# The harmonic conjugate of the function

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?

• Yuriy S Is my answer correct or not? – BalancedTryteOperators Nov 7 '18 at 22:39

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:

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

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