The harmonic conjugate of the function

Question

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?

Answer 1

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:

$v(x,y) = \sinh x \sin y + C.$

Answer 2

I'm not sure whether I did this incorrectly or whether I'm just writing the correct answer in a very weird way. This is the most straight-forward definition of a harmonic function.

Definition. Given $u(x, y)$, the harmonic conjugate of $u$ is the function $v(x, y)$ such that $v_x = -u_y$ and $v_y = u_x$.

So we need a function so that $v_x = \sin{(y)}\cosh{(x)}$ and $v_y = \cos{(y)}\sinh{(x)}$. Integrating both sides gives $v(x, y) = \sin{(y)}\sinh{(x)} + C$.