Proving :$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$ [duplicate]

Possible Duplicate:
Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$?

How to prove $$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$$

Once I have seen a very nice proof of this: your claim is equivalent to proving that the sum of red, green and blue angles is $\pi$. Note that in the second picture, the blue-green triangle is right and isosceles (that is 45-45-90 triangle and thus similar to small red-black triangle in the first diagram).

Cheers!

Consider $(1+i)$, $(1+2i)$ and $(1+3i)$ which have respective arguments of $\arctan 1$, $\arctan 2$ and $\arctan 3$. Their product equals the sum of the arguments. Thus

$$\arctan 1+\arctan 2+\arctan 3 = \operatorname {Arg} ((1+i)(1+2i)(1+3i)) = \operatorname {Arg} (-10) = \pi$$