If $\arctan x + \arctan y + \arctan z = \pi$, prove that $x+y+z= xyz$.
Please let me know if this is solvable through complex numbers?
I tried to define three complex numbers in the argand plane and then use the property $\arg(x)+ \arg(y)+\arg(z)= \arg(xyz)$ but it didn't work.
First note that since $\arctan x + \arctan y + \arctan z = \pi$ and range of $\arctan$ is $(-\pi/2, \pi / 2) $ so we must have that $\arctan x, \arctan y, \arctan z \gt 0$ or that $x,y,z$.
Consider complex numbers $z_1(1,x), z_2(1,y), z_3(1,z)$ on Argand plane. Now these equivalent complex numbers must lie in quadrant I. Then $\arg(z_1) = \arctan(x)$ etc. meaning $$\arg(z_1 ) + \arg(z_2) + \arg(z_3) = \arctan(x)+\arctan(y)+\arctan(z) = \pi \tag{1}$$
Now consider the following expression $\zeta$: $$\zeta=\arg(z_1 ) + \arg(z_2) + \arg(z_3) = \arg(z_1 z_2 z_3) \\ = \arg((1+xi)(1+yi)(1+zi))\\=\arg(1-(xy+yz+zx)+i(x+y+z-xyz))$$
And this is equal to $\pi$ implying $\Im(\zeta) =0$ or that $x+y+z -xyz =0 \,\blacksquare$