t$an \Bigg[ ilog\bigg( \dfrac{a-ib}{a+ib} \bigg) \Bigg] \;=\; tan[ i\{ log(a-ib)-log(a+ib) \} ] \\ \; \\ \; \\ = tan[ ilog(a-ib)-ilog(a+ib) ] \\ \; \\ \;\\ = tan[ i \{ log\sqrt{a^2+b^2} \;-\; itan^{-1} \Big( \dfrac{b}{a} \Big) \} - i \{ log\sqrt{a^2+b^2} \;+\; itan^{-1} \Big( \dfrac{b}{a} \Big) \} ] \\ \{ \because log(a+ib) \;=\; log\sqrt{a^2+b^2} \;+\; itan^{-1} (\dfrac{b}{a}) \} \\ \; \\ \; \\ tan[ ilog\sqrt{a^2+b^2} \;-\; i^2tan^{-1} \Big( \dfrac{b}{a} \Big) - i log\sqrt{a^2+b^2} \;-\; i^2tan^{-1} \Big( \dfrac{b}{a} \Big) ] \\ \; \\ \; \\ tan[ tan^{-1} \Big( \dfrac{b}{a} \Big) + tan^{-1} \Big( \dfrac{b}{a} \Big) ] \\ \; \\ \; \\ = \dfrac{tan[ tan^{-1} \Big( \dfrac{b}{a} \Big)] + tan[tan^{-1} \Big( \dfrac{b}{a} \Big) ] } {1-tan[ tan^{-1} \Big( \dfrac{b}{a} \Big)] tan[tan^{-1} \Big( \dfrac{b}{a} \Big) ] } \\ \; \\ \; \\ = \dfrac{ \dfrac{b}{a} + \dfrac{b}{a} } {1- \dfrac{b}{a} \cdot \dfrac{b}{a} } \; =\; \dfrac{ \dfrac{2b}{a} } {1- \dfrac{b^2}{a^2} } \;=\; \dfrac{2ba^2}{a(a^2-b^2)} \;=\; \dfrac{2ba}{(a^2-b^2)} \\ \; \\ \; \\ \; \\ \; \\ tan \Bigg[ ilog\bigg( \dfrac{a-ib}{a+ib} \bigg) \Bigg] \;=\; \dfrac{2ab}{a^2-b^2} \; \; \; \; \; Hence \; Proved. $