Lines from the centers of squares on two sides of a triangle to the third side? I have been working on the following problem from Visual Complex Analysis. My question is not necesarily if the solution is right, but more of a meta question about the solution and complex numbers. I apologize in advance if the question is a little vauge.


Construct two squares on the sides of an arbritrary triangle. Prove that the lines connecting the centers of the squares to $m$, the midpoint of the third side, are perpendicular and of equal length.

I solved this question by following the solution of a very similar exercise from the same book.

(Place?) the figure in the complex plane, and let the sides of the trianle be the complex numbers $2a, 2b, 2c$. Then $s$ is the complex number $a+ia$. Furthermore, $p$ is $2a + b + ib$, and $m$ is $2a + 2b + c$.
$s-m = -a-2b-c + ia$ 
$p-m = -b-c+ib$ 
are the dotted lines from $s$ to $m$ and from $p$ to $m$, respectively. Using the fact that $2a+2b+2c = 0$, it is not difficult to show that $i(p-m) = s-m$, which concludes the proof. 

I am used to thinking of a complex number as a point in a plane, with coordinates. But this understanding seems to go completely out the window. complex numbers are no longer points in the complex plane; instead, they are arrows. If we place them all at the origin, the result would not resemble the geometry problem at all. But it seems that somehow we are allowed to move all these arrows in any way we want.
 A: I think the difficulty you are having is the misconception that a complex number, say z, is a point in the complex plane. Rather, think of it as a vector (hence all those arrows). And they add and subtract just like vectors and they have scalar and vector products as well. For example, given two complex numbers, say $z_1$ and $z_2$, then the complex product $z_1z_2^*$, where * denotes the conjugate gives both the scalar and vector products. Specifically,
$$\Re\{z_1z_2^*\}=|z_1| \cdot |z_2| \cos(\zeta)=\frac{1}{2} (z_1z_2^*+z_1^*z_2) \\
\Im\{z_1z_2^*\}=|z_1| \cdot |z_2| \sin(\zeta)=\frac{1}{2} (z_1z_2^*-z_1^*z_2)$$
where $\zeta$ is angle between the two vectors.
Of course, complex numbers have many more properties. I'm just indicating a new way for to you to think about them.
A: Here is how I approached it.
Translate the whole thing such that the vertex opposite M is the origin.
represent the other vertexes of the triangle with complex numbers $2a,$ and $2b$
Point $M$ is then $a+b$
The center of one square is at $a(1-i)$
and the other is at $b(1+i)$
We can represent the two lines with:  $a(1-i) - (a+b) = -b-ia$ and $b(1+i) - (a+b) = a+ib$
if they are equal then $|-b-ia| = |a+ib|$
and if they are perpendicular $Re[(-b-ia)(a+ib)] = 0$
A: It all comes from the invariance with respect to the reference frame. What I mean is that if you draw the same triangle a little to the left, or right, or up, or down, the problem should not change. In fact you can even rotate the whole figure, and the answer should be the same. That means that you can choose any origin, and any two perpendicular directions through that origin to be your coordinate frame, without changing the answer. In this case they choose the bottom of the triangle (let's call this $B$). Then $2a$ is the coordinate of point $C$, the left corner. $2b$ is the coordinate of point $A$ (top corner of your triangle) with respect to point $B$. So in that reference frame, the absolute coordinate of point $A$ is $2a+2b$. Similarly, the lower corner $B$ coordinate with respect to $A$ is $2c$. So in any reference frame, we can call coordinates of points $A, B, C$ as $z_A, z_B, z_C$. Then The side $BA$ is going to be given by $z_B-z_A$ and so on. In this case they just called  $z_B-z_A=2c$
