Question

Consider the real function of two real variables given by
\[ u(x, y) = e^{2x}[\sin 3x \cos 2y \cosh 3y - \cos 3x \sin 2y \sinh 3y]. \] 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 \( 4 + 2i f(i\pi) \) is

• A. \( e^{3\pi} + e^{-3\pi} \)
• B. \( e^{3\pi} - e^{-3\pi} \)
• C. \( -e^{3\pi} + e^{-3\pi} \)
• D. \( -e^{3\pi} - e^{-3\pi} \)

Hint:

For solving such problems, it's helpful to use the Cauchy-Riemann equations to find the harmonic conjugates of real functions.

Answer 1

The Correct Option is C

The given function is: \[ u(x, y) = e^{2x}[\sin 3x \cos 2y \cosh 3y - \cos 3x \sin 2y \sinh 3y]. \] To find \( f(z) \), we first compute \( u(x, y) \) and its harmonic conjugate \( v(x, y) \), which are related by the Cauchy-Riemann equations. We are asked to find the value of \( 4 + 2i f(i\pi) \). First, evaluate \( u(x, y) \) and \( v(x, y) \) at \( z = i\pi \). After calculations, the value of \( 4 + 2i f(i\pi) \) results in: \[ - e^{3\pi} + e^{-3\pi}. \] Thus, the correct answer is \( \boxed{-e^{3\pi} + e^{-3\pi}} \).