Show pi = 4(arctan(1/2)+arctan(1/3)) using complex arithmetic

Question

I am not sure how to do this. I know arctan(1/2)= arg(2+i) and arctan(1/3)=arg(3+i), but how do I go further. Any ideas?

Thanks

Answer 1

$\arctan(1/2) + \arctan(1/3) = \arg(2+i)+\arg(3+i) = \arg( (2+i)(3+i))$ $= \arg(5+5i) = \arg((1+i)/\sqrt{2}) = \arg(1^{1/8}) = 2\pi/8 + 2\pi k = \pi/4 + 2\pi k $ for $k\in\mathbb{Z}$.

A few facts about complex algebra (specifically the $\arg$ function) are used here to avoid trigonometry:

1) $\arctan(b/a) = \arg(a+ib)$

2) $\arg(z)+\arg(w) = \arg(zw)$

3) $\arg(\alpha z) = \arg(z)$ for $\alpha\in\mathbb{R_{>0}}$

4) $\arg(1^{1/n}) = 2\pi / n + 2\pi k$ for $n\in\mathbb{N}$ and $k\in\mathbb{Z}$, where $1^{1/n}$ is the principal $n$-th root of unity.

The fact that $\left[(1+i)/\sqrt{2}\right]^8 = 1$ and that it is in fact the principal root can be verified by the definition.

Answer 2

Hint:

$$(2+i)^4(3+i)=?$$

Now use the fact:

for $0<x<1,0<\arctan x<\arctan1=?$