$\alpha + i\beta = \tanh \left ( x+i \dfrac {\pi}{4}\right )= - i \tan \left [ i \left ( x+ i \dfrac {pi}{4} \right ) \right ] = - i \tan \left [ix - \dfrac {\pi}{4} \right ] $ $\tan \left [ ix - \dfrac {\pi}{4} \right ]= \dfrac {\tan ix - \tan \left [\frac {\pi}{4} \right ]}{1-\tan ix \tan \left [ \frac {\pi}{4} \right ]} = \dfrac {\tan ix -1}{1-\tan ix}= - 1 $ $\alpha + i \beta =-i (-1) =i$ Taking modulus on both sides, $|\alpha + i \beta | =|i|$ $\sqrt{\alpha^2+\beta^2}= \sqrt{0^2+1^2}$ Squaring both sides, $\alpha^2+\beta^2=1$