On the picture, we see three squares: $ABGH$, $BCFG$ and $CDEF$. Prove that the sum of angles: $\angle DAE$, $\angle CAF$ and $\angle BAG$ is equal to $90°$.
The real problem is that we have to prove it using complex geometry. The hint given in the textbook is:
Consider the product $(1+i)(2+i)(3+i)$
I have never done any exercised involving applications of complex numbers in geometry, so any help is greatly appreciated.
When we multiply complex numbers we do add their arguments. Identifying $\;\Bbb C\sim\Bbb R^2\;$ and$\;A,B,C,D\;$ with $\;(0,0), (1,0), (2,0), (3,0)\;$ respectively, we then have that $\;G,F,E\;$ are resp. $\;1+i\sim(1,1)\,,\,\,2+i\sim(2,1)\,,\,\,3+i\sim(3,1)$ , so
$$(1+i)(2+i)(3+i)=(1+3i)(3+i)=10i$$
so the sum of the angles is $\;\arg(10i)=...$