If \(\alpha+i\beta=\tanh\left(x+\frac{i\pi}{4}\right)\), prove that \(\alpha^2+\beta^2=1\).

Separate real and imaginary parts: \(\tanh(x+iy)=\frac{\sinh(x+iy)\cosh(x-iy)}{\cosh(x+iy)\cosh(x-iy)}\).

Numerator \(=\frac12[\sinh2x+\sinh2iy]=\frac12[\sinh2x+i\sin2y]\); denominator \(=\frac12[\cosh2x+\cosh2iy]=\frac12[\cosh2x+\cos2y]\).

So \(\tanh(x+iy)=\frac{\sinh2x+i\sin2y}{\cosh2x+\cos2y}\).

With \(y=\frac\pi4\): \(\sin2y=1,\ \cos2y=0\) \(\Rightarrow\alpha+i\beta=\frac{\sinh2x+i}{\cosh2x}\).

\(\alpha=\tanh2x,\quad\beta=\mathrm{sech}\,2x\).

\(\alpha^2+\beta^2=\tanh^22x+\mathrm{sech}^22x=1\) (proved).