Geometry through Complex Numbers

Help can come from the most surprising places. Case in point -- sometimes proving things in geometry can be made vastly simpler by appealing to complex numbers! Consider the following problem...

An arbitrary quadrilateral is drawn, with four squares built on it's outside edges as shown below. Prove that the two segments that join the opposite squares' centers (i.e., the red ones) are both perpendicular and of the same length.

Suppose the four vertices $A$, $B$, $C$, and $D$, correspond to complex numbers $a$, $b$, $c$, and $d$. We would like to get our hands on the complex numbers that represent the four centers of the squares. We can work this out explicitly for $R$, the center of the square with segment $\overline{AB}$ as one of its sides, and then argue the rest from symmetry.

One approach would be to notice that the center of this square, $R$, is the midpoint of a diagonal that contains $A$. We know the complex value that corresponds to $A$ is $a$, but getting our hands on the opposite corner's complex value is a bit trickier. Note that the difference $(b-a)$ gives a complex number matching the displacement from $A$ to $B$. Now here's the cool part... If we multiply a complex number by $i$, the result is a complex number that has been rotated $90\,^{\circ}$ counter-clockwise about the origin. This is a straight-forward application of De Moivre's Theorem, which students should be familiar with from precalculus. So if we start at $b$ and add a $90\,^{\circ}$-rotated version of the displacement from $A$ to $B$, we arrive at the complex number of the vertex opposite $A$. This must then correspond to the complex number $b + i(b-a)$.

Remember, we are after the complex number that corresponds to $R$ (let's call it $r$), and $R$ is the midpoint of this aforementioned diagonal. Consequently, we take the average of the complex numbers that correspond to the diagonal's endpoints, which yields $$r = {a+b+i(b-a) \over 2}$$

Now we can appeal to symmetry to do the rest. If we want the complex number that corresponds to $T$ (let's call it $t$), which is the center of the square that has $\overline{CD}$ as an edge. Here $C$ is playing the role of $A$ from the previous step, and $D$ is playing the role of $B$. As such, $$t = {c+d+i(d-c) \over 2}$$

Note the difference $(t-r)$ gives the complex number (let's call it $\alpha$) matching the displacement from $R$ to $T$, and $$\begin{array}{rcl} \alpha &=& \frac{c+d+i(d-c)}{2} - \frac{a + b + i(b-a)}{2}\\\\ &=& \frac{(c+d-a-b) + i(d-c-b+a)}{2}\\\\ \end{array}$$

To find the complex number (let's call it $\beta$) matching the displacement from $Q$ to $S$, we again appeal to symmetry. The calculation of $\alpha$ was based on segments $\overline{AB}$ and $\overline{CD}$, while the calculation of $\beta$ must be based on the segments $\overline{DA}$ and $\overline{BC}$. Thus we can make the replacements $a \rightarrow d$, $b \rightarrow a$, $c \rightarrow b$, and $d \rightarrow c$, yielding... $$\beta = {(b+c-d-a) +i(c-b-a+d) \over 2}$$

Finally, to prove $\overline{RT}$ and $\overline{QS}$ are both perpendicular and of the same length, we show that one of the corresponding displacements, $\beta$, is just a $90\,^{\circ}$ rotation of $\alpha$. In other words, we must show that $i\alpha = \beta$. But this is easy: $$\begin{array}{rcl} i\alpha &=& i \cdot \frac{(c+d-a-b) + i(d-c-b+a)}{2}\\\\ &=& \frac{-(d-c-b+a) + i(c+d-a-b)}{2}\\\\ &=& \frac{(b+c-d-a) +i(c-b-a+d)}{2}\\\\ &=& \beta\\ \end{array}$$

QED.

And there we have it: Geometry made simple... by making it complex!