= Solution
The given <complex numbers> have positive <real parts>, so $\arg z_1=\arctan A$ and $\arg z_2=\arctan B$. Their product is
$$
z_1z_2=(1-AB)+i(A+B).
$$
By the assumed <complex argument> ranges, this product also has positive <real part>, hence $1-AB>0$. Its <complex argument> is consequently $\arctan[(A+B)/(1-AB)]$, with no extra multiple of $\pi$. Since multiplication adds <complex arguments>,
$$
\boxed{\arctan A+\arctan B=\arctan\left(\frac{A+B}{1-AB}\right).}
$$
This proves the <arctangent addition formula> under the stated branch restriction. For arbitrary real $A,B$, the principal <inverse tangent> formula can require adding or subtracting $\pi$ when $AB>1$.
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