How can I prove $\frac{\pi}{4}=\arctan\frac{1}{2}+\arctan\frac{1}{3}$ by using the product $(2+i)(3+i)$?

Question

How can I prove that $$\frac{\pi}{4}=\arctan\frac{1}{2}+\arctan\frac{1}{3}$$ by using the product $(2+i)(3+i)$?

What I noticed is that for $2+i$ in polar form, $\arctan\theta_1=\frac{1}{2}$ and for $3+i$ in polar form, $\arctan\theta_2=\frac{1}{3}$. However, I don't know how to proceed from here.

Answer 1

hint...use $\arg(zw)=\arg(z)+\arg(w)$