Arctan Identities Via Exponentiation

[4dhayman]

Is it possible to prove identities involving arctan by complex exponentiation?

I had in mind something like the following for the arctan angle addition formula, but I feel there is something not quite right in the argument.

$$\arctan{(a)}+\arctan{(b)}= \arctan{\left(\dfrac{a+b}{1-ab}\right)} \implies e^{i\left(\arctan{(a)}+\arctan{(b)}\right)}=e^{i \arctan{\left(\frac{a+b}{1-ab}\right)}} \implies (ai+1)(bi+1) \propto ((a+b)i+(1-ab)) \implies True$$

Is this argument valid? If not, can it be modified to make it correct?

 

[fresh_42]

Your formula is only valid for ##ab<1##. And yes, the arcus tangent function is related to the complex logarithm but one has to take care of the complex area for which equations hold, because the complex logarithm splits into a Riemann surface.