To Show

$\tan^{-1}{\dfrac{x+\iota y}{x-\iota y}} = \dfrac{\pi}{4} + \iota \ln{\left( \dfrac{x+y}{x-y}\right)}$

Let, $ \tan^{-1}{\dfrac{x+\iota y}{x-\iota y}} =z $

where $z $ is complex number.

$\therefore \dfrac{x+\iota y}{x-\iota y}=\tan{z} \\\therefore \dfrac{x+\iota y}{x-\iota y}=\dfrac{ e^{\iota z}-e^{-\iota z}}{ \iota \left( e^{\iota z}+e^{-\iota z} \right)} \\ \therefore \dfrac{\iota x- y}{x-\iota y}=\dfrac{ e^{\iota z}-e^{-\iota z}}{ e^{\iota z}+e^{-\iota z}} \\ \therefore \dfrac{\iota x- y+ x-\iota y}{\iota x- y -x+\iota y}=\dfrac{ e^{\iota z}-e^{-\iota z}+ e^{\iota z}+e^{-\iota z}}{ e^{\iota z}-e^{-\iota z} -e^{\iota z}-e^{-\iota z}} $

$ \therefore \dfrac{\left( 1+\iota \right)x-\left( 1+\iota \right)y}{\left( \iota-1\right)x-\left(1-i\right)y}=\dfrac{2e^{\iota z}}{-2e^{-\iota z}} \\ \therefore \dfrac{\left(1+\iota \right)\left( x-y\right)}{\left(1-\iota\right)\left( x+y\right)}=e^{2\iota z} \\\therefore \ln{ \dfrac{\left(1+\iota \right)\left( x-y\right)}{\left(1-\iota\right)\left( x+y\right)} } = 2\iota z$

$\\ \therefore \ln{ \dfrac{\sqrt{2} \left(\dfrac{1}{\sqrt{2}}+\dfrac{\iota}{\sqrt{2}} \right)\left( x-y\right)}{\sqrt{2} \left(\dfrac{1}{\sqrt{2}}-\dfrac{\iota}{\sqrt{2}} \right)\left( x+y\right)} } = 2\iota z \\ \therefore \ln{ \dfrac{ \left(\cos{\dfrac{\pi}{4}}+\iota \sin{\dfrac{\pi}{4}} \right)\left( x-y\right)}{ \left(\cos{\dfrac{\pi}{4}}-\iota \sin{\dfrac{\pi}{4}} \right)\left( x+y\right)} } = 2\iota z $

$\\ \therefore \ln{ \dfrac{ \left( e^{\iota\dfrac{\pi}{4}} \right)\left( x-y\right)}{ \left(e^{-\iota\dfrac{\pi}{4}} \right)\left( x+y\right)} } = 2\iota z \\ \therefore \ln{\left(e^{\iota\dfrac{\pi}{2}} \right) \dfrac{\left(x-y\right)}{ \left( x+y\right)} } = 2\iota z $

$ \therefore \ln{\left(e^{\iota\dfrac{\pi}{2}} \right)}+\ln{\dfrac{\left(x-y\right)}{ \left( x+y\right)} } = 2\iota z \\ \therefore \iota\dfrac{\pi}{2}+\ln{\dfrac{\left(x-y\right)}{ \left( x+y\right)} } = 2\iota z \\ \therefore \dfrac{\pi}{4}+\dfrac{1}{2\iota}\ln{\dfrac{\left(x-y\right)}{ \left( x+y\right)} } = z$

$ \therefore \dfrac{\pi}{4}+\dfrac{-\iota}{2}\ln{\dfrac{\left(x-y\right)}{ \left( x+y\right)} } = z \\ \therefore \dfrac{\pi}{4}+\dfrac{\iota}{2}\ln{\dfrac{\left(x+y\right)}{ \left( x-y\right)} } = z$

i.e.

$ \tan^{-1}{\dfrac{x+\iota y}{x-\iota y}} = \dfrac{\pi}{4}+\dfrac{\iota}{2}\ln{\dfrac{\left(x+y\right)}{ \left( x-y\right)} } $

Hence Shown