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Arctan Identities Via Exponentiation

  1. May 29, 2012 #1
    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?
  2. jcsd
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