t$an \Big[ log(x+iy) \Big] \;=\; a+ib \\ \; \\ \; \\ \therefore log(x+iy) \;=\; tan^{-1} (a+ib) \; \; \; \; \ldots (i) \\ \; \\ \; \\ log(x-iy) \;=\; tan^{-1} (a-ib) \; \; \; \; \ldots (ii) \\ \; \\ \; \\ $

Adding equations (i) \& (ii)

$ log(x+iy) +log(x-iy) \;=\; tan^{-1} (a+ib) + tan^{-1} (a-ib) \\ \; \\ \; \\ \therefore log[(x+iy)(x-iy) ]\;=\; tan^{-1} \bigg[ \dfrac{a+ib+a-ib}{1-(a+ib)(a-ib)} \bigg] \\ \; \\ \; \\ \therefore log[(x^2-i^2y^2) ]\;=\; tan^{-1} \bigg[ \dfrac{2a}{1-(a^2-i^2b^2)} \bigg] \\ \; \\ \; \\ \therefore log[(x^2+y^2) ]\;=\; tan^{-1} \bigg[ \dfrac{2a}{1-(a^2+b^2)} \bigg] \\ \; \\ \; \\ tan[log(x^2+y^2)] \;=\; \dfrac{2a}{1-a^2-b^2} \; \; \; where \; a^2+b^2 \ne 1 \; \; \; \; \; \; \; \ldots Proved. \\ \; \\ \; \\ \; \\ \; \\ $

Proving $log(x-iy) \;=\; tan^{-1} (a-ib) $

$ \\ \; \\ log(x+iy) \;=\; tan^{-1} (a+ib) \\ \; \\ \; \\ \therefore log \sqrt{x^2+y^2} + i tan^{-1} \Big( \dfrac{y}{x} \Big) \;=\; \dfrac{tan^{-1}a + tan^{-1}(ib)}{ 1 - tan^{-1}a \cdot tan^{-1}(ib)} \\ \; \\ \; \\ Let \; tan^{-1}(ib) \;=\; iz \\ \; \\ \; \\ \therefore log \sqrt{x^2+y^2} + i tan^{-1} \Big( \dfrac{y}{x} \Big) \;=\; \dfrac{tan^{-1}a + iz}{ 1 - tan^{-1}a \cdot iz} \\ \; \\ \; \\ \therefore tan^{-1} a + iz \;=\; log \sqrt{x^2+y^2} + i tan^{-1} \Big( \dfrac{y}{x} \Big) \; -ilog \sqrt{x^2+y^2} \cdot tan^{-1}a \cdot z + tan^{-1} \Big( \dfrac{y}{x} \Big) \cdot tan^{-1}a \cdot z \\ \; \\ \; \\ \therefore tan^{-1} a + iz \;=\; \bigg[ log \sqrt{x^2+y^2} + tan^{-1} \Big( \dfrac{y}{x} \Big) \cdot tan^{-1}a \cdot z \bigg] + i \bigg[ tan^{-1} \Big( \dfrac{y}{x} \Big) - log \sqrt{x^2+y^2} \cdot tan^{-1}a \cdot z \bigg] \\ \; \\ \; \\ \therefore tan^{-1}a \;=\; \bigg[ log \sqrt{x^2+y^2} + tan^{-1} \Big( \dfrac{y}{x} \Big) \cdot tan^{-1}a \cdot z \bigg] \\ \; \\ \; \\ z \;=\; tan^{-1} \Big( \dfrac{y}{x} \Big) - log \sqrt{x^2+y^2} \cdot tan^{-1}a \cdot z \\ \; \\ \; \\ Now, \; tan^{-1} a - iz \;=\; \bigg[ log \sqrt{x^2+y^2} + tan^{-1} \Big( \dfrac{y}{x} \Big) \cdot tan^{-1}a \cdot z \bigg] -i \bigg[ tan^{-1} \Big( \dfrac{y}{x} \Big) - log \sqrt{x^2+y^2} \cdot tan^{-1}a \cdot z \bigg] \\ \; \\ \; \\ \therefore tan^{-1} a - iz \;=\; log \sqrt{x^2+y^2} \Big[ 1+itan^{-1} a \cdot (z) \Big] \;-\; i tan^{-1} \Big( \dfrac{y}{x} \Big) \Big[ 1+itan^{-1} a \cdot (z) \Big] \\ \; \\ \; \\ \therefore log \sqrt{x^2+y^2} \;-\; tan^{-1} \dfrac{y}{x} \; = \; \dfrac{tan^{-1}a - iz}{1+itan^{-1}a.z} \\ \; \\ \; \\ \therefore log[x-iy] \;=\; \dfrac{tan^{-1}a - tan^{-1}a}{1+tan^{-1}a tan^{-1}b} \;=\; tan^{-1}(a-ib) $