Discussion
Consider the real function of two real variables given by
\[ \begin{aligned} u(x,y) = e^{2x}[&\sin 3x \cos 2y \cosh 3y \\ &- \cos 3x \sin 2y \sinh 3y]. \end{aligned} \]
Let v(x, y) be the harmonic conjugate of u(x, y) such that v(0, 0) = 2. Let z = x+iy and f(z) = u(x, y) + iv(x, y), then the value of \(4 + 2if(i\pi)\) is
Answer: C
The value is \(-e^{3\pi} + e^{-3\pi}\).
Recognise u as the real part of a product. With z = x + iy:
\[ \begin{aligned} e^{2z} &= e^{2x}(\cos 2y + i\sin 2y) \\ \sin 3z &= \sin 3x\cosh 3y + i\cos 3x\sinh 3y \end{aligned} \]
The real part of \(e^{2z}\sin 3z\) is exactly the given u. So f(z) = \(e^{2z}\sin 3z\) + ic for a real constant c. At z = 0 the product is 0, so v(0, 0) = c = 2:
\[ \begin{aligned} f(i\pi) &= e^{2\pi i}\sin(3\pi i) + 2i \\ &= i\sinh 3\pi + 2i \\ 4 + 2if(i\pi) &= 4 - 2\sinh 3\pi - 4 \\ &= -(e^{3\pi} - e^{-3\pi}) \end{aligned} \]
That is \(-e^{3\pi} + e^{-3\pi}\). The option \(e^{3\pi} - e^{-3\pi}\) has the sign lost, and the two with a plus between the exponentials would come from cosh rather than sinh.
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