Inverse Trigonometry of Complex Numbers? Really?

It is a very easy problem in disguise. First of all it is rated in Geometry which is truly a disguise.

We will use the Taylor series for $\displaystyle {tan}^{-1}(x) = \sum_{k=0}^{\infty}{\dfrac{(-1)^k {x}^{2k+1}}{2k+1}}$

We will replace $\alpha, \beta, \gamma$ with $1, \omega, {\omega}^{2}$ [just a little easy to move with them)

Then,

$\displaystyle P = \sum_{k=0}^{\infty}{\dfrac{(-1)^k {1}^{2k}}{2k+1}} + \sum_{k=0}^{\infty}{\dfrac{(-1)^k {\omega}^{2k+1}}{\omega (2k+1)}} + \sum_{k=0}^{\infty}{\dfrac{(-1)^k {\omega^2}^{2k+1}}{\omega^2 (2k+1)}}$

$\displaystyle P = \sum_{k=0}^{\infty}{\dfrac{(-1)^k}{2k+1} (1 + {\omega}^{2k} + {\omega}^{k})}$

$\displaystyle (1 + {\omega}^{2k} + {\omega}^{k})$ will give a nonzero answer(equal to $1 + 1 + 1 = 3$ ) only for $k = 3n$ for some integer $n$ , rest for all integers it becomes equal to $0$ [You can prove that by induction and even by De Moivre]

Hence, our sum would become

$\displaystyle P = 3 \sum_{n=0}^{\infty}{\dfrac{(-1)^k}{2(3n)+1}} = 3 \sum_{n=0}^{\infty}{\dfrac{(-1)^k}{6n+1}}$

And as a result, $\frac{P}{Q} = 3$ .