Discussion
\(\sin(x + iy)\) is equal to :
Answer: B
\(\sin(x + iy) = \)\(\sin x \cosh y + i \cos x \sinh y\).
Use the addition formula, which holds for complex arguments:
\[ \sin(x + iy) = \sin x \cos(iy) + \cos x \sin(iy) \]
From the exponential definitions, \(\cos(iy) = \dfrac{e^{-y} + e^{y}}{2} = \cosh y\) and \(\sin(iy) = \dfrac{e^{-y} - e^{y}}{2i} = i\sinh y\). Substituting gives \(\sin x \cosh y + i\cos x \sinh y\).
\(\sin x \cos y + i\cos x \sin y\) treats \(y\) as a real angle, forgetting that the imaginary argument turns cos and sin into cosh and sinh. The other two options pair the functions wrongly; at \(y = 0\) they do not even reduce to \(\sin x\).
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