Let $u + iv$ be analytic, and $u(x, y) = \cosh{(x)}\cos{(y)}$. Find the harmonic conjugate function $v(x, y)$.
The harmonic conjugate function is given by
$
\begin{align}
v(z) &= \int_{z_0}^z u_x dy + \int_{z_0}^z u_y dx \\
&= \int_{z_0}^z \sinh{(x)}\,\cos{(y)}\,dy – \int_{z_0}^z \cosh{(x)}\,\sin{(y)}\,dx\\
&= \int_{z_0}^z \sinh{(x)}\,\cos{(y)}\,dy – \int_{z_0}^z \cosh{(x)}\,\sin{(y)}\,dx \\
&= \sinh{(x)}\int_{z_0}^z \cos{(y)}\,dy – \sin{(y)}\int_{z_0}^z \cosh{(x)}\,dx \\
&= \sinh{(x)}\left(\sin{(z)} – \sin{(z_0)}\right) – \sin{(y)}\left(\sinh{(z)} – \sinh{(z_0)}\right)
\end{align}
$
To get it in correct form, we then take the imaginary part.
$v(x, y) = \Im{(v(z))}$.
Is my answer correct or not?
Best Answer
From the first Cauchy-Riemann condition we have:
$\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = \sinh x \cos y$
$v(x,y) = \int \sinh x \cos y \mathrm{d}y = \sinh x \sin y + F(x)$
$\frac{\partial v}{\partial x} = \cosh x \sin y + F'(x)$
$-\frac{\partial u}{\partial y} = \cosh x \sin y.$
Since the second Cauchy-Riemann condition requires:
$\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y},$
we have:
$F'(x) = 0,$
$F(x) = constant.$
The required function is then:
$v(x,y) = \sinh x \sin y + C.$