TITLE: $\arctan x=\frac{1}{2}i[\ln(1-ix)-\ln(1+ix)]$
QUESTION [7 upvotes]: In wikipedia it says,
$$\arctan x=\frac{1}{2}i[\ln(1-ix)-\ln(1+ix)]$$
I want to now why is this true and what does a logarithm of a complex number even mean. I'm guessing that if I use the Taylor series I can prove this result. 
Thanks

REPLY [2 votes]: If $\arctan x=y,$ 
$$\ln(1-ix)-\ln(1+ix)=\ln\dfrac{1-ix}{1+ix}=\ln\dfrac{1-i\tan y}{1+i\tan y}=\ln(e^{-2iy})=-2iy$$