How do you solve $\arctan(n)$ to radians by hand?
I. e. $\arctan(1)$ >> process >> $\pi/4$

I have this Taylor expansion that allows me to calculate an approximate value for arctan, but am wondering if there's a closed-form solution (Or a more general formula than below):

$$\begin{align}\arctan z&=z-\frac{z^3}3+\frac{z^5}5-\frac{z^7}7+\cdots\\&=\sum\_{n=0}^\infty\frac{(-1)^nz^{2n+1}}{2n+1};\quad\|z\|\leq1\quad z\ne i,-i\end{align}$$