i} = a+bi$, then $74i = a^2-b^2 + 2abi \implies a=b=\sqrt{37}$, so $\sqrt{74i} = \sqrt{37} + i\sqrt{37}$. Hence, $z_3 = \frac{1 + 5i \pm (\sqrt{37} + i \sqrt{37})}{2}$ $=\boxed{\frac{1 \pm \sqrt{37}}{2} + \frac{5 \pm \sqrt{37}}{2}i}$, so there are $\boxed{2}$ solutions. EDIT: Fixed it!

modularmarc101 wrote: $\implies \boxed{z_3 = \frac{(1+5i) \pm \sqrt{74i}}{2}}$, so we have $2$ solutions. Thank you for your reply, but this solution is wrong. I think you didn't understand the question properly, or I didn't write it correctly. If the triangle say $\triangle ABC$ we have $z_1=B=(-1,1)$ and $z_2=C=(2,4)$ so we must find $z_3=A=(x,y)$ such that $\triangle ABC$ is equilateral.

I solved it and my results are Z3=(1±√27)/2 + i(5±√27)/2 I hope it's right

Obel1x wrote: The complex numbers $z_1$ and $z_2$ are given such that $z_1=-1+i$ and $z_2=2+4i$. Find the complex number $z_3$ such that $z_1,z_2,z_3$ are the points of an equilateral triangle. How many solutions do we have ? Moreover, for any two different points $A, B$ from the complex plane $\mathcal{P},$ there is exactly two different points $C, D$ in $\mathcal{P}$ such the triangles $ABC$ and $ABD$ are equilateral triangles, which is easy to prove without any calculation method.