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

$$provethatforz\in Carctanz=\frac{1}{2i}ln\left(\frac{1+iz}{1-iz}\right)$$$$$$

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

$$let{\tan }^{-1}z=u$$$$\tan u=z$$$$\frac{\sin u}{\cos u}=z$$$$\frac{\frac{e^{iu}-e^{-iu}}{2i}}{\frac{e^{iu}+e^{-iu}}{2}}=z$$$$\frac{e^{iu}-e^{-iu}}{e^{iu}+e^{-iu}}=iz$$$$\frac{{\left(e^{iu}\right)}^2-1}{{\left(e^{iu}\right)}^2+1}=iz$$$$e^{2iu}(1-iz)=1+iz$$$$e^{2iu}=\frac{1+iz}{1-iz}$$$$u=\frac{1}{2i}\ln \left(\frac{1+iz}{1-iz}\right)$$$$\Rightarrow {\tan }^{-1}z=\frac{1}{2i}\ln \left(\frac{1+iz}{1-iz}\right)$$

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

$$thankxsirmrw.$$

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