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.

 A: 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.
A: We have the following expansion :
$$(1+\tfrac13i)^{\color{red}{2}}(1+\tfrac17i)=\frac{50}{63}(1+i)\tag{1}$$

from which we retrieve at once the following relationship on the arguments (remember : argument of a product of complex numbers = sum of their arguments mod. $2 \pi$):
$$ \color{red}{2} \operatorname{atan} \tfrac13 + \operatorname{atan} \tfrac17 = \underbrace{\operatorname{atan} 1}_{\pi/4} + 2k \pi$$
for a certain integer $k$ which is necessarily $0$.
Remark: (1) can be seen as successive similitudes applied to the hypotenuses of the right triangles:
$$(1+\tfrac13i)\xrightarrow{\times (1+\tfrac13i)}(\tfrac89+\tfrac23i)\xrightarrow{\times (1+\tfrac17i)}\frac{50}{63}(1+i)$$
A: When we convert $z=x+yi\ne0$ to $z=re^{it}$, polar form, then it satisfies these two equation $x=r\cos t, y=r\sin t$, so $\frac yx=\tan t$.
Then, $t=\tan^{-1}\frac yx$.
So...
$$50\sqrt2e^{\frac{i\pi}4}=50(1+i)=(3+i)^2(7+i)=(r_1e^{i\tan^{-1}\frac13})^2(r_2e^{i\tan^{-1}\frac1t})=r_1r_2e^{i(2\tan^{-1}\frac13+\tan^{-1}\frac17)}$$
Then following is true:
$$\frac{\pi}{4}= 2 \arctan\frac13+\arctan\frac17$$
A: 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
$$
2\tan \frac{1}{3}+\tan^{-1}\frac{1}{7}=\tan^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1- \frac{3}{28}}\right)=\tan^{-1}\frac{\frac{25}{28}}{\frac{25}{28}}=\tan^{-1}(1)=\frac{\pi}{4}
$$
