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.

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    $\begingroup$ $$ (2+i)(3+i) = 5+5i \text{ and } \frac 5 5 = \tan\frac\pi4. $$ $\endgroup$ Mar 14 '20 at 16:52

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


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