Let, $ x+iy=tan^{-1} (e^{i\theta}) \; \; \therefore e^{i\theta} \;=\; cos\theta+isin\theta \;=\; tan(x+iy) \\ \; \\ \; \\ Similarly, \; let \; x+-y=tan^{-1} (e^{-i\theta}) \; \; \therefore e^{-i\theta} \;=\; cos\theta-isin\theta \;=\; tan(x-iy) \\ \; \\ \; \\ \therefore x+iy+x-iy \;=\; 2x \;= tan^{-1} (e^{i\theta}) + tan^{-1} (e^{-i\theta}) \\ \; \\ \; \\ = tan^{-1} \bigg[ \dfrac{e^{i\theta}+e^{-i\theta}}{1-e^{i\theta}e^{-i\theta}} \bigg] \; \; \; \ldots \{ tan^{-1}A + tan^{-1}B \;=\; ta^{-1} \bigg[ \dfrac{A+B}{1-AB} \bigg] \} \\ \; \\ \; \\ = tan^{-1} \bigg[ \dfrac{e^{i\theta}+e^{-i\theta}}{1-1} \bigg] \\ \; \\ \; \\ =tan^{-1} (\inf) \\ \; \\ \; \\ = n\pi + \dfrac{\pi}{2} \\ \; \\ \; \\ x \;=\; \dfrac{1}{2} \bigg( n\pi + \dfrac{\pi}{2} \bigg) \\ \; \\ \; \\ \; \\ \; \\ Similarly, \; x+iy -(x-iy) \;=\; 2iy \;=\; tan^{-1} (e^{i\theta}) - tan^{-1} (e^{-i\theta}) \\ \; \\ \; \\ = tan^{-1} \bigg[ \dfrac{e^{i\theta}-e^{-i\theta}}{1+e^{i\theta}e^{-i\theta}} \bigg] \\ \; \\ \; \\ = tan^{-1} \bigg[ \dfrac{2isin\theta}{1+1} \bigg] \; = \;tan (isin\theta ) \\ \; \\ \; \\ \therefore tan(2iy) \;=\; isin\theta \; or \; itanh(2iy) \;=\; isin\theta \; \; \; \ldots \{ \because tan(i\beta) \;=\; tanh \beta \} \\ \; \\ \; \\ \therefore tanh(2y)=sin\theta \; \; \therefore 2y \;=\; tanh^{-1} (sin \theta) \\ \; \\ \; \\ \therefore 2y \;=\; \dfrac{1}{2} log \bigg[ \dfrac{1+sin\theta}{1-sin\theta} \bigg] \; \; \; \; \{ \because tanh^{-1}(x) \;=\; \dfrac{1}{2} \Big( \dfrac{1+x}{1-x} \Big) \} \\ \; \\ \; \\ \dfrac{1}{2} log \Bigg[ \dfrac{1+cos(\pi/2-\theta)}{1-cos(\pi/2-\theta)} \Bigg] \; = \; \dfrac{1}{2} log \Bigg[ \dfrac{cos^2(\pi/4-\theta/2)}{cos^2(\pi/4-\theta/2)} \Bigg] \\ \; \\ \; \\ \dfrac{1}{2} log \Bigg[ \dfrac{cos(\pi/4-\theta/2)}{cos(\pi/4-\theta/2)} \Bigg]^2 \;=\; log \bigg[ cot \Big( \dfrac{\pi}{4} - \dfrac{\theta}{2} \Big) \bigg] \\ \; \\ \; \\ \therefore 2y \;=\;log \bigg[ cot \Big( \dfrac{\pi}{4} - \dfrac{\theta}{2} \Big) \bigg] \;=\; -log \bigg[ tan \Big( \dfrac{\pi}{4} - \dfrac{\theta}{2} \Big) \bigg] \\ \; \\ \; \\ \therefore y= \dfrac{-1}{2} log \bigg[ tan \Big( \dfrac{\pi}{4} - \dfrac{\theta}{2} \Big) \bigg] \\ \; \\ \; \\ \therefore tan^{-1} e^{i\theta} \;=\; \bigg( n+\dfrac{1}{2} \bigg) \dfrac{\pi}{2} - \dfrac{-i}{2} log \bigg[ tan \Big( \dfrac{\pi}{4} - \dfrac{\theta}{2} \Big) \bigg] $