Square in the complex plane given three vertices. Find the fourth complex number vertice. There is a square in the complex plane. Four complex numbers form the four vertices of this square. Three of the complex numbers are $-19 + 32i,$ $-5 + 12i,$ and $-22 + 15i$. Find the fourth complex number. 
If this is a square, wouldn't the sides/distances between points have to be the same length? The points given don't seem like they would all be the same length when calculating the distance. Plotting points $(-19,32), (-5,12), (-22,15)$ shows a different graph entirely. 
 A: Let $x+yi$ be the fourth complex number. Then, one has
$$(-22+15i)-(-5+12i)=\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)((x+yi)-(-5+12i))$$
$$\Rightarrow -17+3i=-y+12+(x+5)i\Rightarrow x=-2\ \text{and}\ y=29.$$
A: The point $D := (-19,32) + (-5, 12) - (-22, 15) = (-2,29)$ has the property that (With the other points named $A$, $B$ and $C$ resp.)
$$|D-A| = |B-C| = |D-C| = |B-A|$$
All you must check now is that, for example $\angle CBA = \frac\pi2$
A: Another method:
Maybe the simplest math would be to plot the points and then see what transform makes (-22,15) into (-19,32) and then apply that transform (+3,+17) to (-5,12).  This would give you (-2,29)

A: One of the three complex numbers is going to be the vertex of the right angle formed by the three numbers. Call that complex number $z_0$. Then if the other complex numbers are $z_1$ and $z_2$, you'll have $z_1 - z_0$ is perpendicular to $z_2 - z_0$. You can figure out what $z_0$ is by trying all three possibilities. Then you can figure out the fourth vertex $z_3$ through the equation $z_3 + z_0 = z_1 + z_2$.
