<s>[INST] 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.

 [/INST] This question is equivalent to finding the product
$$(3+i) \cdot (3+i) \cdot (7+i)$$
and expecting it in the form $n(1+i)$.
Do you see why?
This is because $\arg(3+i) = \tan^{-1} \frac{1}{3}$, $\arg(7+i) = \tan^{-1} \frac{1}{7}$ and $\arg \lambda(1+i)=\frac{\pi}{4}$. On complex multiplication, principal arguments add up. So it can be concluded that
$$2\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{7}=n\pi + \frac{\pi}{4}$$
As noted by @MarkSaving, complex multiplication in general helps only upto multiples of $\pi$. We need to take into account $0 < \tan^{-1} \frac{1}{7} < \frac{\pi}{4}$ and  $0 < \tan^{-1} \frac{1}{3} < \frac{\pi}{4}$ so that the sum
$$0 < 2\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{7}<\frac{3\pi}{4}$$
from where it can be seen that our sum can be equal to $\pi/4$ only.
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