Third point of a triangle in the complex plane I have an equilateral triangle with two points equal to $(2+2i)$ and $(5+i)$. I want to find the third point(s) (there are $2$ of these). I have that the side length of the triangle is $\sqrt{10}$.  
 A: A different Hint:  if $A,B\in\mathbb{C}$ and $\omega=e^{\frac{\pi i}{3}}=\frac{1+\sqrt{-3}}{2}$, the points you are looking for are given by:
$$ A+(B-A)\omega, \qquad A+(B-A)\omega^{-1}.$$
This happens because the multiplication by $\omega$ ($\omega^{-1}$) acts like a $60^\circ$ ($-60^\circ$) rotation.
A: Hint: You already  know that $\big|z-(5+i)\big|=\big|z-(2+2i)\big|=\ldots$ (you are to fulfil the gap), where $z=x+yi$ the complex number(s) you are looking for. Plus, you have an equilateral triangle, which means all sides are equal!
A: Yet another hint: Why not work in $\Bbb R^2$ with vectors? You have the points $A = (2,2)$ and $B = (5,1)$, and the length $\ell = \sqrt{10}$. Find the perpendicular bisector of $\overline{AB}$ and go $h = \frac{\ell\sqrt{3}}{2} = \frac{\sqrt{30}}{2}$ in both senses.
A: (I'm purposely not trying 
to be clever here
and doing all calculations in my head.)
The distance
$d$
between the two known points
satisfies
$d^2
=(2-5)^2+(2-1)^2
=10
$.
If the point is
$(x, y)$,
then
$(x-2)^2+(y-2)^2
=(x-5)^2+(y-1)^2
=10
$,
or
$x^2-4x+4+y^2-4y+4
=x^2-10x+25+y^2-2y+1
=10
$,
or
$x^2-4x+y^2-4y+8
=x^2-10x+y^2-2y+26
=10
$.
Subtracting the first two,
$6x-2y-18 = 0$,
or
$y = 3x-9$.
Substituting this
in the first equation,
$10
=x^2-4x+(3x-9)^2-4(3x-9)+8
=x^2-4x+9x^2-54x+81-12x+36+8
=10x^2-70x+125
$
or
$10x^2-70x+115=0$
or
$2x^2-14x+23=0$
Using the good old
quadratic equation formuls,
$D^2 
= b^2-4ac
= 14^2-4\cdot 2 \cdot 23
=196-184
=12
$
so
$D = 2\sqrt{3}$.
The roots are therefore
$x
=\dfrac{-b \pm D}{2 a }
=\dfrac{14\pm 2\sqrt{3}}{2\cdot 2}
=\dfrac{7\pm \sqrt{3}}{2}
$
and
$y
=3x-9
=\dfrac{21\pm 3\sqrt{3}-18}{2}
=\dfrac{3\pm 3\sqrt{3}}{2}
$.
Note:
I find it somewhat 
embarrassing and dishearting
how many errors
I made
while doing this
all in my head.
