Problem: Consider the real function of two variables given by
\[
u(x,y) = e^{2x} \left[ \sin 3x \cos 2y \cosh 3y- \cos 3x \sin 2y \sinh 3y \right].
\]
Let $v(x,y)$ be the harmonic conjugate of $fu(x,y)$ such that $v(0,0) = 2$. Let $z = x + \iota y$ and $f(z) = u(x,y) + \iota v(x,y)$, then the value of $4 + 2\iota f(\iota \pi )$ is
- $e^{3\pi } + e^{-3\pi }$
- $e^{3\pi } - e^{-3\pi }$
- $-e^{3\pi } + e^{-3\pi }$
- $-e^{3\pi } - e^{-3\pi }$
Solution: We will use
Milne-Thomson method for finding a holomorphic function to find the corresponding holomorphic function. We can do it by using the Cauchy-Riemann equations, but it will be too tedious. Let us give a brief overview of this method.
-
Problem 1:
Both \( u(x, y) \) and \( v(x, y) \) are given. Determine \( f(z) \).
Solution:
\( f(z) = u(z, 0) + i v(z, 0) \).
-
Problem 2:
Only \( u(x, y) \) is known, \( v(x, y) \) is not provided, and \( f(x + i0) \) is real. Determine \( f(z) \).
Solution:
\( f(z) = u(z, 0) \).
-
Problem 3:
\( u(x, y) \) is available, but \( v(x, y) \) is unknown. Find \( f(z) \).
Solution:
\( f(z) = u(z, 0) - i \int u_y(z, 0) \, \mathrm{d} z \), where \( u_y(x, y) \) denotes the partial derivative of \( u(x, y) \) with respect to \( y \).
-
Problem 4:
\( u(x, y) \) is not given, but \( v(x, y) \) is known. Find \( f(z) \).
Solution:
\( f(z) = \int v_x(z, 0) \, \mathrm{d} z + i v(z, 0) \), where \( v_x(x, y) \) represents the partial derivative of \( v(x, y) \) with respect to \( x \).
Here we are in the third category. So, corresponding holomorphic function will be
\[
f(z) = u(z,0) - \iota \int u_y(z,0)\mathrm{d} z.
\]
Note that
\begin{align*}
u_y(x,y) & = e^{2x}\left[ -2\sin 3x \sin 2y \cosh 3y + 3\sin 3x \cos 2y \sinh 3y\right. \\
& \kern 1cm \left.-2\cos 3x\sin 2y \sinh 3y - 3 \cos 3x\sin 2y\cosh 3y\right].
\end{align*}
Thus,
\begin{align*}
f(z) & = e^{2z} \sin 3z - C,
\end{align*}
where $C$ is constant of integration that will be determined by the condition given. Since, $v(0,0) = 2$, we have
\begin{align*}
f(0) = u(0,0) + \iota v(0,0) & \implies -C = 0 + 2 \iota \\
& \implies C = -2\iota .
\end{align*}
Thus,
\[
f(z) = e^{2z}\sin 3z + 2\iota .
\]
Now, we need to determine the value of $4 + 2\iota f(\iota \pi )$. We have
\begin{align*}
f(\iota \pi ) & = e^{2\iota \pi } \sin (3\iota \pi ) + 2\iota \\
& = \frac{e^{(3\iota\pi )\iota} - e^{(-3\iota \pi )\iota}}{2\iota } + 2\iota \\
& = \frac{e^{-3\pi} - e^{3\pi }}{2\iota } + 2\iota .
\end{align*}
Thus,
\begin{align*}
4 + 2\iota f(\iota \pi ) & = 4 + 2\iota \left( \frac{e^{-3\pi} - e^{3\pi }}{2\iota } + 2\iota \right) = e^{-3\pi} - e^{3\pi }.
\end{align*}
Thus, the correct answer is option (C).