prove-that-for-z-C-arctanz-1-2i-ln-1-iz-1-iz-

Question Number 73243 by mathmax by abdo last updated on 09/Nov/19

p r o v e t h a t f o r z \in C a r c t a n z = \frac{1}{2 i} l n (\frac{1 + i z}{1 - i z})

Answered by mr W last updated on 09/Nov/19

l e t \tan^{- 1} z = u

\tan u = z

\frac{\sin u}{\cos u} = z

\frac{\frac{e^{i u} - e^{- i u}}{2 i}}{\frac{e^{i u} + e^{- i u}}{2}} = z

\frac{e^{i u} - e^{- i u}}{e^{i u} + e^{- i u}} = i z

\frac{{(e^{i u})}^{2} - 1}{{(e^{i u})}^{2} + 1} = i z

e^{2 i u} (1 - i z) = 1 + i z

e^{2 i u} = \frac{1 + i z}{1 - i z}

u = \frac{1}{2 i} \ln (\frac{1 + i z}{1 - i z})

\Rightarrow \tan^{- 1} z = \frac{1}{2 i} \ln (\frac{1 + i z}{1 - i z})

Commented by mathmax by abdo last updated on 09/Nov/19

t h a n k x s i r m r w .