If \(\sin(\alpha+i\beta)=x+iy\), prove that (i) \(\dfrac{x^2}{\cosh^2\beta}+\dfrac{y^2}{\sinh^2\beta}=1\) and (ii) \(\dfrac{x^2}{\sin^2\alpha}-\dfrac{y^2}{\cos^2\alpha}=1\).

Answer: Separating real and imaginary parts gives \(x=\sin\alpha\cosh\beta,\ y=\cos\alpha\sinh\beta\), from which both results follow.

\(\sin(\alpha+i\beta)=\sin\alpha\cos(i\beta)+\cos\alpha\sin(i\beta)\). Using \(\cos(i\beta)=\cosh\beta\) and \(\sin(i\beta)=i\sinh\beta\):

\[x+iy=\sin\alpha\cosh\beta+i\cos\alpha\sinh\beta.\]

Equating real and imaginary parts: \(x=\sin\alpha\cosh\beta\), \(y=\cos\alpha\sinh\beta\).

(i) Eliminate \(\alpha\): \(\sin\alpha=\dfrac{x}{\cosh\beta}\), \(\cos\alpha=\dfrac{y}{\sinh\beta}\). Since \(\sin^2\alpha+\cos^2\alpha=1\),

\[\frac{x^2}{\cosh^2\beta}+\frac{y^2}{\sinh^2\beta}=1.\]

(ii) Eliminate \(\beta\): \(\cosh\beta=\dfrac{x}{\sin\alpha}\), \(\sinh\beta=\dfrac{y}{\cos\alpha}\). Since \(\cosh^2\beta-\sinh^2\beta=1\),

\[\frac{x^2}{\sin^2\alpha}-\frac{y^2}{\cos^2\alpha}=1.\]