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.
Best Answer
Let $x+yi$ be the fourth complex number. Then, one has $$(-22+15i)-(-5+12i)=\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)((x+yi)-(-5+12i))$$ $$\Rightarrow -17+3i=-y+12+(x+5)i\Rightarrow x=-2\ \text{and}\ y=29.$$