Showing $\frac{\pi}{4}= 2 \arctan\frac13+\arctan\frac17$ by using multiplication of complex numbers Imagine three right triangles on top of each other. The legs of the first two triangles have the ratio $1/3$ and the third triangle $1/7$. The total angle of all three angles is $\pi/4$.
I can get that with the arctan, of course, but I don't know how to use the complex numbers. Because the task is:

Show that
$$\frac{\pi}{4}= 2 \arctan\frac13+\arctan\frac17$$
by using multiplication of complex numbers.


Since
$$
\tan^{-1}x+\tan^{-1}y =\tan^{-1}[\tan(\tan^{-1}x+\tan^{-1}y )]
$$
i.e.
$$
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right)
$$
we have
$$
2\tan \frac{1}{3}=\tan^{-1}\left(\frac{\frac{2}{3}}{1- \frac{1}{9}}\right)=\tan^{-1}\frac{6}{8}=\tan^{-1}\frac{3}{4}
$$
and
$$
2\tan \frac{1}{3}+\tan^{-1}\frac{1}{7}=\tan^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1- \frac{3}{28}}\right)=\tan^{-1}\frac{\frac{25}{28}}{\frac{25}{28}}=\tan^{-1}(1)=\frac{\pi}{4}
$$
